NCERT Solutions for Class 10 Maths Chapter 14 Probability

Looking for NCERT solutions for Class 10 Maths Chapter 14? Probability is the final chapter of the Class 10 course, and it has a shape that surprises most students: a single exercise, Exercise 14.1, holding 25 questions. Everything in the chapter lives inside that one exercise.

This page is your route map, not the answer sheet. Each solved question sits one level down, on the Exercise 14.1 solutions page.

Here you learn what each band of questions tests, which of the chapter’s three methods it demands, and the mistakes the exam actually penalises — so you can pick the right question and know its method before you open the solution.

The whole chapter runs on one formula plus one assumption: the theoretical probability\( P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes of the experiment}} \), which is valid only when every outcome is equally likely. Keep that gate in mind and most of the chapter becomes counting.

Exercise 14.1 Question Map: Which Method Each Question Needs

Chapter 14 has exactly one exercise — Exercise 14.1 — and the full solutions for all 25 questions live on that exercise page. The directory below links each question to its solution; the list that follows tells you what each band demands before you open it.

• Q1 — pure vocabulary: fill-in-the-blank definitions of complement, impossible event, sure event, elementary events, and the range any probability must lie in. Method: recall definitions, no arithmetic.
• Q2–Q3 — the equally likely judgement. Decide which real experiments really give each outcome the same chance, and say why a coin toss is fair. Method: physical reasoning, not a formula.
• Q4 — which values can be a probability at all. Method: the rule that every probability lies in \( 0 \leq P(E) \leq 1 \), with percentages converted to decimals before you compare.
• Q5 and Q7 — the complement rule in both directions: given \( P(E) \), find \( P(\bar{E}) \); given “not same birthday”, find “same birthday”.
• Q6 — impossible and sure events in a bag holding only lemon-flavoured candies: one event can never happen, the other always will.
• Q8–Q10 — a single draw from a mixed group: balls, marbles, and a piggy bank of coins. Each asks one direct probability and one “not X” that is faster by complement.
• Q11–Q12 — questions pegged to textbook figures: the fish tank in Fig. 14.4 and the spinner in Fig. 14.5. Keep the figure open; the counts come straight from it.
• Q13 — one throw of a die: test your mental list of primes between 1 and 6, plus “between 2 and 6” and “odd”.
• Q14 — a full deck of 52 cards, six sub-parts that demand exact card counts: suits, face cards, red face cards, and specific named cards.
• Q15 — five diamond cards, one set aside. After the queen is drawn and put aside, the sample space drops from 5 down to 4.
• Q16, Q17 and Q21 — defective versus good items: pens, bulbs, ball pens. Q17(ii) is the trap — the first non-defective bulb is not replaced, so the total falls from 20 to 19.
• Q18 — discs numbered 1 to 90: count the favourable numbers for a two-digit number, a perfect square, and a multiple of 5. Remember that 1 is a perfect square.
• Q19 — a die whose faces carry letters, read from the figure in the book: find the probability of A and of D straight from the faces shown.
• Q20* — area as probability, using Fig. 14.6. The textbook itself marks it as not from the examination point of view; the method is favourable area divided by total area.
• Q22 — the two-dice sum problem from Example 13: complete the sum table, then judge a student’s claim that the 11 possible sums are each equally likely. The method is counting ordered pairs — some sums are reachable in more ways than others.
• Q23–Q24 — repeated trials: three coin tosses (8 outcomes) and a die thrown twice (36 outcomes). Both ask for “at least once”-type events, which are fastest as \( 1 – P(\text{none}) \).
• Q25 — two flawed arguments to judge: that two coins give three outcomes each with probability \( \frac{1}{3} \), and that a die gives two outcomes each with probability \( \frac{1}{2} \). The method is the equally likely test — list the real outcomes with order.

The Three Methods This Chapter Runs On

Three techniques carry every question in Exercise 14.1. Master these and you can route any probem by its shape.

• Method 1 — Count and divide. \( P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes of the experiment}} \). Valid only when outcomes are equally likely — that gate is the first thing to check.
• Method 2 — Complement. \( P(\bar{E}) = 1 – P(E) \). The moment a question says not, at least one, or does not, try the complement first. It nearly always shortens the count.
• Method 3 — Build the full sample space. Two coins give 4 ordered outcomes, two dice give 36, three tosses give 8. Order matters, so \( (H,T) \) and \( (T,H) \) are different outcomes.

The key results to recall are few and fixed: a sure event has probability 1, an impossible event 0, every probability obeys \( 0 \leq P(E) \leq 1 \), and the elementary events of any experiment sum to 1.

The deck facts for Q14 come from Example 4 and are worth memorising as a block.

Which method goes with which question band is a compact route table of its own.

The official NCERT Class 10 Mathematics Chapter 14 PDF reproduces the text, examples and figures exactly — keep it open to verify Fig. 14.4, Fig. 14.5 and Fig. 14.6 while you work.

Common Mistakes in Probability 14.1

• Treating outcomes as equally likely when they are not. The 1/11 argument in Q22 and the 1/3 argument in Q25 both assume equal chances for outcomes with different weights. Always list the real outcomes before dividing.
• Forgetting that order matters with two dice. \( (1,4) \) and \( (4,1) \) are different outcomes, so two dice give 36 outcomes, not 21 and not 11.
• Ignoring a shrinking sample. When an item is set aside, the total drops by one — Q15 falls from 5 cards to 4, and Q17(ii) from 20 bulbs to 19.
• Miscounting a deck. There are 12 face cards, the ace is not a face card, and the deck splits 26 red and 26 black. One slip here ruins all six parts of Q14.
• Accepting a probability outside 0 to 1. A negative number can never be a probability, and a percentage like 15% must be converted (15% = 0.15) before the rule \( 0 \leq P(E) \leq 1 \) is applied. Q4 punishes exactly this.
• Missing that 1 is a perfect square. When Q18 asks for perfect-square discs between 1 and 90, forgetting that \( 1 = 1^2 \) loses a favourable outcome.
• Counting “at least once” the hard way. Q24’s “at least once” is far shorter as \( 1 – P(\text{none}) \) than by adding every case with a 5.

How to Use This Page

1. Find your homework question number in the route table above.
2. Read the method column so you know what the question wants before you open the solution.
3. If the row names a figure — Q11, Q12, Q19, Q20* — keep the NCERT figure open because the solution page renders it.
4. Revising? Work the bands in order: definitions and rules (Q1–Q7), single draws (Q8–Q18), repeated trials (Q22–Q24), then argument checking (Q25).

Probability is the last chapter of the course, arriving right after Chapter 13 Statistics, and the two share a counting mindset. For the wider picture, browse the Class 10 Maths notes, or step back to the Class 10 notes hub and the full notes library.

Reference: NCERT textbooks (CBSE).

FAQs on NCERT Solutions for Class 10 Maths Chapter 14 Probability

How many exercises are there in Class 10 Maths Chapter 14 Probability?

One. Exercise 14.1 is the chapter’s only exercise and it holds 25 questions. Definitions, the complement rule, dice, cards, coins and repeated trials all sit inside that single exercise.

Which questions in Exercise 14.1 use the complement rule \( P(\bar{E}) = 1 – P(E) \)?

Q5 and Q7 use it directly, and it is the fastest route in every “not” or “at least” question — Q8(ii), Q9(iii), Q10(ii), Q17(ii), Q21(ii) and Q24. Whenever a question says not, at least one, or does not, reach for the complement first.

Why is the answer 1/11 wrong for the sums of two dice in Question 22?

Because the 11 sums are not equally likely. Two dice give 36 ordered outcomes — \( (1,4) \) and \( (4,1) \) are different — and the sums spread unevenly across those 36 pairs, so no sum other than 2 and 12 sits on a single pair. The faulty 1/11 argument treats 11 distinct sums as if each had the same chance, which the 36-pair table shows is false.

Which questions in Exercise 14.1 need a figure from the textbook?

Q11 (the fish tank, Fig. 14.4), Q12 (the spinner, Fig. 14.5), Q19 (the die whose faces carry letters) and Q20* (the rectangle and circle, Fig. 14.6). Keep the figure open when you work these — the counts and outcomes come from it.

Do starred questions such as Question 20 come in the board exam?

The textbook itself marks Q20* (and Examples 10* and 11*) as not from the examination point of view, so treat them as optional for exam practice but useful for understanding area-based probability. Follow your teacher’s guidance on what the current board pattern includes.