Probability Class 10 Previous Year Questions (Chapter 14)

Probability in CBSE board papers

CBSE asked 187 different Probability questions in the Class 10 Mathematics board papers of 2023, 2024, 2025 and 2026 (main and compartment exams, Standard and Basic courses). 5 of these questions have a worked solution on this page. Most of them test Section 14.1, Probability — A Theoretical Approach (187 of 187).

Where Probability questions came from, 2023–2026

A question printed in several question-paper sets counts once in each table. A question asked in two exams counts in both rows of the first table.

Questions by exam

Year	Exam	Questions asked	Marks	Question types
2026	Main	23	13 × 1 mark · 1 × 2 marks · 7 × 3 marks · 2 × 4 marks	12 MCQs, 1 Assertion–Reason, 2 case studies and 8 written answers
2025	Main	48	26 × 1 mark · 12 × 2 marks · 8 × 3 marks · 1 × 4 marks · 1 × 5 marks	23 MCQs, 3 Assertion–Reason, 1 case study and 21 written answers
2025	Compartment	8	7 × 1 mark · 1 × 4 marks	7 MCQs and 1 case study
2024	Main	50	34 × 1 mark · 8 × 2 marks · 7 × 3 marks · 1 × 4 marks	30 MCQs, 4 Assertion–Reason, 1 case study and 15 written answers
2024	Compartment	12	9 × 1 mark · 2 × 2 marks · 1 × 3 marks	8 MCQs, 1 Assertion–Reason and 3 written answers
2023	Main	43	32 × 1 mark · 3 × 2 marks · 3 × 3 marks · 5 × 4 marks	30 MCQs, 2 Assertion–Reason, 5 case studies and 6 written answers
2023	Compartment	11	9 × 1 mark · 2 × 2 marks	6 MCQs, 3 Assertion–Reason and 2 written answers

Questions by NCERT section and marks

NCERT section	1 mark	2 marks	3 marks	4 marks	5 marks	Total
14.1 Probability — A Theoretical Approach	125	28	23	10	1	187
All questions	125	28	23	10	1	187

Most-tested sections

14.1 Probability — A Theoretical Approach: 187 questions

From 2023 to 2026, CBSE took most Probability questions from Section 14.1 (187 of 187). These counts describe past papers; they are not predictions.

Topics in Probability

Each NCERT section of the chapter, in book order: what CBSE asked from it and where to practise it.

14.1 Probability — A Theoretical Approach

Theoretical probability of E = number of favourable outcomes ÷ number of all possible outcomes, assuming equally likely outcomes. The complement of E, written not E, has probability 1 - P(E). Every probability lies between 0 and 1; impossible is 0, sure is 1.

How to answer questions on this section

List all possible outcomes first, e.g. 1, 2, 3, 4, 5, 6 for a die.
Count favourable outcomes separately, like 2 numbers greater than 4.
Use complement when the question says 'not', e.g. P(not ace) = 1 - 4/52.
Simplify fractions, for example 4/52 becomes 1/13.
Check your answer lies between 0 and 1.

Mistakes to avoid in this section

Treating outcomes as equally likely when they are not, e.g. 4 red and 1 blue are not equally likely.
Using number of trials from repeated experiments instead of counting favourable outcomes.
For 'not an ace', writing 48 as favourable but not dividing by 52.

CBSE asked 187 questions from this section in the 2023–2026 papers (125 of 1 mark; 28 of 2 marks; 23 of 3 marks; 10 of 4 marks; 1 of 5 marks): 111 MCQs, 14 Assertion–Reason, 10 case studies and 52 written answers. 101 were printed in Standard papers and 86 in Basic papers.

Marks	Questions asked	Years	Asked in Standard	Asked in Basic
1 mark	125	2023, 2024, 2025, 2026	67	58
2 marks	28	2023, 2024, 2025, 2026	15	13
3 marks	23	2023, 2024, 2025, 2026	12	11
4 marks	10	2023, 2024, 2025, 2026	6	4
5 marks	1	2025	1	0

Solved on this page: question 1, question 2, question 3, question 4, question 5.

Key ideas used on this page

Theoretical probability
Used in questions 3 4 5
Complementary events
Used in question 1
Theoretical Probability (Classical Probability)
Used in question 2

3 solved questions last asked in 2025

Q1. The probability of guessing the correct answer of a certain test question is x/12. If the probability of not guessing the correct answer is 5/6, then find the value of x.

Mark practised

Marks: 2
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/3/1, Q22, 2 marks)
Status: Verified solution

Step-by-step solution

Step 1: Write the complement relation
1 mark (suggested)
For the event E = 'guessing the correct answer', P(E) + P(not E) = 1, so x/12 + 5/6 = 1. Hence x/12 = 1 − 5/6 = 1/6.
Why? For any event E, the sum of the probabilities of E and its complement 'not E' is 1. Here 'not guessing the correct answer' is the complement of 'guessing the correct answer'.
Step 2: Solve for x
1 mark (suggested)
Thus, x = 12 × 1/6 = 2.
Why? Multiply both sides of x/12 = 1/6 by 12 to isolate x.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

Answer:

x = 2

✓ Answer matches the CBSE marking scheme, 2025 Main, Paper 30/3/1.

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

Taking the given 5/6 as the probability of guessing correctly instead of the probability of not guessing correctly.
Comparing x/12 directly with 5/6, without using the fact that the two probabilities add to 1.
Stopping after finding x/12 as 1/6 and not multiplying by 12 to get x.

Why this works

Guessing correctly and not guessing correctly are complementary events, so their probabilities add to 1. The question gives P(not correct) = 5/6, so P(correct) = 1 − 5/6 = 1/6. Since P(correct) = x/12, x/12 = 1/6 and x = 12 × 1/6 = 2. Quick check: with this value, P(correct) = 2/12 = 1/6, and 1/6 + 5/6 = 1, so the complement rule holds.

Key idea to remember

Complementary events: see Key ideas used on this page.

Explanation

Two events are complementary when one is 'not E' of the other. NCERT states that for any event E, P(E) + P(not E) = 1, so P(not E) = 1 − P(E). This solution treats 'guessing the correct answer' and 'not guessing the correct answer' as complementary events and uses their sum to find x.

Important formulas

P(E) + P(not E) = 1 · E and 'not E' are complementary events.
P(not E) = 1 − P(E) · Used to find P(correct) from the given P(not correct).

Q2. From a pack of 52 cards, all aces and all kings are removed. A card is drawn at random from the remaining cards. Find the probability that the card so drawn is

Mark practised

(i) a face card.
(ii) a card of red colour.

Marks: 2
Level: Standard
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/5/3, Q24, 2 marks)
Status: Verified solution

Step-by-step solution

(i) · 1 mark

Step 1: Probability of a face card
1 mark (suggested)
Remaining cards = 52 − (4 + 4) = 44. Face cards left = 2 × 4 = 8, so P(face card) = 8/44 = 2/11.
Why? The theoretical probability of an event is the number of favourable equally likely outcomes divided by the total number of equally likely outcomes; after removing aces and kings, queens and jacks are the only face cards left.

(ii) · 1 mark

Step 2: Probability of a red card
1 mark (suggested)
Red cards left = 26 − (2 + 2) = 22, so P(red card) = 22/44 = 1/2. Hence (i) P(face card) = 2/11 and (ii) P(red card) = 1/2.
Why? Red colour cards are hearts and diamonds; the red aces and red kings are removed, so the favourable count is 22 out of 44 equally likely cards.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

Answer:

(i) 2/11, (ii) 1/2

✓ Answer matches the CBSE marking scheme, 2025 Main, Paper 30/5/3.

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 the four kings among the face cards even though kings were removed, and writing 3/11 instead of 2/11.
Using 52 as the total number of cards for both parts, forgetting that 8 cards were removed.
For the red colour part, using 26 red cards out of 44 and writing 13/22, without subtracting the 2 red aces and 2 red kings.

Why this works

A deck has 4 suits of 13 cards each. All 4 aces and all 4 kings are removed, so 4 + 4 = 8 cards go out and 52 − 8 = 44 cards remain, all equally likely. In each suit the face cards are king, queen and jack; since kings are removed, only the queen and jack remain, so face cards left are 2 × 4 = 8. Therefore P(face card) = 8/44 = 2/11. The red suits are hearts and diamonds, together 26 cards. The red aces and red kings removed are 2 + 2 = 4, so red cards left are 26 − 4 = 22. Therefore P(red card) = 22/44 = 1/2. Another correct method also earns full marks.

Key idea to remember

Theoretical Probability (Classical 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. After the removals, the 44 remaining cards are equally likely, so each probability is the favourable count out of 44.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Use 44 as the number of all possible outcomes after removing aces and kings.

Q3. The number of red balls in a bag is three more than the number of black balls. If the probability of drawing a red ball at random from the given bag is 12/23, find the total number of balls in the given bag.

Mark practised

Marks: 2
Level: Standard
Years: 2025
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2025 Mathematics Standard, Set 30/6/1, Q25, 2 marks)
Status: Verified solution

Step-by-step solution

Step 1: Let the number of black balls be x
½ mark
Let the number of black balls be x. Then the number of red balls is x + 3.
Why? The statement 'three more' means the red count is the black count increased by 3.
Step 2: Write the total number of balls
½ mark
Total balls = x + (x + 3) = 2x + 3.
Why? The total is the sum of the black balls and the red balls.
Step 3: Apply the given probability
½ mark
P(red) = (x + 3)/(2x + 3) = 12/23. So 23(x + 3) = 12(2x + 3), i.e. 23x + 69 = 24x + 36, which gives x = 33.
Why? Theoretical probability compares favourable outcomes (red balls) with all possible outcomes (total balls). Cross-multiplying removes the fractions, and solving gives the number of black balls.
Step 4: Find the total number of balls
½ mark
Total balls = 2(33) + 3 = 69. The total number of balls in the bag is 69.
Why? Substitute x = 33 into the total expression to get the required answer.

Marks: ½ + ½ + ½ + ½ = 2 · Split as in the CBSE marking scheme, 2025 · Standard 30/6/1 Q25

Answer:

69 balls

✓ Answer matches the CBSE marking scheme, 2025 Main, Paper 30/6/1.

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

Using x as the number of red balls and forgetting that red balls are x + 3, so the probability is written wrongly.
Stopping at 33, the number of black balls, and giving that as the total number of balls.
Cross-multiplying with 12 on the left and 23 on the right, which reverses the proportion.

Why this works

Drawing a ball at random makes every ball equally likely, so the theoretical probability formula applies. Let the black balls be x. Then red balls are x + 3, and the total is x + (x + 3) = 2x + 3. The given probability gives (x + 3)/(2x + 3) = 12/23. Cross-multiplying gives 23(x + 3) = 12(2x + 3), i.e. 23x + 69 = 24x + 36. So 69 − 36 = x, hence x = 33. Total balls = 2(33) + 3 = 69. Quick check: red balls = 33 + 3 = 36, and 36/69 = 12/23, 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 is the ratio of the number of outcomes favourable to the event to the number of all possible outcomes, when the outcomes are equally likely. Here each ball is equally likely to be drawn, so the probability of a red ball is the number of red balls divided by the total number of balls.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment · Used to write P(red) as red balls divided by total balls.

2 solved questions last asked in 2024

Q4. 15 defective pens are accidentally mixed with 145 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Mark practised

Marks: 2
Level: Basic
Asked in: 2 paper sets
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Basic, Set 430/2/1, Q23, 2 marks)
Status: Answer checked step by step

Step-by-step solution

Step 1: Total pens and good pens
1 mark
Total pens = 15 + 145 = 160. Good pens = 160 − 15 = 145.
Why? The total number of pens is the number of all possible outcomes, because any pen can be chosen. The good pens are the outcomes favourable to the event.
Step 2: Probability of a good pen
1 mark
P(good pen) = 145/160 = 29/32. Therefore, the probability that the pen taken out is good is 29/32.
Why? Theoretical probability is the number of favourable outcomes divided by the number of all possible outcomes. Dividing numerator and denominator by 5 simplifies the fraction.

Marks: 1 + 1 = 2 · Split as in the CBSE marking scheme, 2024 · Basic 430/2/3 Q21 (the marks split is taken from the 2024 Basic paper, where this question also appeared)

Answer:

29/32

Common mistakes to avoid

Taking the total number of pens as 145 and forgetting to add the 15 defective pens.
Writing 15/145, which compares defective pens with good pens, instead of 145/160.
Leaving the answer as 145/160 without simplifying it to 29/32.
Using 15/160 for a good pen, which is actually the probability of a defective pen.

Why this works

Drawing one pen at random makes every pen equally likely to be chosen, so the theoretical probability formula applies. The total number of pens is 15 + 145 = 160. Of these, 160 − 15 = 145 are good. Therefore P(good pen) = 145/160. Dividing the numerator and denominator by 5 gives 29/32. Since 29/32 is between 0 and 1, this is a valid probability.

Key idea to remember

Theoretical probability: see Key ideas used on this page.

Explanation

The theoretical probability of an event E is the number of outcomes favourable to E divided by the number of all possible outcomes, assuming all outcomes are equally likely. Here, drawing a pen at random makes all pens equally likely, so the favourable outcomes are the good pens and all possible outcomes are the total pens.

Important formulas

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

Q5. All the kings and queens are removed from a deck of 52 playing cards.
Remaining cards are well shuffled and then a card is drawn at random.
Find the probability that the drawn card is
1+1+1

Mark practised

(A) an ace of hearts
(B) a black card
(C) a jack of spades

Marks: 3
Level: Basic
Years: 2024
Topic: 14.1 Probability — A Theoretical Approach
Source: (CBSE 2024 Mathematics Basic, Set 430/2/3, Q28, 3 marks)
Status: Verified solution

Step-by-step solution

(i) · 1 mark

Step 1: Ace of hearts
1 mark (suggested)
Remaining cards = 52 − 8 = 44. P(ace of hearts) = 1/44.
Why? Theoretical probability divides the number of favourable outcomes by the number of all possible outcomes. Only kings and queens are removed, so the ace of hearts is still in the deck.

(ii) · 1 mark

Step 2: Black card
1 mark (suggested)
Black cards remaining = 26 − 4 = 22. P(black card) = 22/44 = 1/2.
Why? A black card is a spade or a club. Each black suit lost its king and queen, so only the remaining black cards are favourable.

(iii) · 1 mark

Step 3: Jack of spades
1 mark (suggested)
P(jack of spades) = 1/44. Hence the required probabilities are 1/44, 1/2 and 1/44.
Why? Jacks are not kings or queens, so the jack of spades is not removed. It is exactly one favourable card among the 44 remaining cards.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

Answer:

1/44, 1/2, 1/44

✓ Answer matches the CBSE marking scheme, 2024 Main, Paper 430/2/3.

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

Using 52 as the denominator instead of 44, forgetting that 4 kings and 4 queens have been removed.
Counting all 26 black cards as favourable, forgetting that the black king and queen of spades and the black king and queen of clubs are removed.
Thinking the jack of spades is removed because jacks are face cards; only kings and queens are removed.

Why this works

Removing the kings and queens changes the sample space. There are 4 kings and 4 queens, so 8 cards are removed and 52 − 8 = 44 cards remain. Well-shuffling makes these 44 cards equally likely. The ace of hearts and the jack of spades are not kings or queens, so each is one favourable card, giving 1/44. For black cards, the black suits are spades and clubs. Each suit has 13 cards, and after its king and queen are removed, 13 − 2 = 11 cards remain. The two black suits give 2 × 11 = 22 favourable cards, so P(black card) = 22/44 = 1/2. A quick check: each of the 4 suits leaves 11 cards, so 4 × 11 = 44 cards remain.

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. After the kings and queens are removed, the 44 remaining cards are equally likely, so each required probability is the number of favourable cards divided by 44.

Important formulas

P(E) = Number of outcomes favourable to E / Number of all possible outcomes · Use this definition after finding the new total number of remaining cards.
Cards left in a suit = 13 − 2 = 11 · After removing the king and queen from that suit.

Most repeated Probability questions

The questions CBSE printed in the most question-paper sets, 2023–2026. Each one is solved where the line points.

Printed in 3 question-paper sets (2025) · 2 marks · Standard · question 1 on this page
Printed in 2 question-paper sets (2024) · 2 marks · Basic · question 4 on this page

Official papers behind this chapter

Probability was asked in 120 of the 120 question-paper sets of 2023–2026. The table names the official CBSE files for each exam in which it was asked.

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	Sets with a Probability question	Question paper	Marking scheme
2026	Main	Standard (041)	10 of 10 question-paper sets had a Probability question	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.
2026	Main	Basic (241)	10 of 10 question-paper sets had a Probability question	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 29.1 MB, Class 10 Maths Basic (241)) to see how marks are given.
2025	Main	Standard (041)	17 of 17 question-paper sets had a Probability question	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)	12 of 12 question-paper sets had a Probability question	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	Standard (041)	3 of 3 question-paper sets had a Probability question	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 3.4 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 8.1 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Compartment	Basic (241)	3 of 3 question-paper sets had a Probability question	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)	14 of 14 question-paper sets had a Probability question	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)	14 of 14 question-paper sets had a Probability question	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.
2024	Compartment	Standard (041)	3 of 3 question-paper sets had a Probability question	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 3.2 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 7.1 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Compartment	Basic (241)	4 of 4 question-paper sets had a Probability question	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 2.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 5.6 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Main	Standard (041)	12 of 12 question-paper sets had a Probability question	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)	12 of 12 question-paper sets had a Probability question	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)	3 of 3 question-paper sets had a Probability question	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.
2023	Compartment	Basic (241)	3 of 3 question-paper sets had a Probability question	Open CBSE's official question paper for the 2023 compartment exam (one ZIP file of about 7.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 compartment exam (one ZIP file of about 7.5 MB, Class 10 Maths Basic (241)) to see how marks are given.

Continue studying

Other chapters: Statistics previous year questions.

To revise the lessons behind these questions, read NCERT Chapter 14 (official PDF), the chapter exactly as NCERT publishes it on ncert.nic.in.

Sources and data verification

The counts on this page come from 120 CBSE Class 10 Mathematics question-paper sets: 62 Standard (041) and 58 Basic (241), from the main exams of 2023, 2024, 2025 and 2026 and the compartment exams of 2023, 2024 and 2025. A question printed in several sets counts once; these counts describe past papers and are not predictions.

CBSE settles the question papers and the marking schemes: each question is reproduced as printed in its CBSE paper, with its year, paper code and question number, and each answer shown here matches the official marking scheme or answer key. NCERT settles the textbook: chapter and section names follow NCERT Mathematics Class X (reprint 2026-27).

Every solution is original. 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.

The official question-paper and marking-scheme links on this page were checked and open on 9 October 2026.

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.

Probability in CBSE board papers

Where Probability questions came from, 2023–2026

Questions by exam

Questions by NCERT section and marks

Most-tested sections

Topics in Probability

14.1 Probability — A Theoretical Approach

How to answer questions on this section

Mistakes to avoid in this section

Key ideas used on this page

3 solved questions last asked in 2025

Q1. The probability of guessing the correct answer of a certain test question is x/12. If the probability of not guessing the correct answer is 5/6, then find the value of x.

Step-by-step solution

Step 1: Write the complement relation

Step 2: Solve for x

Answer:

Common mistakes to avoid

Q2. From a pack of 52 cards, all aces and all kings are removed. A card is drawn at random from the remaining cards. Find the probability that the card so drawn is

Step-by-step solution

(i) · 1 mark

Step 1: Probability of a face card

(ii) · 1 mark

Step 2: Probability of a red card

Answer:

Common mistakes to avoid

Q3. The number of red balls in a bag is three more than the number of black balls. If the probability of drawing a red ball at random from the given bag is 12/23, find the total number of balls in the given bag.

Step-by-step solution

Step 1: Let the number of black balls be x

Step 2: Write the total number of balls

Step 3: Apply the given probability

Step 4: Find the total number of balls

Answer:

Common mistakes to avoid

2 solved questions last asked in 2024

Q4. 15 defective pens are accidentally mixed with 145 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Step-by-step solution

Step 1: Total pens and good pens

Step 2: Probability of a good pen

Answer:

Common mistakes to avoid

Q5. All the kings and queens are removed from a deck of 52 playing cards.Remaining cards are well shuffled and then a card is drawn at random.Find the probability that the drawn card is1+1+1

Step-by-step solution

(i) · 1 mark

Step 1: Ace of hearts

(ii) · 1 mark

Step 2: Black card

(iii) · 1 mark

Step 3: Jack of spades

Answer:

Common mistakes to avoid

Most repeated Probability questions

Official papers behind this chapter

Continue studying

Sources and data verification

More from this section

Q5. All the kings and queens are removed from a deck of 52 playing cards.
Remaining cards are well shuffled and then a card is drawn at random.
Find the probability that the drawn card is
1+1+1