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NCERT Solutions · Class 10 Maths · Chapter 14

Chapter 14: Probability (Probability)

Step-by-step NCERT solutions for Class 10 Maths Chapter 14, Probability (2026-27 reprint): all 25 questions of Exercise 14.1, covering complementary events, coins, dice, a deck of cards, bags of balls and marbles, two dice and three coin tosses. All 25 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-14-probability

$P (E) = \frac{number of outcomes favourable to E}{number of all possible outcomes}$ , when the outcomes are equally likely. $0 \leq P (E) \leq 1$ ; a sure event has probability 1 and an impossible event 0. For the complement, $P (not E) = 1 - P (E)$ . A deck has 52 cards: 4 suits (♠ ♣ black, ♥ ♦ red) of 13 cards each, and 12 face cards (J, Q, K of each suit).

Exercise 14.1

Complete: (i) P(E) + P(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 (ii) 0, an impossible event (iii) 1, a sure (certain) event (iv) 1 (v) 0, 1

Which of these experiments have equally likely outcomes? (i) A car starts or does not start. (ii) A player shoots or misses a basketball shot. (iii) A true-false answer is right or wrong. (iv) A baby is born a boy or a girl.

Solution

(i) Not equally likely: whether a car starts depends on its condition, fuel, battery and so on.
(ii) Not equally likely: it depends on the player's skill.
(iii) Equally likely: a guessed answer is equally likely to be right or wrong.
(iv) Equally likely: a newborn is (taken to be) equally likely to be a boy or a girl.

(iii) and (iv)

Why is tossing a coin a fair way of deciding which team gets the ball at the start of a football game?

Solution

A fair coin has two equally likely outcomes, head and tail, each with probability $\frac{1}{2}$ . The result of a single toss cannot be predicted or influenced, so neither team has an advantage.

Because heads and tails are equally likely and the result is unpredictable, so both teams have the same chance.

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

Solution

A probability always lies between 0 and 1. $\frac{2}{3}$ , $15% = 0.15$ and $0.7$ do; $- 1.5$ is negative.

(B) −1.5

If P(E) = 0.05, what is P(not E)?

Solution

$P (not E) = 1 - 0.05 = 0.95$

0.95

A bag contains only lemon-flavoured candies. Malini takes one out without looking. Find the probability that it is (i) orange-flavoured (ii) lemon-flavoured.

Solution

There is no orange candy, so this is an impossible event: probability 0.
Every candy is lemon, so this is a sure event: probability 1.

(i) 0 (ii) 1

The probability that 2 students of a group do not have the same birthday is 0.992. What is the probability that they have the same birthday?

Solution

The two events are complementary: $1 - 0.992 = 0.008$ .

0.008

A bag has 3 red and 5 black balls. One ball is drawn at random. Find P(i) red (ii) not red.

Solution

Total $= 8$ balls.

$P (red) = \frac{3}{8}$
$P (not red) = 1 - \frac{3}{8} = \frac{5}{8}$

(i) 3/8 (ii) 5/8

A box has 5 red, 8 white and 4 green marbles. One is taken out at random. Find P(i) red (ii) white (iii) not green.

Solution

Total $= 5 + 8 + 4 = 17$ .

$\frac{5}{17}$
$\frac{8}{17}$
$1 - \frac{4}{17} = \frac{13}{17}$

(i) 5/17 (ii) 8/17 (iii) 13/17

A piggy bank has a hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. One coin falls out. Find the probability that it (i) is a 50p coin (ii) is not a ₹5 coin.

Solution

Total $= 100 + 50 + 20 + 10 = 180$ coins.

$\frac{100}{180} = \frac{5}{9}$
$1 - \frac{10}{180} = \frac{170}{180} = \frac{17}{18}$

(i) 5/9 (ii) 17/18

A shopkeeper takes out one fish at random from a tank with 5 male and 8 female fish. Find the probability that it is male.

Solution

$P (male) = \frac{5}{5 + 8} = \frac{5}{13}$

5/13

An arrow spun on a dial stops at one of 1, 2, …, 8 (equally likely). Find the probability that it points at (i) 8 (ii) an odd number (iii) a number greater than 2 (iv) a number less than 9.

Solution

There are 8 outcomes.

$\frac{1}{8}$
Odd numbers 1, 3, 5, 7: $\frac{4}{8} = \frac{1}{2}$
3 to 8 (six numbers): $\frac{6}{8} = \frac{3}{4}$
All 8 numbers are less than 9: $\frac{8}{8} = 1$

(i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1

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

Solution

The outcomes are 1, 2, 3, 4, 5, 6.

Primes 2, 3, 5: $\frac{3}{6} = \frac{1}{2}$
Between 2 and 6: 3, 4, 5, so $\frac{3}{6} = \frac{1}{2}$
Odd 1, 3, 5: $\frac{3}{6} = \frac{1}{2}$

(i) 1/2 (ii) 1/2 (iii) 1/2

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

Solution

Red kings (♥K, ♦K): $\frac{2}{52} = \frac{1}{26}$
Face cards: $\frac{12}{52} = \frac{3}{13}$
Red face cards (J, Q, K of ♥ and ♦): $\frac{6}{52} = \frac{3}{26}$
$\frac{1}{52}$
Spades: $\frac{13}{52} = \frac{1}{4}$
$\frac{1}{52}$

(i) 1/26 (ii) 3/13 (iii) 3/26 (iv) 1/52 (v) 1/4 (vi) 1/52

The ten, jack, queen, king and ace of diamonds are shuffled face down and one is picked. (i) Find P(queen). (ii) If the queen is drawn and put aside, find the probability that the second card is (a) an ace (b) a queen.

Solution

$\frac{1}{5}$
Now 4 cards remain (10, J, K, A). (a) $P (ace) = \frac{1}{4}$ . (b) There is no queen left, so $P (queen) = 0$ .

(i) 1/5 (ii) (a) 1/4 (b) 0

12 defective pens are mixed with 132 good ones. One pen is taken at random. Find the probability that it is good.

Solution

$P (good) = \frac{132}{132 + 12} = \frac{132}{144} = \frac{11}{12}$

11/12

(i) A lot of 20 bulbs has 4 defective ones. Find the probability that a bulb drawn at random is defective. (ii) If that bulb is not defective and is not replaced, find the probability that the next bulb drawn is not defective.

Solution

$\frac{4}{20} = \frac{1}{5}$
19 bulbs remain, of which $16 - 1 = 15$ are good: $\frac{15}{19}$

(i) 1/5 (ii) 15/19

A box has 90 discs numbered 1 to 90. One is drawn at random. Find the probability that it bears (i) a two-digit number (ii) a perfect square (iii) a number divisible by 5.

Solution

Two-digit numbers 10 to 90: $81$ , so $\frac{81}{90} = \frac{9}{10}$
Perfect squares 1, 4, 9, …, 81: $9$ , so $\frac{9}{90} = \frac{1}{10}$
Multiples of 5 up to 90: $18$ , so $\frac{18}{90} = \frac{1}{5}$

(i) 9/10 (ii) 1/10 (iii) 1/5

A die has the letters A, B, C, D, E, A on its faces. It is thrown once. Find P(i) A (ii) D.

Solution

Two faces show A: $\frac{2}{6} = \frac{1}{3}$
$\frac{1}{6}$

(i) 1/3 (ii) 1/6

A die is dropped at random on a 3 m × 2 m rectangle containing a circle of diameter 1 m. Find the probability that it lands inside the circle.

Solution

Here the probability is the ratio of areas.

Rectangle $= 3 \times 2 = 6$ m²
Circle $= π \times (\frac{1}{2})^{2} = \frac{π}{4}$ m²

$P = \frac{π /4}{6} = \frac{π}{24}$

π/24

A lot of 144 ball pens has 20 defective ones. Nuri buys a pen only if it is good. The shopkeeper draws one at random. Find the probability that (i) she buys it (ii) she does not buy it.

Solution

Good pens $= 144 - 20 = 124$ .

$\frac{124}{144} = \frac{31}{36}$
$1 - \frac{31}{36} = \frac{5}{36}$

(i) 31/36 (ii) 5/36

(i) Complete the table of probabilities for the sum on two dice. (ii) A student argues that the 11 sums 2 to 12 each have probability 1/11. Do you agree?

Solution

Two dice give $6 \times 6 = 36$ equally likely outcomes. Count the pairs for each sum (for 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)).

Sum	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

(ii) No. The 11 sums are not equally likely: a sum of 2 happens only as (1,1), but a sum of 7 happens in 6 ways. So they cannot each have probability $\frac{1}{11}$ .

(i) As in the table. (ii) No: the sums are not equally likely outcomes.

Hanif tosses a coin 3 times and wins if all three results are the same (three heads or three tails). Find the probability that he loses.

Solution

The outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT: 8 in all. He wins in 2 (HHH, TTT).

$P (lose) = 1 - \frac{2}{8} = \frac{3}{4}$

3/4

A die is thrown twice. Find the probability that (i) 5 does not come up either time (ii) 5 comes up at least once.

Solution

There are 36 outcomes. 5 comes up at least once in (5,1)…(5,6) and (1,5)…(6,5), counting (5,5) once: $6 + 6 - 1 = 11$ outcomes.

$1 - \frac{11}{36} = \frac{25}{36}$ (or: 5 choices each time, $\frac{5 \times 5}{36}$ )
$\frac{11}{36}$

(i) 25/36 (ii) 11/36

Which arguments are correct? (i) Tossing two coins has three outcomes (two heads, two tails, one of each), so each has probability 1/3. (ii) A die gives an odd or an even number, so P(odd) = 1/2.

Solution

Incorrect. The equally likely outcomes are HH, HT, TH and TT. "One of each" happens in 2 of the 4 ways, so its probability is $\frac{1}{2}$ , while two heads and two tails each have $\frac{1}{4}$ .
Correct. Of the 6 equally likely outcomes, 3 are odd (1, 3, 5), so $P (odd) = \frac{3}{6} = \frac{1}{2}$ .

(i) Incorrect (ii) Correct

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