Probability Class 10 Previous Year Questions: 1 Mark (Chapter 14)

1-mark questions on Probability

This page has 46 1-mark questions on Probability from CBSE Class 10 board papers of 2023, 2024, 2025 and 2026. Of these, 38 appeared only in Standard papers and 8 appeared only in Basic papers. Question formats: 46 MCQs. 25 of them were asked in two or more paper sets.

0 of 46 practised

All 46 Probability 1-mark questions

Q1. A pair of dice is thrown simultaneously. Let E denote the event that “The sum of numbers obtained on both dice is at least 9.” The number of outcomes in favour of event E is :

Mark practised

(A) 4
(B) 6
(C) 10
(D) 26

Marks: 1
Level: Basic
Asked in: 6 paper sets
Years: 2025, 2026
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Basic, Set 430/3/2, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 10, because the sums 9, 10, 11 and 12 give 4 + 3 + 2 + 1 = 10 favourable ordered pairs.
Why? An ordered pair records the number on the first die and the number on the second die, so (3,6) and (6,3) are different outcomes. Counting these equally likely outcomes gives the number of outcomes favourable to E.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/3/2: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting only the four ordered pairs with sum 9 — (3,6), (4,5), (5,4), (6,3) — and giving 4, forgetting sums 10, 11 and 12.
Treating (3,6) and (6,3) as the same outcome and counting only the unordered pairs {3,6}, {4,5}, {4,6}, {5,5}, {5,6}, {6,6}, giving 6 instead of 10.
Subtracting 10 from 36 and giving 26, the number of outcomes with sum less than 9, instead of the favourable outcomes themselves.

Why this works

When two dice are thrown, the first die can show any number from 1 to 6 and the second die can show any number from 1 to 6, so the experiment has 6 × 6 = 36 equally likely ordered pairs. Count only the pairs whose sum is at least 9: sum 9 gives (3,6), (4,5), (5,4), (6,3); sum 10 gives (4,6), (5,5), (6,4); sum 11 gives (5,6), (6,5); sum 12 gives (6,6). Total = 4 + 3 + 2 + 1 = 10. Option A leaves out sums 10, 11 and 12; option B gives 6 by treating (3,6) and (6,3) as the same; option D is 36 − 10 = 26, the number of outcomes with sum less than 9.

Key idea to remember

Favourable outcomes: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes of the experiment). This question asks only for the numerator of that fraction. Here E is 'sum at least 9', so an outcome is favourable exactly when the sum of the two numbers is 9, 10, 11 or 12.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · The question asks only for the numerator of this fraction.

Q2. The total number of outcomes in the experiment of simultaneous throw of three dice is :

Mark practised

(A) 6
(B) 18
(C) 36
(D) 216

Marks: 1
Level: Basic
Asked in: 4 paper sets
Years: 2025, 2026
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Basic, Set 430/2/2, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 216, because each die has 6 possible outcomes, so total outcomes = 6 × 6 × 6 = 216.
Why? When an experiment is made of stages, the total number of outcomes is the product of the number of outcomes of each stage; for three dice this gives 6 × 6 × 6.

The full 1 mark is for the correct option.

Answer:

(D) 216

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/2/2: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 18 by adding 6 + 6 + 6 instead of multiplying 6 × 6 × 6.
Choosing 36, which is the number of outcomes for two dice, not three dice.

Why this works

Each die is a separate object, so the outcome on the first die can combine with each outcome on the second die, and each of those combinations can combine with each outcome on the third die. That is why the total is 6 × 6 × 6 = 216. For two dice, the total would be 6 × 6 = 36, as in the NCERT text; three dice give one more multiplication by 6. Another correct method also earns full marks.

Key idea to remember

Number of all possible outcomes: see Key ideas used on this page.

Explanation

The theoretical probability of an event is the number of outcomes favourable to the event divided by the number of all possible outcomes of the experiment. To count all possible outcomes when several dice are thrown, multiply the number of outcomes of each die.

Important formulas

Total outcomes = 6 × 6 × 6 · For three dice, each die has 6 equally likely outcomes.

Q3. A number is chosen from the numbers 1, 2, 3 and denoted as x, and a number is chosen from the numbers 1, 4, 9 and denoted as y. Then \(P(xy \lt 9)\) is :

Mark practised

(A) \(\frac{1}{9}\)
(B) \(\frac{3}{9}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{7}{9}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Compartment: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/2, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 5/9, because the favourable pairs (1, 1), (1, 4), (2, 1), (2, 4), (3, 1) are 5 out of 9 equally likely pairs.
Why? By the definition of theoretical probability, P(E) = number of outcomes favourable to E ÷ number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/2: option (C).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Adding 3 and 3 to get 6 total outcomes instead of multiplying 3 by 3 to get 9.
Missing the pair (2, 4), whose product 8 is less than 9, and writing the probability as 4/9.
Counting (1, 9) as favourable even though its product 9 is not less than 9.

Why this works

x has 3 choices and y has 3 choices, so there are 3 × 3 = 9 equally likely ordered pairs. The favourable pairs are (1, 1), (1, 4), (2, 1), (2, 4) and (3, 1); their products are 1, 4, 2, 8 and 3, all less than 9. The other four pairs give products 9, 18, 12 and 27, which are not less than 9. Therefore the required probability is five out of nine.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes, where the outcomes are equally likely. Here the experiment has 3 × 3 = 9 equally likely outcomes (the ordered pairs), and 5 of them are favourable, so the required probability is five out of nine.

Important formulas

P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes of the experiment · Use this only when all outcomes are equally likely.

Q4. In a family of two children, the probability of having at least one girl is :

Mark practised

(A) \(\frac{1}{2}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{3}{4}\)
(D) \(\frac{1}{4}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Compartment: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 3/4, because out of the four equally likely outcomes (G, G), (G, B), (B, G), (B, B), three have at least one girl.
Why? The theoretical probability of an event is the number of favourable equally likely outcomes divided by the number of all possible outcomes.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/1: option (C).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/4, because that is the probability of both children being girls, not of at least one girl.
Merging (G, B) and (B, G) into one outcome and getting 2/3, forgetting that the elder and younger child are different.
Saying 1/2 because each child is a girl with probability 1/2, without counting the four two-child outcomes.

Why this works

In a two-child family, the elder and the younger child are different, so the equally likely outcomes are (G, G), (G, B), (B, G), (B, B). The event 'at least one girl' means one girl or two girls; only (B, B) is left out. So the favourable outcomes are 3 out of 4, and P = 3/4. Quick check: the complementary event 'no girl' has probability 1/4, and 1 − 1/4 = 3/4.

Key idea to remember

Theoretical Probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the ratio of the number of outcomes favourable to E to the number of all possible outcomes, provided all outcomes are equally likely. Here the four ordered pairs of the two children's genders are equally likely, so the required probability is 3/4.

Important formulas

\(P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes}}\) · Used when all outcomes of the experiment are equally likely.

Q5. Probability of happening of an event is denoted by \(p\) and probability of non-happening of the event is denoted by \(q\) . Relation between \(p\) and \(q\) is

Mark practised

(A) \(p + q = 1\)
(B) \(p = 1, q = 1\)
(C) \(p = q - 1\)
(D) \(p + q + 1 = 0\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/1/1, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) \(p + q = 1\), because happening and non-happening are complementary events, so \(P(E) + P(\text{not } E) = 1\) (for a coin, \(1/2 + 1/2 = 1\)).
Why? NCERT states that for any event E, \(P(E) + P(\bar{E}) = 1\). Happening and non-happening of an event are complementary, so the two probabilities add to 1.

The full 1 mark is for the correct option.

Answer:

(A) p + q = 1

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/1: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing B because both p and q are at most 1, without remembering that p and q must add up to 1.
Choosing D because it contains p + q, without noticing that it makes the sum of two non-negative probabilities negative, which is impossible.
Choosing C because it looks like a neat relation between p and q, without checking that it makes q one more than p, which is not true for complementary probabilities in general.

Why this works

Here p is the probability that the event happens and q is the probability that it does not happen. For any event, either it happens or it does not happen, so these two are complementary events. NCERT gives \(P(E) + P(\bar{E}) = 1\), hence \(p + q = 1\). Quick check with a coin: \(P(\text{head}) = 1/2\), so \(P(\text{not head}) = 1 - 1/2 = 1/2\), and \(1/2 + 1/2 = 1\). The other options fail: B would make the sum of the two probabilities 2, but the total probability of an experiment is 1; D would make the sum of two non-negative probabilities negative, which is impossible; C would make q one more than p, which is not the general relation between a probability and its complement.

Key idea to remember

Complementary events: see Key ideas used on this page.

Explanation

An event E either happens or it does not happen. The event 'not E', written as \(\bar{E}\), is called the complement of E, and E and \(\bar{E}\) are complementary events. NCERT states that for any event E, \(P(E) + P(\bar{E}) = 1\). In this question, p is P(E) and q is P(\(\bar{E}\)), so p + q = 1.

Important formulas

P(E) + P(\(\bar{E}) = 1\) · For any event E and its complement; equivalently, P(\(\bar{E}) = 1 - P(E)\).

Q6. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought ?

Mark practised

(A) 40
(B) 240
(C) 480
(D) 750

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/1/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 480, because 0.08 × 6000 = 480.
Why? By the definition of theoretical probability, the probability of an event equals the number of favourable outcomes divided by the number of all possible outcomes; solving for the favourable number gives the tickets she bought.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/1: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing 48000 by multiplying 6000 by 8 and forgetting that 0.08 means 8/100.
Dividing 6000 by 0.08 to get 75000, instead of multiplying 6000 by 0.08.
Choosing 750 by dividing 6000 by 8, because 0.08 was read as 8.

Why this works

Each of the 6000 tickets is equally likely to be the winning ticket. If she bought x tickets, then the number of outcomes favourable to her winning is x, and the number of all possible outcomes is 6000. By the definition of theoretical probability, P(winning) = x/6000. The question gives this probability as 0.08, so x/6000 = 0.08. Multiplying both sides by 6000 gives x = 0.08 × 6000 = 480. Quick check: 480 tickets out of 6000 gives 480/6000 = 0.08, which matches the given probability.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = number of outcomes favourable to E divided by number of all possible outcomes, assuming equally likely outcomes. In this lottery every ticket has the same chance, so if she bought x tickets, P = x/6000. Equating this to 0.08 and solving gives x = 480.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · The definition of theoretical probability for equally likely outcomes.
Number of favourable outcomes = P(E) × Number of all possible outcomes · Rearranged form used to find how many tickets she bought.

Q7. A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is

Mark practised

(A) \(\frac{1}{2}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{4}{13}\)
(D) \(\frac{1}{13}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/6/1, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 3/13, because a deck has 12 face cards, so P = 12/52 = 3/13.
Why? Theoretical probability is the number of favourable outcomes divided by the total number of equally likely outcomes.

The full 1 mark is for the correct option.

Answer:

(B) 3/13

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the ace as a face card, which gives 16 face cards and the wrong probability 16/52 (4/13).
Counting only kings and queens and forgetting the jacks, which gives 8 face cards and the wrong probability 8/52 (2/13).

Why this works

A well-shuffled deck makes all 52 cards equally likely. Each suit has 13 cards, and in each suit the king, queen and jack are face cards. So each suit gives 3 face cards, and 4 suits × 3 = 12 face cards. Therefore P(face card) = 12/52 = 3/13. Quick check: 12 ÷ 4 = 3 and 52 ÷ 4 = 13, so the fraction reduces to 3/13.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The NCERT defines theoretical probability as P(E) = Number of outcomes favourable to E divided by Number of all possible outcomes of the experiment, assuming all outcomes are equally likely. Here the favourable outcomes are the 12 face cards and the total possible outcomes are the 52 cards.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this when all outcomes are equally likely.

Q8. From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :

Mark practised

(A) \(\frac{2}{5}\)
(B) \(\frac{1}{5}\)
(C) \(\frac{1}{7}\)
(D) \(\frac{2}{7}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/1/1, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
After removing the even numbers 4 and 16, the remaining numbers are 1, 7, 9, 21, 25; only 7 is prime, so P(prime) = 1/5. Hence option (B) 1/5.
Why? The theoretical probability of an event is the number of favourable outcomes divided by the number of all possible outcomes, when the outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 1/5

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/1/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting 1 as a prime number, which would wrongly give 2/5 instead of 1/5.
Forgetting to remove 16, so the remaining list is taken as 1, 7, 9, 21, 25, 16 with 6 numbers, giving a wrong denominator.
Thinking 25 is prime because it is odd, though 25 = 5 × 5.

Why this works

First remove all even numbers from 1, 4, 7, 9, 16, 21, 25. The even numbers are 4 and 16, so 7 − 2 = 5 numbers remain: 1, 7, 9, 21, 25. Among these, the only prime number is 7; 1 is not prime, and 9, 21, 25 are composite. So there is 1 favourable outcome out of 5 possible outcomes, giving P(prime) = 1/5.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes), assuming all outcomes are equally likely. Here the experiment is choosing one number at random from the remaining list, so each remaining number is equally likely to be chosen.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this after removing the even numbers and counting the remaining numbers.

Q9. A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is

Mark practised

(A) \(\frac{7}{50}\)
(B) \(\frac{7}{55}\)
(C) \(\frac{1}{10}\)
(D) \(\frac{5}{49}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/2/1, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1/10, because total cards = 55 − 6 + 1 = 50, the 5 perfect squares are 9, 16, 25, 36, 49, so P = 5/50 = 1/10.
Why? A random draw makes every card equally likely, so the theoretical probability is the ratio of favourable outcomes to all possible outcomes.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/2/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting the total cards as 55 instead of 50, forgetting that both 6 and 55 are included (the count is 55 − 6 + 1).
Counting perfect squares from 1 to 55 and getting 7, even though 1 and 4 are not on the cards.
Taking 64 as a perfect square, even though the cards end at 55.
Leaving the answer as 5/50 without reducing it to 1/10, so the option that matches is not seen.

Why this works

Since the draw is random, each of the 50 cards is equally likely. The perfect squares between 6 and 55 are 3², 4², 5², 6² and 7², that is, 9, 16, 25, 36 and 49, so 5 cards are favourable. Therefore P = 5/50 = 1/10. The correct option is C. The wrong options come from taking the total as 55, taking 49 as the total, or counting perfect squares from 1 instead of from 6.

Key idea to remember

Theoretical Probability: see Key ideas used on this page.

Explanation

Theoretical probability of an event E is P(E) = number of outcomes favourable to E divided by the number of all possible outcomes, assuming the outcomes are equally likely. Here the random draw makes all cards equally likely, so we count the perfect-square cards and divide by the total number of cards.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · Used after assuming equally likely outcomes; here E is the event that the card shows a perfect square.

Q10. Two dice are rolled together. The probability of getting the sum of the two numbers to be more than 10, is

Mark practised

(A) 1/9
(B) 1/6
(C) 7/12
(D) 1/12

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/2/1, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 1/12, because the favourable outcomes are (5, 6), (6, 5) and (6, 6), which are 3 out of 36, so 3/36 = 1/12.
Why? By the definition of theoretical probability, P(E) = number of outcomes favourable to E divided by number of all possible outcomes; two dice give 6 × 6 = 36 equally likely ordered outcomes.

The full 1 mark is for the correct option.

Answer:

(D) 1/12

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/2/1: option (D).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting only (5, 6) and (6, 6) and leaving out (6, 5), because the two dice are different objects even when rolled together.
Treating the 11 possible sums 2, 3, ..., 12 as equally likely and writing 2/11; the sums are not equally likely because each ordered pair of dice has the same chance.
Using total outcomes as 12 by adding 6 + 6 instead of multiplying 6 × 6.

Why this works

When two dice are rolled, the first die can show any of 6 numbers and the second die can show any of 6 numbers, so the total number of equally likely ordered outcomes is 6 × 6 = 36. A sum greater than 10 means the sum is 11 or 12. The ordered pairs giving 11 are (5, 6) and (6, 5), and the pair giving 12 is (6, 6). So there are 3 favourable outcomes. Therefore P(sum > 10) = 3/36 = 1/12. A quick check: the number of favourable outcomes is 2 + 1 = 3, and 3/36 simplifies to 1/12.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

Theoretical probability, also called classical probability, of an event E is P(E) = number of outcomes favourable to E divided by number of all possible outcomes, assuming equally likely outcomes. In this solution the experiment is rolling two dice; the 36 ordered pairs are equally likely, so the probability is found by counting the favourable ordered pairs.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this when all outcomes are equally likely.

Q11. The number of red balls in a bag is 10 more than the number of black balls. If the probability of drawing a red ball at random from this bag is \(\frac{3}{5}\), then the total number of balls in the bag is :

Mark practised

(A) 50
(B) 60
(C) 80
(D) 40

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/4/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 50, because if black balls = x, then red balls = x + 10, and (x + 10)/(2x + 10) = 3/5, giving x = 20, so total = 2(20) + 10 = 50.
Why? By the definition of theoretical probability, P(red) is red balls divided by total balls; substituting the relation red = black + 10 turns the given 3/5 into an equation that gives x = 20.

The full 1 mark is for the correct option.

Answer:

(A) 50

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Finding x = 20 and then giving the number of red balls (30) as the total instead of adding black and red balls to get 50.
Writing the total as 2x instead of 2x + 10, so the total comes out 40 and option D is chosen.
Cross-multiplying with 3 on the red-ball side and 5 on the total side, which reverses the ratio and gives a negative value of x.

Why this works

Let the number of black balls be x. Then red balls are x + 10, and the total is x + (x + 10) = 2x + 10. Put these into the definition of probability: P(red) = (x + 10)/(2x + 10) = 3/5. Cross-multiply: 5(x + 10) = 3(2x + 10), so 5x + 50 = 6x + 30, giving x = 20. Now total balls = 2(20) + 10 = 50. Quick check: red balls = 20 + 10 = 30, and 30/50 = 3/5, so the answer matches the given probability.

Key idea to remember

Theoretical Probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the ratio of the number of outcomes favourable to E to the number of all possible outcomes, assuming equally likely outcomes. In this question, the event is drawing a red ball. We express the number of favourable outcomes (red balls) and the number of all possible outcomes (total balls) using one unknown, then equate this ratio to the given probability 3/5 and solve.

Important formulas

P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes · This definition sets up (x + 10)/(2x + 10) = 3/5.

Q12. The probability of getting a composite number greater than 3 on throwing a die is

Mark practised

(A) \(\frac{1}{6}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{2}{3}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/5/2, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1/3, because the composite numbers greater than 3 on a die are 4 and 6, so P = 2/6 = 1/3.
Why? By the definition of theoretical probability, P(E) is the number of favourable outcomes divided by the number of all possible outcomes, and the six faces of a fair die are equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 1/3

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/2: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting 1 as a composite number; 1 has only one factor, so it is neither prime nor composite.
Counting 5 as composite because it is greater than 3; 5 is prime, so only 4 and 6 are favourable.
Forgetting that 6 is composite and taking only 4 as favourable, which gives 1/6.

Why this works

A composite number has more than two factors. On a die, the numbers are 1, 2, 3, 4, 5 and 6. Among these, 4 and 6 are composite, and both are greater than 3. So there are 2 favourable outcomes out of 6 equally likely outcomes. Therefore P = 2/6 = 1/3. The other options would need 1, 3 or 4 favourable outcomes, which do not match the composite numbers greater than 3 on a die.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = number of outcomes favourable to E divided by the number of all possible outcomes of the experiment, assuming the outcomes are equally likely. Here the die is fair, so 1, 2, 3, 4, 5 and 6 are equally likely outcomes.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this when all outcomes of the experiment are equally likely.

Q13. A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that drawn card shows number '9' is :

Mark practised

(A) \(\frac{1}{26}\)
(B) \(\frac{4}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{13}\)

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Basic, Set 430/4/2, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 1/13, because 4 cards in a deck of 52 show the number 9, so the probability is 4/52 = 1/13.
Why? The theoretical probability of an event is the number of outcomes favourable to the event divided by the number of all possible outcomes, and well-shuffling makes all 52 outcomes equally likely.

The full 1 mark is for the correct option.

Answer:

(D) 1/13

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/2: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/52 by counting only one particular 9, such as the 9 of hearts, instead of all four 9s in the deck.
Choosing 4/13 by taking the total as 13 cards, which is the size of one suit, instead of the whole deck of 52 cards.

Why this works

A well-shuffled deck has 52 equally likely cards. A standard deck has 4 suits: spades, hearts, diamonds and clubs, and each suit has exactly one card showing 9. So the number of outcomes favourable to 'the card shows 9' is 4. By the definition of theoretical probability, P(9) = 4/52 = 1/13. Quick check: 52 ÷ 4 = 13, so 4 cards out of 52 is 1 card out of every 13.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment, assuming equally likely outcomes. Here E is 'the card drawn shows 9'. There are 4 favourable outcomes, the four 9s, and 52 possible outcomes, so P(E) = 4/52 = 1/13.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Used with equally likely outcomes; here E is 'the card drawn shows 9'.

Q14. Three coins are tossed together. The probability that only one coin shows tail, is :

Mark practised

(A) \(\frac{1}{2}\)
(B) \(\frac{3}{8}\)
(C) \(\frac{7}{8}\)
(D) 1

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Basic, Set 430/4/2, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 3/8, because the 8 equally likely outcomes of three coins include exactly 3 with one tail (HHT, HTH, THH), so P = 3/8.
Why? The theoretical probability of an event is the ratio of the number of favourable outcomes to the total number of equally likely outcomes.

The full 1 mark is for the correct option.

Answer:

(B) 3/8

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/2: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 7/8 by counting every outcome that has at least one tail; the question says only one coin shows tail.
Writing 1/3 because there are three coins and one tail, instead of counting the 8 equally likely outcomes.
Listing HHT, HTH and THH as one outcome because the coins look alike; the three orders are different outcomes of the experiment.

Why this works

The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 2 × 2 × 2 = 8 of them, and each is equally likely. The phrase 'only one coin shows tail' means exactly one T appears, which happens in HHT, HTH and THH, so 3 outcomes are favourable. Therefore P = 3/8. Option (B) is correct.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = number of outcomes favourable to E divided by number of all possible outcomes, when all outcomes are equally likely. For three coins, list all 8 outcomes and count the ones with exactly one tail.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Definition of theoretical probability; use it when all outcomes are equally likely.

Q15. A die is thrown once. Find the probability of getting a number less than 7.

Mark practised

(A) \(\frac{5}{6}\)
(B) 1
(C) \(\frac{1}{6}\)
(D) 0

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Basic, Set 430/1/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1, because every face of the die shows a number less than 7, so all 6 outcomes are favourable and P = 6/6 = 1.
Why? By the definition of theoretical probability, P(E) = number of favourable outcomes ÷ total number of outcomes. When every outcome is favourable, the event is sure to happen, so its probability is 1.

The full 1 mark is for the correct option.

Answer:

(B) 1

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 5/6 because you think the number 6 is not less than 7, but 6 is less than 7.
Choosing 1/6 because you think 'less than 7' means getting the number 6 only.
Choosing 0 because you think a number less than 7 cannot appear on a die.

Why this works

A fair die has six equally likely outcomes: 1, 2, 3, 4, 5, 6. Every one of these numbers is less than 7, so the number of favourable outcomes is 6 and the total number of outcomes is 6. Therefore P(getting a number less than 7) = 6/6 = 1. This is called a sure event because it is certain to happen.

Key idea to remember

Sure event: see Key ideas used on this page.

Explanation

An event that is certain to happen is called a sure event or a certain event. Since every number on a die is less than 7, the event 'getting a number less than 7' is sure, so its probability is 1.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the definition of theoretical probability.
P(sure event) = 1 · An event that is certain to happen has probability 1.

Q16. If three coins are tossed simultaneously, what is the probability of getting at most one tail ?

Mark practised

(A) \(\frac{3}{8}\)
(B) \(\frac{4}{8}\)
(C) \(\frac{5}{8}\)
(D) \(\frac{7}{8}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/2/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 4/8, because total outcomes are 2³ = 8 and favourable outcomes (0 tail or 1 tail) are 1 + 3 = 4, so P = 4/8.
Why? By the definition of theoretical probability, P(E) is the number of outcomes favourable to E divided by the number of all possible outcomes, when the outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 4/8

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/2/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting only HHT, HTH, THH (exactly one tail) and leaving out HHH; that gives option A, 3/8.
Taking 'at most one tail' as 'not all tails'; that also includes HTT, THT, TTH and gives option D, 7/8.
Listing only HHH, HHT, HTT, TTT as the outcomes and using 2 out of 4, without listing all 8 equally likely outcomes.

Why this works

When three coins are tossed, the sample space (all possible outcomes) is HHH, HHT, HTH, THH, HTT, THT, TTH, TTT — 8 outcomes in all, since 2³ = 8 and each coin has 2 faces. 'At most one tail' means 0 tails or exactly 1 tail. The outcomes with 0 tails: HHH (1 outcome). The outcomes with exactly 1 tail: HHT, HTH, THH (3 outcomes). So favourable outcomes = 1 + 3 = 4. Therefore P(at most one tail) = 4/8 = 1/2. The option is (B). Another quick check: the complement 'at least two tails' has outcomes HTT, THT, TTH, TTT, also 4 outcomes, so the 8 outcomes split equally.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

Theoretical probability of an event E is P(E) = (number of outcomes favourable to E)/(number of all possible outcomes), assuming equally likely outcomes. Here the event is 'at most one tail', and there are 4 favourable outcomes out of 8 equally likely outcomes.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use when all outcomes of the experiment are equally likely.

Q17. A bag contains 100 cards numbered 1 to 100. A card is drawn at random from the bag. What is the probability that the number on the card is a perfect cube ?

Mark practised

(A) \(\frac{1}{20}\)
(B) \(\frac{3}{50}\)
(C) \(\frac{1}{25}\)
(D) \(\frac{7}{100}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/2/1, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1/25, because the perfect cubes from 1 to 100 are 1, 8, 27, 64, so P = 4/100 = 1/25.
Why? A random draw makes all 100 cards equally likely, so the theoretical probability of an event is the number of outcomes favourable to it divided by the total number of possible outcomes.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/2/1: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting that 1 is a perfect cube and counting only 8, 27 and 64, which gives 3/100 instead of 1/25.
Counting perfect squares such as 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 instead of perfect cubes, giving 10 favourable outcomes.
Taking 125 as a possible card because 5³ = 125, without noticing that the cards stop at 100.

Why this works

A perfect cube is a number that can be written as n³ for some whole number n. Checking the whole numbers in order: 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, and 5³ = 125, which is already more than 100. So the cards with perfect cube numbers are 1, 8, 27 and 64: that is 4 favourable outcomes. There are 100 cards, so the number of all possible outcomes is 100. Therefore P(perfect cube) = 4/100 = 1/25. Quick check: 1/25 means 4 out of 100, and the four cubes are exactly 1, 8, 27, 64.

Key idea to remember

Theoretical Probability: see Key ideas used on this page.

Explanation

The theoretical probability P(E) of an event E is defined as the number of outcomes favourable to E divided by the number of all possible outcomes of the experiment, assuming the outcomes are equally likely. Here the event is 'the number on the card is a perfect cube', and its favourable outcomes are 1, 8, 27 and 64 out of 100 equally likely outcomes.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the definition of theoretical probability, used when all outcomes are equally likely.

Q18. If the probability of a player winning a game is 0.79, then the probability of his losing the same game is :

Mark practised

(A) 1.79
(B) 0.31
(C) 0.21%
(D) 0.21

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/1/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 0.21, because P(losing) = 1 − P(winning) = 1 − 0.79 = 0.21.
Why? Winning and losing are complementary events, so their probabilities add to 1; hence P(losing) = 1 − P(winning).

The full 1 mark is for the correct option.

Answer:

(D) 0.21

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/1/1: option (D).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1.79, forgetting that no probability can be greater than 1.
Choosing 0.31, which is 1 − 0.69 instead of 1 − 0.79.
Choosing 0.21%, because 0.21% is 0.0021, not 0.21.

Why this works

Winning and losing a game are complementary events: exactly one of them happens. So their probabilities add up to 1. Use P(losing) = 1 − P(winning) = 1 − 0.79 = 0.21. Quick check: 0.79 + 0.21 = 1. Option A (1.79) is impossible because no probability can exceed 1; option B (0.31) would be the answer if the given probability were 0.69; option C (0.21%) is a percentage, and 0.21% is 0.0021, not 0.21.

Key idea to remember

Complementary events: see Key ideas used on this page.

Explanation

For any event E, the event 'not E' is called the complement of E. Together, E and 'not E' cover every possible outcome, so P(E) + P(not E) = 1, giving P(not E) = 1 − P(E). Here, E = 'the player wins' and 'not E' = 'the player loses'.

Important formulas

P(not E) = 1 − P(E) · For complementary events E and not E.

Q19. If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9; then the probability that this digit is an odd prime number is :

Mark practised

(A) \(\frac{1}{3}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{4}{9}\)
(D) \(\frac{5}{9}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/3/1, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 1/3, because the odd prime digits among 1 to 9 are 3, 5 and 7, so P = 3/9 = 1/3.
Why? The theoretical probability of an event is the ratio of the number of favourable outcomes to the number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(A) 1/3

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/1: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting 2 as an odd prime and getting 4/9, although 2 is prime but not odd.
Counting all odd digits 1, 3, 5, 7, 9 and getting 5/9, although 1 and 9 are not prime.
Treating 1 as a prime number, which wrongly adds one more favourable outcome.

Why this works

A digit is chosen at random, so each of the nine digits 1 to 9 has the same chance of being chosen. The odd prime numbers in this list are 3, 5 and 7: each is a prime number, and each is odd. So there are 3 favourable outcomes out of 9 equally likely outcomes. The probability is 3/9, which simplifies to 1/3. A quick check: including 2 would give 4 favourable outcomes and 4/9, but 2 is a prime that is not odd; counting all odd digits would give 5/9, but 1 and 9 are not prime.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (Number of outcomes favourable to E) / (Number of all possible outcomes of the experiment), assuming equally likely outcomes. Here the experiment is choosing one digit from 1 to 9, so there are 9 possible outcomes. The favourable outcomes for the event 'odd prime number' are 3, 5 and 7, so P = 3/9 = 1/3.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this formula only when all outcomes of the experiment are equally likely.

Q20. Two dice are thrown together. The probability that they show different numbers is :

Mark practised

(A) \(1/6\)
(B) \(5/6\)
(C) \(1/3\)
(D) \(2/3\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/3/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5/6, because the 36 equally likely outcomes include 6 with the same number, so favourable outcomes = 36 − 6 = 30 and P = 30/36 = 5/6.
Why? Theoretical probability is P(E) = number of favourable outcomes ÷ number of all possible outcomes, assuming equally likely outcomes; here the favourable event is 'different numbers'.

The full 1 mark is for the correct option.

Answer:

(B) 5/6

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/6, the probability that both dice show the same number, instead of the probability that they show different numbers.
Counting (1,2) and (2,1) as the same outcome, which gives 15 favourable outcomes instead of 30.

Why this works

When two dice are thrown, each die has 6 faces, so the total number of equally likely outcomes is 6 × 6 = 36. The outcomes with the same number are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) — 6 outcomes. So the number of outcomes with different numbers is 36 − 6 = 30. Therefore P(different numbers) = 30/36 = 5/6. As a quick check, P(same numbers) = 6/36 = 1/6, and P(different) = 5/6, since 1 − 1/6 = 5/6. Option B is correct.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes, when all outcomes are equally likely. Here the total possible outcomes are the 36 ordered pairs from two dice, and the favourable outcomes are the 30 pairs with different numbers.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Used with equally likely outcomes; here 30 favourable outcomes out of 36 total.
P(not E) = 1 − P(E) · Alternative check: same-number and different-number outcomes are complementary events.

Q21. The probability of guessing the correct answer to a certain test question is
\(\frac{x}{6}\). If the probability of not guessing the correct answer to this question is
\(\frac{2}{3}\), then the value of \(x\) is :

Mark practised

(A) 2
(B) 3
(C) 4
(D) 6

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/3/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 2, because P(correct) + P(not correct) = 1 gives x/6 = 1 − 2/3 = 1/3, so x = 2.
Why? Guessing correctly and not guessing correctly are complementary events, so their probabilities add to 1.

The full 1 mark is for the correct option.

Answer:

(A) 2

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/1: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 4, by taking P(correct) as 2/3 instead of using P(correct) + P(not correct) = 1.
Choosing 6, by taking x/6 as the whole probability 1 without subtracting P(not correct) = 2/3 from 1.

Why this works

The event of guessing the correct answer and the event of not guessing it are complementary, so exactly one of them happens. Therefore their probabilities add to 1: x/6 + 2/3 = 1. Then x/6 = 1 − 2/3 = 1/3, so x = 6 × 1/3 = 2. Quick check: when x = 2, P(correct) = 2/6 = 1/3, and 1/3 + 2/3 = 1. If x were 4, then P(correct) = 4/6 = 2/3, and 2/3 + 2/3 = 4/3, which is more than 1, so that value is impossible for a probability.

Key idea to remember

Complementary events: see Key ideas used on this page.

Explanation

The complement of an event E is the event 'not E'. NCERT states that for any event E, P(not E) = 1 − P(E), so P(E) + P(not E) = 1. Here, 'guessing the correct answer' and 'not guessing the correct answer' are complementary events, so their probabilities add to 1.

Important formulas

P(E) + P(not E) = 1 · For any event E and its complement 'not E'.
P(not E) = 1 − P(E) · Used to find the missing probability.

Q22. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 7 is :

Mark practised

(A) \(\frac{1}{7}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{1}{5}\)
(D) \(\frac{7}{40}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/5/1, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1/8, because the multiples of 7 from 1 to 40 are 7, 14, 21, 28, 35, so 5 favourable outcomes out of 40 give 5/40 = 1/8.
Why? The theoretical probability of an event is the number of outcomes favourable to the event divided by the number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 1/8

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 7/40 by using the number 7 as the count of favourable outcomes instead of listing 7, 14, 21, 28, 35.
Choosing 1/7 by assuming that every 7th ticket is favourable, without checking that 40 is not a multiple of 7; the correct count is 5, so the probability is 5 out of 40.
Leaving the answer as 5 out of 40 without simplifying to 1 out of 8, so not recognising option B.

Why this works

Each ticket has the same chance of being drawn, so the 40 outcomes are equally likely. List the multiples of 7 that lie between 1 and 40: 7, 14, 21, 28, 35. There are 5 of them. By the definition of theoretical probability, P(multiple of 7) = 5/40 = 1/8. Quick check: 35 is the largest multiple of 7 not more than 40, and 35 ÷ 7 = 5, so the count is 5. Option A (1/7) would mean 40/7 favourable outcomes, which is not a whole number; option C (1/5) would need 8 favourable tickets; option D (7/40) would need 7 favourable tickets.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E)/(number of all possible outcomes), assuming all outcomes are equally likely. Here E is 'the number is a multiple of 7'. The favourable outcomes are 7, 14, 21, 28, 35, so the numerator is 5; the denominator is 40 because the tickets are numbered 1 to 40. Therefore P(E) = 5/40 = 1/8.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the theoretical (classical) probability; it assumes all outcomes are equally likely.

Q23. In a cricket match, a batsman hits the boundary 7 times out of the 42 balls he plays. The probability of his not hitting a boundary is :

Mark practised

(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{5}{6}\)
(D) \(\frac{1}{6}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/2/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 5/6, because the balls without a boundary are 42 − 7 = 35, so P(not hitting a boundary) = 35/42 = 5/6.
Why? The theoretical probability of an event is the number of outcomes favourable to the event divided by the number of all possible outcomes; not hitting a boundary is the event whose favourable outcomes are the non-boundary balls.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/2/1: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/6, which is the probability of hitting a boundary, instead of the probability of not hitting a boundary.
Simplifying 7/42 to 1/7 instead of 1/6.
Taking the favourable count as 42, which would give probability 1 instead of 5/6.

Why this works

Out of the 42 balls, 7 are boundaries, so the other 42 − 7 = 35 balls are not boundaries. The probability of not hitting a boundary is the number of favourable outcomes divided by the total number of outcomes: 35/42 = 5/6. A quicker check: P(hitting a boundary) = 7/42 = 1/6. Hitting and not hitting are complementary events, so P(not hitting) = 1 − 1/6 = 5/6. Another correct method also earns full marks.

Key idea to remember

Complement of an event: see Key ideas used on this page.

Explanation

For an event E, the event 'not E' is called the complement of E, and P(not E) = 1 − P(E). Here E is 'hitting a boundary', with P(E) = 7/42 = 1/6. So P(not E) = 1 − 1/6 = 5/6.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · The theoretical probability of an event, assuming equally likely outcomes.
P(not E) = 1 − P(E) · The probability of the complement of an event.

Q24. The probability of drawing an even prime number out of numbers from 1 to 30 is :

Mark practised

(A) \(\frac{1}{30}\)
(B) \(\frac{4}{15}\)
(C) \(\frac{7}{30}\)
(D) 0

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/2/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 1/30, because the only even prime number is 2, so 1 of the 30 numbers is favourable.
Why? The theoretical probability of an event is the ratio of favourable outcomes to all possible outcomes; the only even prime number is 2, so exactly one outcome is favourable.

The full 1 mark is for the correct option.

Answer:

(A) 1/30

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/2/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting all 15 even numbers as favourable, because 2 is the only even prime; the other 14 even numbers have more than two factors.
Including 1 as a prime number; 1 has only one factor, so it is not prime.
Choosing 0 because 2 is even and prime but thinking it is not in the list; 2 is between 1 and 30.

Why this works

Numbers from 1 to 30 give 30 equally likely outcomes. A prime number has exactly two factors: 1 and itself. The even numbers from 1 to 30 are 2, 4, 6, ..., 30, which are 15 numbers. Every even number except 2 is divisible by 2 and by itself, so it has more than two factors and is not prime. Therefore the only even prime number is 2, so the number of favourable outcomes is 1. By the definition of theoretical probability, P = 1/30.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E)/(number of all possible outcomes), when all outcomes are equally likely. Here the experiment is picking one number from 1 to 30, so there are 30 equally likely outcomes, and the event 'even prime number' has exactly one favourable outcome, 2.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · Used when all outcomes are equally likely.

Q25. A die is thrown once. The probability of getting a number which is not a factor of 36, is :

Mark practised

(A) 1/2
(B) 2/3
(C) 1/6
(D) 5/6

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/3/1, Q14, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1/6, because among the six outcomes 1, 2, 3, 4, 5, 6, only 5 is not a factor of 36, so P = 1/6.
Why? The theoretical probability of an event is the ratio of favourable outcomes to all equally likely outcomes; a die has six equally likely outcomes.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/3/1: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 5/6 by counting the five factors 1, 2, 3, 4, 6 as favourable, instead of counting the one number that is not a factor.
Forgetting that 6 is a factor of 36 and counting both 5 and 6 as not factors, which gives 2/6.
Thinking 36 is divisible by every number from 1 to 6, so the probability becomes 0.

Why this works

A fair die has six equally likely outcomes: 1, 2, 3, 4, 5, 6. A number is a factor of 36 when 36 divided by it leaves no remainder. Among these outcomes, 1, 2, 3, 4 and 6 divide 36 exactly, while 5 does not (5 × 7 = 35, so the remainder is 1). So the only outcome that is not a factor of 36 is 5. Hence P(not a factor of 36) = 1/6, which is option C.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E) / (number of all possible outcomes of the experiment), assuming equally likely outcomes. In this question, throwing a die once gives 6 equally likely outcomes, and the event 'not a factor of 36' has exactly 1 favourable outcome, namely 5.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the theoretical probability formula; it needs equally likely outcomes.

Q26. Lali tosses two different coins simultaneously. The probability that she gets at most one head is :

Mark practised

(A) 1
(B) 3/4
(C) 1/2
(D) 1/7

Marks: 1
Level: Standard
Compartment: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 3/4, because the four equally likely outcomes are HH, HT, TH, TT, and the three outcomes HT, TH, TT have at most one head, so P = 3/4.
Why? By the definition of theoretical probability, P(E) = number of outcomes favourable to E divided by the number of all possible outcomes, and all four outcomes here are equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 3/4

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/1: option (B).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/2 by counting only HT and TH as favourable and leaving out TT, which also has at most one head.
Choosing 1 by thinking at most one head is certain, and forgetting that HH is a possible outcome.
Treating the two different coins as identical and taking HH, HT, TT as three equally likely outcomes, which would give 2/3.

Why this works

When two different coins are tossed together, each coin can show H or T, so the possible outcomes are HH, HT, TH and TT. Because the coins are different, HT and TH are two separate outcomes. 'At most one head' means 0 heads or exactly 1 head, so the favourable outcomes are TT, HT and TH, which is 3 outcomes out of 4. Therefore P(at most one head) = 3/4. Quick check: the only outcome that is not favourable is HH, so P = 1 − 1/4 = 3/4. Another correct method also earns full marks.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the ratio of the number of outcomes favourable to E to the number of all possible outcomes, assuming all outcomes are equally likely. Here the experiment of tossing two different coins has four equally likely possible outcomes, and the event 'at most one head' has three favourable outcomes, so P = 3/4.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the definition of theoretical probability, used when all outcomes are equally likely.

Q27. In a single throw of two dice, the probability of getting a sum of 10 is :

Mark practised

(A) 1/12
(B) 1/36
(C) 1/6
(D) 1/4

Marks: 1
Level: Standard
Compartment: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/2, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 1/12, because out of 6 × 6 = 36 equally likely outcomes of throwing two dice, the favourable outcomes (4, 6), (5, 5), (6, 4) are 3, so P = 3/36 = 1/12.
Why? By the definition of theoretical probability, P(E) is the number of outcomes favourable to E divided by the number of all possible equally likely outcomes.

The full 1 mark is for the correct option.

Answer:

(A) 1/12

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/2: option (A).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/36 because only (5, 5) is counted, leaving out (4, 6) and (6, 4).
Counting (4, 6) and (6, 4) as one outcome, although the two dice are different, so both ordered pairs are distinct.
Assuming the sums 2 to 12 are equally likely and giving 1/11.

Why this works

When two dice are thrown, the first die has 6 outcomes and the second die has 6 outcomes, so the total number of equally likely ordered pairs is 6 × 6 = 36. The pairs that give sum 10 are (4, 6), (5, 5), (6, 4), that is 3 favourable outcomes. So P(sum 10) = 3/36 = 1/12. Option B (1/36) is the probability of one particular ordered pair such as (5, 5); option C (1/6) and option D (1/4) are not supported by the count. A quick check: 3 favourable outcomes out of 36 means 1 out of 12, so 1/12 is correct.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is defined as P(E) = (number of outcomes favourable to E)/(number of all possible outcomes of the experiment), assuming equally likely outcomes. For two dice, the 36 ordered pairs are all equally likely; the event 'sum is 10' has 3 favourable outcomes, so P = 3/36 = 1/12.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this only when the outcomes are equally likely, as in a throw of two dice.

Q28. In a group of 20 people, 5 can't swim. If one person is selected at random, then the probability that he/she can swim, is

Mark practised

(A) \(\frac{3}{4}\)
(B) \(\frac{1}{3}\)
(C) 1
(D) \(\frac{1}{4}\)

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/1/2, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 3/4, because 20 − 5 = 15 people can swim, so P = 15/20 = 3/4.
Why? Theoretical probability is the number of favourable outcomes divided by the number of all equally likely outcomes.

The full 1 mark is for the correct option.

Answer:

(A) 3/4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Giving 1/4, the probability of not swimming, because 5 of the 20 people cannot swim.
Choosing 1 because one person is selected, without using the ratio of favourable outcomes to total outcomes.

Why this works

Selecting one person at random makes all 20 people equally likely outcomes. Five cannot swim, so the number who can swim is 20 − 5 = 15. Therefore P(can swim) = 15/20 = 3/4. Quick check: 3/4 + 1/4 = 1, and 1/4 is the probability of not swimming, so the two probabilities add to 1.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

For an experiment with equally likely outcomes, P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes). Here the total outcomes are 20 people and the favourable outcomes are 20 − 5 = 15 people who can swim, so P = 15/20 = 3/4.

Important formulas

\(P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes of the experiment}}\) · Use this when all outcomes are equally likely.
\(P(\bar{E}) = 1 - P(E)\) · Not needed for the answer, but it confirms the result.

Q29. In a single throw of two dice, the probability of getting 12 as a product of two numbers obtained is :

Mark practised

(A) \(\frac{1}{9}\)
(B) \(\frac{2}{9}\)
(C) \(\frac{4}{9}\)
(D) \(\frac{5}{9}\)

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/2/2, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 1/9, because the favourable outcomes are (2, 6), (3, 4), (4, 3), (6, 2), so P = 4/36 = 1/9.
Why? Theoretical probability is the ratio of the number of outcomes favourable to the event to the number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(A) 1/9

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/2/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting only (2, 6) and (3, 4), ignoring the reverse orders (6, 2) and (4, 3), and writing 2/36 instead of 4/36.
Using sums equal to 12, such as (6, 6), instead of products, and writing 1/36.
Writing 4/36 and not reducing it to 1/9, then failing to match the printed option.

Why this works

When two dice are thrown, each die has 6 faces, so there are 6 × 6 = 36 equally likely outcomes. We need the product of the two numbers to be 12. The only ordered pairs are (2, 6), (6, 2), (3, 4) and (4, 3), because 2 × 6 = 12, 6 × 2 = 12, 3 × 4 = 12 and 4 × 3 = 12. That gives 4 favourable outcomes, so P = 4/36 = 1/9. No other pair of numbers from 1 to 6 multiplies to 12.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = Number of outcomes favourable to E divided by Number of all possible outcomes of the experiment, assuming the outcomes are equally likely. Here the experiment is throwing two dice, and the event is getting a product of 12.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Used after assuming the 36 outcomes of throwing two dice are equally likely.

Q30. Two dice are rolled together. What is the probability of getting a sum greater than 10 ?

Mark practised

(A) \(\frac{1}{9}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{12}\)
(D) \(\frac{5}{18}\)

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/5/2, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1/12, because out of 36 equally likely outcomes, only (5, 6), (6, 5) and (6, 6) give a sum greater than 10, so P = 3/36 = 1/12.
Why? By the definition of theoretical probability, P(E) = number of outcomes favourable to E ÷ number of all possible outcomes; fair dice make all 36 outcomes equally likely.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/5/2: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting only (6, 6) as favourable and giving 1/36.
Treating (5, 6) and (6, 5) as the same outcome and giving 1/18.

Why this works

When two dice are rolled, write each result as an ordered pair (first die, second die). The first die has 6 results and the second die has 6 results, so 6 × 6 = 36 equally likely outcomes. The event 'sum greater than 10' means the sum is 11 or 12. The favourable outcomes are (5, 6), (6, 5) and (6, 6), because 5 + 6 = 11, 6 + 5 = 11 and 6 + 6 = 12. So the number of favourable outcomes is 3. Hence P = 3/36 = 1/12. Quick check: 36 ÷ 12 = 3, which is exactly the number of favourable outcomes.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = number of outcomes favourable to E ÷ number of all possible outcomes, when the outcomes are equally likely. For two dice, the 36 ordered pairs are equally likely, and only three ordered pairs give a sum greater than 10. So P = 3/36 = 1/12.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · This is the theoretical probability of an event E; it is valid when all outcomes are equally likely.

Q31. One card is drawn at random from a well shuffled pack of 52 playing cards. The probability that the drawn card is a queen, is :

Mark practised

(A) \(\frac{4}{13}\)
(B) \(\frac{4}{52}\)
(C) \(\frac{2}{13}\)
(D) \(\frac{1}{26}\)

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/5/3, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 4/52, because a well-shuffled pack has 52 equally likely cards and 4 of them are queens, so P(queen) = 4/52.
Why? The theoretical probability of an event is the ratio of favourable outcomes to all possible outcomes when the outcomes are equally likely; a well-shuffled pack makes every card equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 4/52

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/5/3: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 4/13 because there are 4 suits of 13 cards each; this counts suits, not the 4 queens.
Choosing 1/26 because only 2 queens are red; the question asks for any queen, not a red queen.
Rejecting 4/52 because 1/13 is not listed; 4/52 and 1/13 are the same value.

Why this works

There are 4 queens in a pack of 52 cards, so P(queen) = 4/52 = 1/13. Option B prints 4/52, which is the same value as 1/13, so it is the correct choice. You can also use the complement: P(queen) = 1 − P(not queen) = 1 − 48/52 = 4/52. Check the other options: 4/13 of 52 cards means 16 cards, 2/13 of 52 cards means 8 cards, and 1/26 of 52 cards means 2 cards; none of these counts the 4 queens. Another correct method also earns full marks.

Key idea to remember

Theoretical Probability: see Key ideas used on this page.

Explanation

Theoretical probability is defined as P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes), assuming equally likely outcomes. The event here is 'the card is a queen'. The favourable outcomes are the 4 queens and the possible outcomes are the 52 cards, so the ratio 4/52 is the probability.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this when all outcomes of the experiment are equally likely.

Q32. A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball ?

Mark practised

(A) \(\frac{1}{4}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{7}{20}\)
(D) \(\frac{13}{20}\)

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/6/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 7/20, because the 7 yellow balls are the only balls that are neither blue nor pink, out of 20 balls in all.
Why? The theoretical probability of an event is the number of favourable outcomes divided by the number of all possible outcomes, because drawing at random makes every ball equally likely.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/1: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 13/20, the probability of getting a blue or pink ball, because the words 'neither blue nor pink' were not read carefully.
Dividing the 7 yellow balls by only the blue and pink balls (13), instead of by the total 20 balls.

Why this works

Drawing a ball at random makes all 20 balls equally likely. 'Neither blue nor pink' leaves only the 7 yellow balls. So P = 7/20. Quick check: 5 pink + 8 blue = 13 balls are blue or pink, so P(blue or pink) = 13/20, and 1 - 13/20 = 7/20, so the answer is consistent.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E) divided by (number of all possible outcomes), assuming all outcomes are equally likely. Here the total number of balls is the number of all possible outcomes, and the number of yellow balls is the number of outcomes favourable to the event.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · Use this when all outcomes are equally likely, as when a ball is drawn at random.

Q33. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23 is

Mark practised

(A) 7/90
(B) 1/9
(C) 4/45
(D) 9/89

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/6/2, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 4/45, because the prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19, giving 8/90 = 4/45.
Why? The theoretical probability of an event is the ratio of the number of outcomes favourable to the event to the number of all possible outcomes, and drawing at random makes all 90 outcomes equally likely.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/2: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting 1 as a prime number; 1 has only one factor, so it is not prime. This would give 9 favourable outcomes instead of 8.
Including 23 itself, because the question says less than 23, so 23 is not counted.
Using 89 as the total number of discs, forgetting that discs numbered 1 to 90 are 90 discs.
Forgetting the prime 2, the only even prime, and missing one favourable outcome.

Why this works

All 90 discs are equally likely because the disc is drawn at random. The event is 'the disc bears a prime number less than 23'. The prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19, so there are 8 favourable outcomes. Therefore, P = 8/90 = 4/45. A quick check: 4/45 × 90 = 8, so 8 of the 90 discs match the event.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the ratio of the number of outcomes favourable to E to the number of all possible outcomes, assuming all outcomes are equally likely. Here, drawing one disc at random makes all 90 discs equally likely, and the favourable outcomes are the 8 prime-numbered discs less than 23.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the definition of theoretical probability used in NCERT Section 14.1.

Q34. Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out of the bag at random. What is the probability that the number on the card taken out is an even number ?

Mark practised

(A) \(\frac{9}{17}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{7}{18}\)

Marks: 1
Level: Standard
Years: 2023
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2023 Mathematics Standard, Set 30/6/3, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1/2, because the cards numbered 3 to 20 give 18 equally likely outcomes, and the even numbers among them are 4, 6, 8, 10, 12, 14, 16, 18, 20, which are 9, so the probability is 9/18 = 1/2.
Why? By the definition of theoretical probability, P(E) is the number of favourable outcomes divided by the number of all possible outcomes. Thorough mixing makes the cards equally likely, so this definition applies.

The full 1 mark is for the correct option.

Answer:

(B) 1/2

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/3: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting the total cards as 17 by doing 20 − 3, because both 3 and 20 are in the bag; the correct total is 18.
Counting the even numbers as 10 by including 2 or 22, which are outside 3 to 20; the even cards are 4 to 20, which are 9.
Choosing 9/17, because 17 was used as the total number of cards instead of 18.

Why this works

The experiment is drawing one card from the bag. Since the cards are mixed thoroughly, each card is equally likely to be taken out. The cards are numbered 3, 4, 5, ..., 20, so the number of all possible outcomes is 20 − 3 + 1 = 18. The even numbers in this range are 4, 6, 8, 10, 12, 14, 16, 18 and 20, which gives 9 favourable outcomes. Therefore P(even) = 9/18 = 1/2. Option (B) is correct. Quick check: 18 ÷ 2 = 9, so half the cards are even, which matches 1/2.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (Number of outcomes favourable to E) ÷ (Number of all possible outcomes of the experiment), assuming all outcomes are equally likely. Here the event E is 'the number on the card is even'. All 18 cards are equally likely, and 9 of them are favourable, so P(E) = 9/18 = 1/2.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the theoretical probability; it is used only when the outcomes are equally likely.

Q35. If \(P(A)\) denotes the probability of an event \(A\), then

Mark practised

(A) \(P(A) \lt 0\)
(B) \(P(A) \gt 1\)
(C) \(0 \leq P(A) \leq 1\)
(D) \(-1 \leq P(A) \leq 1\)

Marks: 1
Level: Basic
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Basic, Set 430/2/2, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 0 ≤ P(A) ≤ 1, because P(A) = favourable outcomes ÷ total outcomes, so it can never be negative and never exceed 1.
Why? By the definition of theoretical probability, P(A) is the ratio of favourable outcomes to all possible outcomes, so it can never be negative and can never exceed 1.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/2/2: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing D because it includes negative numbers; a probability is a ratio of counts, so it cannot be negative.
Choosing A or B because they forget the bounds; a probability can be 0 or 1, but never less than 0 or more than 1.
Thinking 0 and 1 are not allowed; both are valid because an event can be impossible or certain.

Why this works

Probability is defined as P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes. Favourable outcomes can be zero for an impossible event and can be all the outcomes for a sure event, but they can never be negative and never exceed the total. So every probability lies between 0 and 1, both included. For a die, getting 8 is impossible: P = 0/6 = 0. Getting a number less than 7 is sure: P = 6/6 = 1. For a coin, P(head) + P(tail) = 1/2 + 1/2 = 1. Option D wrongly allows negative values such as -0.5, and options A and B go outside this range.

Key idea to remember

Probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1: see Key ideas used on this page.

Explanation

NCERT's summary says the probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1. This follows from the definition: favourable outcomes are never negative and never more than all possible outcomes. So option C states exactly this condition.

Important formulas

P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes · Since favourable outcomes are never negative and never exceed the total, P(E) lies between 0 and 1.
P(E) + P(not E) = 1 · For example, P(head) + P(tail) = 1/2 + 1/2 = 1, which also shows a probability cannot be negative or greater than 1.

Q36. If P(A) denotes the probability of an event A, then

Mark practised

(A) P(A) < 0
(B) P(A) > 1
(C) 0 ≤ P(A) ≤ 1
(D) -1 ≤ P(A) ≤ 1

Marks: 1
Level: Basic
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Basic, Set 430/2/3, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 0 ≤ P(A) ≤ 1, because P(getting 8) = 0/6 = 0 and P(getting a number less than 7) = 6/6 = 1.
Why? The theoretical probability P(E) is the ratio of favourable outcomes to all possible outcomes, so it is never negative and never more than 1.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/2/3: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing D because it contains 0 and 1, without noticing that it also allows negative probabilities such as -0.5, which can never happen.
Choosing A or B by thinking a probability may be negative or greater than 1, without recalling that the favourable outcomes can never exceed the total outcomes.

Why this works

The theoretical probability of an event E is P(E) = number of outcomes favourable to E ÷ number of all possible outcomes. The favourable outcomes are never fewer than 0 and never more than all possible outcomes, so P(E) can only lie from 0 to 1. In one throw of a die, getting 8 is impossible, so P(getting 8) = 0/6 = 0, and getting a number less than 7 is sure, so P(getting a number less than 7) = 6/6 = 1. These two examples fix the smallest and largest possible probabilities. Option D is wrong because it allows negative values such as -0.5, and options A and B are wrong because probabilities are never negative or greater than 1.

Key idea to remember

Theoretical probability of an event E: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is defined as P(E) = number of outcomes favourable to E ÷ number of all possible outcomes, assuming equally likely outcomes. Since favourable outcomes cannot be fewer than zero or more than all possible outcomes, every probability lies between 0 and 1.

Important formulas

P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes of the experiment · Definition of theoretical probability, assuming equally likely outcomes.
0 ≤ P(E) ≤ 1 · Every probability lies between 0 and 1.

Q37. One card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is a red ace card, is :

Mark practised

(A) \(\frac{1}{13}\)
(B) \(\frac{1}{26}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{2}\)

Marks: 1
Level: Standard
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/1/3, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1/26, because well-shuffling makes all 52 cards equally likely and there are 2 red ace cards, so P = 2/52 = 1/26.
Why? The theoretical probability of an event is the number of favourable outcomes divided by the number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(B) 1/26

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/1/3: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/13 because there are 4 aces, without noticing that the card must also be red.
Choosing 1/2 because half the deck is red, without noticing that the card must also be an ace.
Choosing 1/52 because only one card is drawn, without counting both red aces.

Why this works

A deck has 52 cards in 4 suits. Hearts and diamonds are the red suits, and each suit has one ace, so the red aces are the ace of hearts and the ace of diamonds: exactly 2 cards. Hence P(red ace) = 2/52 = 1/26. The other options are probabilities of different events: 1/13 is the probability of drawing any ace, 1/52 is the probability of drawing one particular card, and 1/2 is the probability of drawing a red card.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

Theoretical probability counts equally likely outcomes. Drawing one card from a well-shuffled deck gives 52 equally likely outcomes. The event 'red ace' has 2 favourable outcomes, so the probability is 2/52, which simplifies to 1/26.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · NCERT definition of theoretical probability. Here favourable outcomes are 2 red aces and all possible outcomes are 52 cards.

Q38. A bag contains 3 red balls, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is :

Mark practised

(A) 1/3
(B) 1/5
(C) 7/15
(D) 8/15

Marks: 1
Level: Standard
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/3/3, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 1/3, because only the 5 white balls are neither red nor black, so 5/15 = 1/3.
Why? The theoretical probability of an event is the ratio of the number of outcomes favourable to the event to the number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

(A) 1/3

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/3: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 7/15 because there are 7 black balls, without noticing that 'nor black' also excludes black balls.
Choosing 8/15 by adding red and white balls, misreading 'neither red nor black' as 'red or white'.
Leaving the answer as 5/15 instead of simplifying it to 1/3.

Why this works

Drawing a ball at random makes all 15 balls equally likely. The only balls that are neither red nor black are the 5 white balls. Therefore the required probability is 5/15 = 1/3. Quick check: 5/15 simplifies to 1/3, and 1/3 lies between 0 and 1, as every probability must.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E)/(number of all possible outcomes), assuming equally likely outcomes. Here the total number of balls is 3 + 5 + 7 = 15. The event 'neither red nor black' means the ball is white, and there are 5 white balls. So P = 5/15 = 1/3.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Theoretical probability, assuming equally likely outcomes.

Q39. The probability of getting a bad egg in a lot of 400 eggs is 0.045. The number of good eggs in the lot is :

Mark practised

(A) 18
(B) 180
(C) 382
(D) 220

Marks: 1
Level: Standard
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/3/3, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 382, because bad eggs = 0.045 × 400 = 18, so good eggs = 400 − 18 = 382.
Why? The theoretical probability of an event is the ratio of favourable outcomes to total outcomes, so the number of bad eggs is the probability times the total number of eggs.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/3: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 18, the number of bad eggs, without subtracting from 400.
Reading 0.045 as 0.45, which shifts the decimal and gives 180 bad eggs instead of 18.

Why this works

The probability of a bad egg tells what fraction of the 400 eggs are bad. So bad eggs = 0.045 × 400 = 18. Every egg is either bad or good, so good eggs = 400 − 18 = 382. Quick check: 382 ÷ 400 = 0.955, and 0.045 + 0.955 = 1, so the two probabilities add to 1.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = number of outcomes favourable to E / number of all possible outcomes. Here the event 'getting a bad egg' has probability 0.045 out of 400 eggs, so the number of bad eggs is 0.045 × 400 = 18. The event 'getting a good egg' is the complement of 'getting a bad egg', so the number of good eggs is 400 − 18 = 382.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · Definition of theoretical probability.
Number of favourable outcomes = P(E) × Total outcomes · Rearranged to find the count from a given probability.
P(not E) = 1 − P(E) · Complementary events; a good egg is 'not a bad egg'.

Q40. From the letters of the word “MOBILE”, a letter is selected at random. The probability that the selected letter is a vowel, is :

Mark practised

(A) \(\frac{3}{7}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{3}\)

Marks: 1
Level: Standard
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Standard, Set 30/4/2, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1/2, because MOBILE has 6 letters, of which O, I, E are 3 vowels, so P = 3/6 = 1/2.
Why? By the definition of theoretical probability, P(E) = number of outcomes favourable to E ÷ number of all possible outcomes, when all outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/4/2: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting only O and I as vowels and forgetting E, which gives 2/6 = 1/3 instead of 1/2.
Taking the total number of letters as 7 and writing 3/7, although MOBILE has only 6 letters.

Why this works

The word MOBILE has 6 letters: M, O, B, I, L, E. Selecting any one letter is equally likely, so the total number of possible outcomes is 6. The vowels in MOBILE are O, I and E, so the number of favourable outcomes is 3. Therefore P(vowel) = 3/6 = 1/2. A quick check: 3/6 = 0.5, which is option C.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the ratio of the number of outcomes favourable to E to the number of all possible outcomes of the experiment, assuming all outcomes are equally likely. Here the experiment is selecting one letter at random, and the event is 'the selected letter is a vowel'.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this when all outcomes are equally likely, as in selecting a letter at random.

Q41. Two different coins are tossed together. The probability of getting at least one head is :

Mark practised

(A) \(\frac{1}{2}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{3}{4}\)
(D) 1

Marks: 1
Level: Basic
Compartment: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Compartment Mathematics Basic, Set 430/S/2, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 3/4, because the four equally likely outcomes (H,H), (H,T), (T,H), (T,T) have three with at least one head.
Why? The theoretical probability is P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes), and it applies only when the outcomes are equally likely.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 430/S/2: option (C).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 1/2 by thinking 'one head' is one of the two coins, without listing the four outcomes (H,H), (H,T), (T,H), (T,T).
Choosing 1/4 by counting only (H,H), which is 'both heads', not 'at least one head'.
Counting (H,T) and (T,H) as the same outcome and then taking 2 out of 3; because the coins are different, these are two different equally likely outcomes.

Why this works

When two different coins are tossed, each coin can show H or T, so the possible outcomes are (H,H), (H,T), (T,H), (T,T). These four outcomes are equally likely. 'At least one head' means one or more heads, so the favourable outcomes are (H,H), (H,T), (T,H): 3 outcomes. Therefore P(at least one head) = 3/4. Quick check: the only outcome with no head is (T,T), so P(no head) = 1/4, and P(at least one head) = 1 − P(no head) = 1 − 1/4 = 3/4.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes), assuming the outcomes are equally likely. Tossing two different coins gives 4 equally likely possible outcomes, and 3 of them contain at least one head, so the required probability is 3/4.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This is the theoretical (classical) probability; it assumes equally likely outcomes.
P(not E) = 1 − P(E) · This is the complement rule, used here as a quick check.

Q42. A pair of dice is thrown once. The probability that sum of numbers appearing on top faces is at least 4 is :

Mark practised

(A) \(\frac{1}{11}\)
(B) \(\frac{10}{11}\)
(C) \(\frac{5}{6}\)
(D) \(\frac{11}{12}\)

Marks: 1
Level: Standard
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/4/2, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 11/12, because of the 36 equally likely outcomes, the 3 outcomes with sum below 4 are (1,1), (1,2), (2,1), so P = (36 − 3)/36 = 11/12.
Why? By the definition of theoretical probability, P(E) = number of outcomes favourable to E ÷ number of all possible outcomes; the 36 ordered outcomes of two dice are equally likely.

The full 1 mark is for the correct option.

Answer:

(D) 11/12

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/2: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Giving 9/11 by treating the 11 possible sums 2 to 12 as equally likely (9 sums, 4 to 12, are at least 4); the sums are not equally likely because (1,2) and (2,1) are two different outcomes while (1,1) is only one.
Counting (1,2) and (2,1) as the same outcome, so the number of outcomes with sum below 4 is taken as 2 instead of 3.

Why this works

When two dice are thrown, the first die can show 6 numbers and the second die can show 6 numbers, so there are 6 × 6 = 36 ordered outcomes, and all are equally likely. The event 'sum at least 4' is the complement of 'sum less than 4'. The only sums less than 4 are 2 and 3. Sum 2 has one outcome: (1,1). Sum 3 has two outcomes: (1,2) and (2,1). So the number of unfavourable outcomes is 3, giving 36 − 3 = 33 favourable outcomes. Therefore P = 33/36 = 11/12. Quick check: the counts of outcomes for sums 4, 5, 6, 7, 8, 9, 10, 11, 12 are 3, 4, 5, 6, 5, 4, 3, 2, 1, and 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 33. Counting the 33 favourable ordered pairs directly also gives the same probability.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the ratio of the number of outcomes favourable to E to the number of all possible outcomes, assuming all outcomes are equally likely. In a two-dice experiment, the 36 ordered pairs are equally likely, so this solution counts the outcomes whose sum is below 4 and subtracts them from 36.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Definition of theoretical probability; assumes equally likely outcomes.
6 × 6 = 36 · Total equally likely outcomes when two dice are thrown; ordered pairs are counted.

Q43. Letters A to F are mentioned on six faces of a die such that each face has a different letter. Two such dice are thrown simultaneously. The probability that vowels turn up on both the dice is :

Mark practised

(A) 1/4
(B) 1/3
(C) 1/9
(D) 1/36

Marks: 1
Level: Standard
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/4/3, Q9, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 1/9, because the 36 equally likely pairs from two dice include only 4 vowel pairs (A,A), (A,E), (E,A), (E,E), so P = 4/36 = 1/9.
Why? By the definition of theoretical probability, P(E) = favourable outcomes ÷ all equally likely outcomes. The two dice are separate, so (A,E) and (E,A) are counted as different outcomes.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/3: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Giving 1/36 for the pair (A,A) only, and missing the other three vowel pairs (A,E), (E,A), (E,E).
Giving 1/3, which is the probability of getting a vowel on one die, instead of finding the probability for both dice together.
Treating (A,E) and (E,A) as the same pair gives three favourable pairs (A,A), (A,E), (E,E); writing 3/36 = 1/12 is wrong because merging the outcomes would also change the total number of outcomes.

Why this works

On a die the faces are A, B, C, D, E, F, so the vowels are A and E. Since the die is fair, each face is equally likely, and P(vowel) = 2/6 = 1/3. When two dice are thrown, each of the 6 faces of the first die can pair with each of the 6 faces of the second, giving 6 × 6 = 36 equally likely ordered pairs. Here 'ordered' means (A,E) is different from (E,A), because the dice are two separate dice. The pairs with a vowel on both dice are (A,A), (A,E), (E,A) and (E,E), which is 4 pairs. So P(both vowels) = 4/36 = 1/9. Quick check: 4/36 simplifies to 1/9 by dividing numerator and denominator by 4, and 2/6 = 1/3 agrees with the probability on one die.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E)/(number of all possible outcomes), when all outcomes are equally likely. In this experiment there are 36 equally likely ordered pairs from two dice, and the event 'vowels on both dice' has 4 favourable pairs, so P = 4/36 = 1/9.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · The definition of theoretical probability, used with equally likely outcomes.

Q44. A piggy bank contains ₹ 1 coins and ₹ 2 coins in the ratio 9 : 11 respectively. The piggy bank is accidentally dropped and a coin pops out of it. The probability that it is a ₹ 2 coin is

Mark practised

(A) \(\frac{9}{11}\)
(B) 0.45
(C) 0.55
(D) \(\frac{1}{11}\)

Marks: 1
Level: Standard
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/5/2, Q9, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 0.55, because P(₹2 coin) = 11/(9+11) = 11/20 = 0.55.
Why? By the definition of theoretical probability, probability is the number of favourable equally likely outcomes divided by the total number of equally likely outcomes; a ratio of 9 : 11 means 11 equal parts out of every 20 equal parts are favourable.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/2: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 9/11, because that is the ratio of ₹1 coins to ₹2 coins, not the probability of getting a ₹2 coin.
Choosing 0.45, because 9/20 is the probability of getting a ₹1 coin, not a ₹2 coin.
Thinking the actual number of coins is needed; the ratio already gives the shares, and the total share 9 + 11 = 20 is enough.
Choosing 1/11, thinking only one of the 11 ₹2 coins can pop out.

Why this works

The ratio 9 : 11 means that for every 9 ₹1 coins there are 11 ₹2 coins, so we can think of the coins as 9 + 11 = 20 equal parts. Each coin is equally likely to pop out, so the theoretical probability of a ₹2 coin is 11/20 = 0.55. A quick check: the probability of a ₹1 coin is 9/20 = 0.45, and 0.45 + 0.55 = 1, which matches the rule that complementary probabilities add to 1.

Key idea to remember

Theoretical Probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment, where the outcomes are equally likely. The piggy bank coins are all equally likely to pop out. The ratio 9 : 11 means there are 9 equal parts of ₹1 coins and 11 equal parts of ₹2 coins, so of 20 equal parts, 11 are favourable to a ₹2 coin. Hence P(₹2 coin) = 11/20 = 0.55.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Used when all outcomes are equally likely; here the favourable outcomes are the ₹2 coins.

Q45. One ticket is drawn at random from a bag containing 50 tickets numbered 1 to 50. The probability that the drawn ticket has a number which is a multiple of 7 is :

Mark practised

(A) \(\frac{1}{5}\)
(B) \(\frac{7}{50}\)
(C) \(\frac{3}{25}\)
(D) \(\frac{4}{25}\)

Marks: 1
Level: Standard
Years: 2026
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2026 Mathematics Standard, Set 30(B), Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 7/50, because 7 of the 50 tickets, namely 7, 14, 21, 28, 35, 42, 49, are multiples of 7.
Why? The theoretical probability of an event is the number of outcomes favourable to the event divided by the number of all possible outcomes, and the 50 tickets are equally likely to be drawn.

The full 1 mark is for the correct option.

Answer:

(B) 7/50

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30(B): option (B).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting 56 as a multiple of 7 that lies on a ticket, which gives 8 favourable outcomes instead of 7.
Counting the ticket numbered 0 as a multiple of 7, although the tickets are numbered only from 1 to 50.
Choosing 1/5 by counting the 10 multiples of 5 (5, 10, 15, ..., 50) instead of the multiples of 7.

Why this works

The tickets are numbered 1 to 50, so drawing any one ticket is equally likely and there are 50 possible outcomes. The multiples of 7 from 1 to 50 are 7, 14, 21, 28, 35, 42 and 49; 7 × 7 = 49 is the last one, and 7 × 8 = 56 is beyond 50. So there are 7 favourable outcomes, giving the probability 7/50, which is option (B).

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = number of outcomes favourable to E divided by number of all possible outcomes of the experiment, assuming all outcomes are equally likely. Here the 50 tickets are equally likely, so the total outcomes are 50, and the favourable outcomes are the 7 tickets whose numbers are multiples of 7.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · This formula assumes that the outcomes of the experiment are equally likely.

Q46. Two dice are thrown simultaneously. The probability of getting a sum of 7 is :

Mark practised

(A) \(\frac{2}{9}\)
(B) \(\frac{1}{9}\)
(C) \(\frac{5}{36}\)
(D) \(\frac{1}{6}\)

Marks: 1
Level: Standard
Years: 2026
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2026 Mathematics Standard, Set 30(B), Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 1/6, because the six ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) give 6/36 = 1/6.
Why? An ordered pair (a, b) records the number on the first die and then the number on the second die, so (1, 6) and (6, 1) are different outcomes. The theoretical probability is the number of favourable outcomes divided by the number of all possible outcomes, assuming equally likely outcomes.

The full 1 mark is for the correct option.

Answer:

(D) 1/6

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30(B): option (D).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Counting (1, 6) and (6, 1) as the same outcome and still using 36 total outcomes gives 1/12 instead of the correct answer.
Choosing 5/36 by confusing the sum 7 with the sum 8, whose favourable outcomes are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2).
Treating the two dice as identical and reducing the total outcomes, instead of counting the ordered outcomes.

Why this works

When two dice are thrown, each die has 6 faces, so the total number of ordered outcomes is 6 × 6 = 36. The favourable ordered pairs for a sum of 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which are six outcomes. So P(sum 7) = 6/36 = 1/6.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is P(E) = (number of outcomes favourable to E) / (number of all possible outcomes of the experiment), and this definition assumes the outcomes are equally likely. Here the experiment has 36 equally likely outcomes, and the event 'sum is 7' has six favourable outcomes.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use this only when the outcomes of the experiment are equally likely.

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Key ideas used on this page

Theoretical probability
Used in questions 3 4 6 7 8 9 10 11 12 13 14 16 17 19 20 22 24 25 26 27 28 29 30 31 32 33 34 37 38 39 40 41 42 43 44 45 46
Complementary events
Used in questions 5 18 21
Favourable outcomes
Used in question 1
Number of all possible outcomes
Used in question 2
Sure event
Used in question 15
Complement of an event
Used in question 23
Probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1
Used in question 35
Theoretical probability of an event E
Used in question 36

How to use these questions

Try each question on paper first. Then read the steps: the mark beside each step shows what the examiner looks for. If a step is new to you, revise Section 14.1 of the NCERT chapter.

To revise the lessons behind these questions, read the NCERT chapter, exactly as NCERT publishes it on its own website.

Official papers behind this page

Every question on this page comes from an official CBSE Class 10 Mathematics board paper. The board publishes the question papers on cbse.gov.in and the marking schemes on cbse.gov.in, where you can see how marks are given.

Year	Exam	Course	Question paper	Marking scheme
2026	Main	Standard (041)	Open CBSE's official question paper for the 2026 main exam (one ZIP file of about 61.9 MB, Class 10 Maths Standard and Basic together) to read the full paper.	Open CBSE's official marking scheme for the 2026 main exam (one ZIP file of about 22.8 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Main	Standard (041)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 54.5 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Main	Basic (241)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 49.9 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Compartment	Basic (241)	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 2.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 18.7 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Main	Standard (041)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 35.3 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Main	Basic (241)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 43.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.8 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Main	Standard (041)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 44 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 17.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Basic (241)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 49.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 26 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Compartment	Standard (041)	Open CBSE's official question paper for the 2023 compartment exam (one ZIP file of about 7.6 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 compartment exam (one ZIP file of about 3.3 MB, Class 10 Maths Standard (041)) to see how marks are given.

Sources and session

Each question is reproduced as printed in its CBSE paper, with its year, paper code and question number; only the layout is tidied. The same question printed in several paper sets counts once, and the card says how often it was printed in 2023–2026. These counts describe past papers; they are not predictions.

Every solution is original and follows NCERT Mathematics Class X (reprint 2026-27). The mark beside each step follows the official CBSE marking scheme where the scheme shows it; a mark labelled “(suggested)” follows the board’s usual pattern for that type of question.

Reference: CBSE Class 10 Mathematics board question papers via cbse.gov.in; all solutions are original. learncbse.net is an independent website and is not affiliated with CBSE or NCERT.

1-mark questions on Probability

All 46 Probability 1-mark questions

Q1. A pair of dice is thrown simultaneously. Let E denote the event that “The sum of numbers obtained on both dice is at least 9.” The number of outcomes in favour of event E is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q2. The total number of outcomes in the experiment of simultaneous throw of three dice is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q3. A number is chosen from the numbers 1, 2, 3 and denoted as x, and a number is chosen from the numbers 1, 4, 9 and denoted as y. Then \(P(xy \lt 9)\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q4. In a family of two children, the probability of having at least one girl is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q5. Probability of happening of an event is denoted by \(p\) and probability of non-happening of the event is denoted by \(q\) . Relation between \(p\) and \(q\) is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q6. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q7. A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q8. From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q9. A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q10. Two dice are rolled together. The probability of getting the sum of the two numbers to be more than 10, is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q11. The number of red balls in a bag is 10 more than the number of black balls. If the probability of drawing a red ball at random from this bag is \(\frac{3}{5}\), then the total number of balls in the bag is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q12. The probability of getting a composite number greater than 3 on throwing a die is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q13. A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that drawn card shows number '9' is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q14. Three coins are tossed together. The probability that only one coin shows tail, is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q15. A die is thrown once. Find the probability of getting a number less than 7.

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q16. If three coins are tossed simultaneously, what is the probability of getting at most one tail ?

Step-by-step solution

Step 1: Choose the option