NCERT Solutions for Class 10 Maths Chapter 14: Probability (NCERT 2026–27)

These Class 10 Maths Chapter 14 solutions cover Probability from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 14.1 is reproduced exactly as in the book and solved step by step using the theoretical (classical) definition of probability, complementary events and equally likely outcomes — so you can revise quickly and score full marks.

Class: 10 Subject: Mathematics Chapter: 14 – Probability Exercises: Exercise 14.1 (25 questions) Topic: Theoretical probability Session: 2026–27
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Chapter 14 Overview

Chapter 14 of Class 10 Maths, Probability, develops the theoretical (classical) approach to probability, building on the experimental (empirical) probability you met in Class 9. Assuming all outcomes are equally likely, the probability of an event E is the ratio of favourable outcomes to all possible outcomes. The chapter explains elementary events, sure and impossible events, the bound 0 ≤ P(E) ≤ 1, and complementary events with the key relation P(E) + P(not E) = 1. Worked examples use coins, dice, playing cards, balls and marbles. The Class 10 Maths Chapter 14 solutions below solve every question of Exercise 14.1 step by step.

Key Concepts & Definitions

Theoretical (classical) probability: for an experiment with equally likely outcomes, P(E) = (Number of outcomes favourable to E) ÷ (Number of all possible outcomes).

Equally likely outcomes: outcomes that each have the same chance of occurring, e.g. heads/tails of a fair coin or the faces of a fair die.

Elementary event: an event having only one outcome of the experiment. The sum of the probabilities of all elementary events of an experiment is 1.

Sure (certain) event: an event that is bound to happen; its probability is 1. Impossible event: an event that cannot happen; its probability is 0.

Complementary event: the event ‘not E’, written E, with P(E) + P(E) = 1.

Range: for any event E, 0 ≤ P(E) ≤ 1.

Important Formulas (Chapter 14)

Probability of an event: P(E) = (favourable outcomes) ÷ (total outcomes).

Complement rule: P(not E) = 1 − P(E), i.e. P(E) + P(E) = 1.

Sure event: P(E) = 1.  •  Impossible event: P(E) = 0.

Bound: 0 ≤ P(E) ≤ 1 for every event E.

Two dice / die thrown twice: total number of equally likely outcomes = 6 × 6 = 36.

Standard deck: 52 cards = 4 suits × 13 cards; 26 red + 26 black; 12 face cards (4 kings, 4 queens, 4 jacks); 4 aces.

Exercise 14.1 — Solutions

Questions are reproduced verbatim from the NCERT Class 10 Maths textbook (Reprint 2026–27); the worked solutions are original and verified.

1. Complete the following statements: (i) Probability of an event E + Probability of the event ‘not E’ = ______. (ii) The probability of an event that cannot happen is ______. Such an event is called ______. (iii) The probability of an event that is certain to happen is ______. Such an event is called ______. (iv) The sum of the probabilities of all the elementary events of an experiment is ______. (v) The probability of an event is greater than or equal to ______ and less than or equal to ______.

SOLUTION (i) 1  [since P(E) + P(not E) = 1]. (ii) 0; such an event is called an impossible event. (iii) 1; such an event is called a sure (or certain) event. (iv) 1. (v) greater than or equal to 0 and less than or equal to 1.

2. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. (iii) A trial is made to answer a true-false question. The answer is right or wrong. (iv) A baby is born. It is a boy or a girl.

SOLUTION (i) Not equally likely — whether the car starts depends on its condition, fuel, etc., so the two results are not equally likely. (ii) Not equally likely — scoring or missing depends on the player’s skill and practice, so the outcomes are not equally likely. (iii) Equally likely — a blind guess on a true-false question is just as likely to be right as wrong. (iv) Equally likely — a newborn is (assumed) equally likely to be a boy or a girl.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

SOLUTION A fair (unbiased) coin is symmetrical, so when tossed at random it is equally likely to land heads or tails — each outcome has probability ½. Since neither team is favoured over the other, tossing a coin gives both teams an equal chance, which is why it is considered a fair method of deciding.

4. Which of the following cannot be the probability of an event? (A) 2/3   (B) −1.5   (C) 15%   (D) 0.7

SOLUTION The probability of an event always satisfies 0 ≤ P(E) ≤ 1. (A) 2/3 ≈ 0.67 — valid. (C) 15% = 0.15 — valid. (D) 0.7 — valid. (B) −1.5 is negative, so it lies outside [0, 1]. ∴ (B) −1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

SOLUTION P(not E) = 1 − P(E) = 1 − 0.05 = 0.95.

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?

SOLUTION (i) There are no orange candies in the bag, so taking out an orange candy is impossible. P(orange) = 0. (ii) Every candy is lemon flavoured, so taking out a lemon candy is a sure event. P(lemon) = 1.

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

SOLUTION ‘Having the same birthday’ is the complement of ‘not having the same birthday’. P(same birthday) = 1 − P(not same birthday) = 1 − 0.992 = 0.008.

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

SOLUTION Total balls = 3 + 5 = 8 (all equally likely). (i) Favourable (red) = 3, so P(red) = 3/8. (ii) P(not red) = 1 − P(red) = 1 − 3/8 = 5/8. (Check: 5 black balls out of 8 also give 5/8.)

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

SOLUTION Total marbles = 5 + 8 + 4 = 17. (i) P(red) = 5/17. (ii) P(white) = 8/17. (iii) P(not green) = 1 − P(green) = 1 − 4/17 = 13/17.

10. A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a ₹ 5 coin?

SOLUTION Total coins = 100 + 50 + 20 + 10 = 180. (i) P(50 p coin) = 100/180 = 5/9. (ii) Number of ₹ 5 coins = 10, so P(₹ 5 coin) = 10/180 = 1/18. Hence P(not a ₹ 5 coin) = 1 − 1/18 = 17/18.

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

SOLUTION Total fish = 5 + 8 = 13. Favourable (male) = 5. P(male fish) = 5/13.

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

SOLUTION Total outcomes = 8 (the numbers 1 to 8, equally likely). (i) Favourable = {8} = 1, so P(8) = 1/8. (ii) Odd numbers = {1, 3, 5, 7} = 4, so P(odd) = 4/8 = 1/2. (iii) Numbers > 2 = {3, 4, 5, 6, 7, 8} = 6, so P(> 2) = 6/8 = 3/4. (iv) Numbers < 9 = {1, 2, 3, 4, 5, 6, 7, 8} = 8 (all of them), so P(< 9) = 8/8 = 1.

13. A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

SOLUTION Total outcomes = 6 (the faces 1, 2, 3, 4, 5, 6). (i) Prime numbers = {2, 3, 5} = 3, so P(prime) = 3/6 = 1/2. (ii) Numbers between 2 and 6 = {3, 4, 5} = 3, so P = 3/6 = 1/2. (iii) Odd numbers = {1, 3, 5} = 3, so P(odd) = 3/6 = 1/2.

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

SOLUTION Total outcomes = 52 (well-shuffled deck, equally likely). (i) Red kings = 2 (king of hearts, king of diamonds), so P = 2/52 = 1/26. (ii) Face cards = 12 (4 kings + 4 queens + 4 jacks), so P = 12/52 = 3/13. (iii) Red face cards = 6 (3 in hearts + 3 in diamonds), so P = 6/52 = 3/26. (iv) Jack of hearts = 1 card, so P = 1/52. (v) Spades = 13, so P = 13/52 = 1/4. (vi) Queen of diamonds = 1 card, so P = 1/52.

15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

SOLUTION (i) Total cards = 5; queens = 1, so P(queen) = 1/5. (ii) After the queen is put aside, 4 cards remain (ten, jack, king, ace). (a) Aces among the remaining 4 = 1, so P(ace) = 1/4. (b) No queen remains, so P(queen) = 0/4 = 0.

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

SOLUTION Total pens = 132 + 12 = 144. Good pens = 132. P(good pen) = 132/144 = 11/12.

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

SOLUTION (i) Total bulbs = 20, defective = 4, so P(defective) = 4/20 = 1/5. (ii) After removing one good bulb, 19 bulbs remain, of which the non-defective ones = 16 − 1 = 15 (originally 20 − 4 = 16 good bulbs, one already drawn). P(not defective) = 15/19 = 15/19.

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

SOLUTION Total discs = 90 (numbers 1 to 90). (i) Two-digit numbers run from 10 to 90 = 81 numbers, so P = 81/90 = 9/10. (ii) Perfect squares from 1 to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9 numbers, so P = 9/90 = 1/10. (iii) Multiples of 5 from 1 to 90: 5, 10, …, 90 = 18 numbers, so P = 18/90 = 1/5.

19. A child has a die whose six faces show the letters A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?

SOLUTION Total faces = 6 (letters A, B, C, D, E, A). (i) Letter A appears on 2 faces, so P(A) = 2/6 = 1/3. (ii) Letter D appears on 1 face, so P(D) = 1/6.

20. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6 (a 3 m × 2 m rectangle containing a circle of diameter 1 m). What is the probability that it will land inside the circle with diameter 1 m?

SOLUTION This is a probability based on area; the die is equally likely to land anywhere in the rectangle. Area of rectangle = 3 × 2 = 6 m2. Radius of circle = 1/2 m, so area of circle = πr2 = π × (1/2)2 = π/4 m2. P(inside circle) = (area of circle) ÷ (area of rectangle) = (π/4) ÷ 6 = π/24.

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?

SOLUTION Total pens = 144; defective = 20; good = 144 − 20 = 124. (i) She buys it only if it is good: P(buy) = 124/144 = 31/36. (ii) P(not buy) = 20/144 = 5/36, or 1 − 31/36 = 5/36.

22. Refer to Example 13. (i) Complete the following table of probabilities for the sum on two dice (sums 2 to 12). (ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11.’ Do you agree with this argument? Justify your answer.

SOLUTION Two dice give 6 × 6 = 36 equally likely outcomes. Counting the number of ways to get each sum gives the table below.
Sum on 2 dice23456789101112
Probability1/362/363/364/365/366/365/364/363/362/361/36
(ii) No, the argument is not correct. The 11 sums are not equally likely. For example, a sum of 2 occurs in only 1 way (1, 1), while a sum of 7 occurs in 6 ways (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). So P(2) = 1/36 but P(7) = 6/36 — they cannot all be 1/11.

23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

SOLUTION Tossing a coin 3 times gives 23 = 8 equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Hanif wins on HHH or TTT — 2 favourable outcomes, so P(win) = 2/8 = 1/4. P(lose) = 1 − P(win) = 1 − 1/4 = 3/4.

24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?

SOLUTION Throwing a die twice gives 6 × 6 = 36 equally likely outcomes. (i) Outcomes where 5 appears at least once: 5 in the first throw (6 outcomes) + 5 in the second throw (6 outcomes) − (5, 5) counted twice = 11. So outcomes with no 5 = 36 − 11 = 25, and P(no 5 either time) = 25/36. (ii) P(5 at least once) = 1 − P(no 5) = 1 − 25/36 = 11/36.

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer. (i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3. (ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

SOLUTION (i) Incorrect. Tossing two coins actually has 4 equally likely outcomes: HH, HT, TH, TT. ‘One of each’ (HT or TH) happens in 2 ways, so P(one of each) = 2/4 = 1/2, while P(two heads) = P(two tails) = 1/4. The three described events are not equally likely, so the probability is not 1/3 for each. (ii) Correct. A die has 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6) out of 6 faces. These are equally likely, so P(odd) = 3/6 = 1/2.

Common Mistakes to Avoid

Watch out for these

  • Assuming all listed events are equally likely — e.g. the sums on two dice (2 to 12) are not equally likely, so you cannot give each 1/11.
  • Forgetting that two coins/two dice give 4 and 36 outcomes — not 3 and 11 — because order matters (HT ≠ TH).
  • Counting endpoints wrongly: ‘between 2 and 6’ on a die means {3, 4, 5}, not {2, 3, 4, 5, 6}.
  • Mishandling ‘without replacement’ questions — reduce both the favourable count and the total after a card/bulb is removed.
  • Writing a probability greater than 1 or negative — always check 0 ≤ P(E) ≤ 1.
  • Forgetting to simplify fractions, e.g. 12/52 should be written as 3/13.

Practice MCQs & Assertion–Reason

1. Which of the following cannot be the probability of an event?

(a) 0    (b) 0.3    (c) 1.2    (d) 1

2. A die is thrown once. The probability of getting a prime number is:

(a) 1/6    (b) 1/3    (c) 1/2    (d) 2/3

3. One card is drawn from a well-shuffled deck of 52 cards. The probability of getting a face card is:

(a) 1/13    (b) 3/13    (c) 1/4    (d) 4/13

4. If P(E) = 0.37, then P(not E) is:

(a) 0.37    (b) 0.63    (c) 0.73    (d) 1.37

5. A bag has 3 red and 5 black balls. The probability of drawing a black ball is:

(a) 3/8    (b) 1/2    (c) 5/8    (d) 5/3

6. Two dice are thrown together. The probability that the sum is 7 is:

(a) 1/6    (b) 5/36    (c) 1/9    (d) 7/36

7. A coin is tossed twice. The probability of getting at least one head is:

(a) 1/4    (b) 1/2    (c) 3/4    (d) 1

8. A box has discs numbered 1 to 90. The probability of drawing a perfect square is:

(a) 1/10    (b) 9/10    (c) 1/9    (d) 1/18

9. The probability of a sure event is:

(a) 0    (b) 1/2    (c) 1    (d) −1

10. The sum of the probabilities of all the elementary events of an experiment is:

(a) 0    (b) 1/2    (c) 1    (d) depends on the experiment

Answer key: 1-(c), 2-(c), 3-(b), 4-(b), 5-(c), 6-(a), 7-(c), 8-(a), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The probability of an impossible event is 0.

Reason: An impossible event has no favourable outcome, so the favourable count is 0.

A-R 2. Assertion: If P(E) = 0.05, then P(not E) = 0.95.

Reason: For any event, P(E) + P(not E) = 1.

A-R 3. Assertion: When two dice are thrown, the 11 sums 2, 3, …, 12 each have probability 1/11.

Reason: There are 11 possible values of the sum.

A-R 4. Assertion: The probability of drawing a red face card from a deck of 52 cards is 3/26.

Reason: There are 6 red face cards in a deck of 52 cards.

A-R 5. Assertion: The probability of any event E satisfies 0 ≤ P(E) ≤ 1.

Reason: The number of favourable outcomes is always less than or equal to the total number of outcomes, and both are non-negative.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Theoretical probability: P(E) = favourable outcomes ÷ total outcomes (for equally likely outcomes).
  • P(sure event) = 1, P(impossible event) = 0, and 0 ≤ P(E) ≤ 1 for every event.
  • Complement rule: P(E) + P(not E) = 1, so P(not E) = 1 − P(E).
  • An elementary event has a single outcome; all elementary-event probabilities of an experiment add up to 1.
  • Two coins → 4 outcomes; two dice (or a die thrown twice) → 36 outcomes; three coins → 8 outcomes.
  • A deck of 52 cards: 26 red + 26 black, 12 face cards, 4 aces; always simplify the final fraction.

How to score full marks in this chapter

Start every problem by writing the total number of equally likely outcomes and the favourable count separately — this earns method marks. Use the complement rule P(not E) = 1 − P(E) to shorten ‘at least one’ and ‘not’ questions. Be careful with ‘without replacement’ (reduce the total), and remember that order matters for two coins/dice, giving 4 and 36 outcomes. Always reduce your final fraction to lowest terms.

Frequently Asked Questions

What is Class 10 Maths Chapter 14 Probability about?

Chapter 14 covers the theoretical (classical) approach to probability: equally likely outcomes, P(E) = favourable ÷ total, elementary events, sure and impossible events, the bound 0 ≤ P(E) ≤ 1, and complementary events with P(E) + P(not E) = 1, applied to coins, dice, cards and balls.

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

There is one exercise, Exercise 14.1, with 25 questions. Every question is reproduced and solved step by step on this page.

What is the formula for the probability of an event?

For equally likely outcomes, P(E) = (number of outcomes favourable to E) ÷ (number of all possible outcomes). The complement is P(not E) = 1 − P(E).

Are these Class 10 Maths Chapter 14 solutions free?

Yes. All solutions are free and follow the official NCERT Class 10 Mathematics textbook for the 2026–27 session, with answers verified against the book.

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