---
title: "NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1"
url: https://www.swavid.com/maths/class/10/chapter/probability/ncert-solutions/exercise-14-1
dateModified: 2026-10-07T15:58:18+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1

Chapter 14: Probability. Every question from Exercise 14.1, with full working and the final answer.

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## EXERCISE 14.1

### Question 1

*1 mark · Fill in the blank*

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 .

**Part (i)**

1. The sum of the probability of an event E and its complementary event not E is always 1.

Answer (i): 1

**Part (ii)**

1. An event that cannot happen has a probability of 0 and is called an impossible event.

Answer (ii): 0, impossible event

**Part (iii)**

1. An event that is certain to happen has a probability of 1 and is called a sure event or a certain event.

Answer (iii): 1, sure event (or certain event)

**Part (iv)**

1. The sum of the probabilities of all the elementary events of an experiment is 1.

Answer (iv): 1

**Part (v)**

1. The probability of any event lies between 0 and 1 inclusive.

Answer (v): 0, 1

**Answer:** Fill in the blanks with 1, 0, impossible event, 1, certain event (or sure event), 1, 0, 1.

> Common mistake: Confusing elementary events with general events.

### Question 2

*3 marks · Short answer*

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.

**Part (i)**

1. A car starting or not starting depends on various factors like mechanical condition and fuel, so the outcomes are not equally likely.

Answer (i): Not equally likely

**Part (ii)**

1. Shooting or missing a basketball depends on the player's skill and practice, so the outcomes are not equally likely.

Answer (ii): Not equally likely

**Part (iii)**

1. A true-false question has two possible answers, right or wrong, which have the same chance, assuming random guessing or equal conditions.

Answer (iii): Equally likely

**Part (iv)**

1. The biological probability of a newborn being a boy or a girl is assumed to be symmetric and equal.

Answer (iv): Equally likely

**Answer:** Only trials (iii) and (iv) have equally likely outcomes.

> Common mistake: Assuming all real-world binary outcomes are equally likely.

### Question 3

*3 marks · Short answer*

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**

1. A coin has only two possible outcomes: head or tail.
2. Both outcomes are symmetrical, fair, and have equal chances of occurrence.
3. Therefore, tossing a coin is considered completely unbiased and a fair way to decide.

**Answer:** Tossing a coin is considered a fair way because its two possible outcomes, head and tail, are equally likely.

> Common mistake: Writing that coins always land on heads or tails 50% of the time in every small sample.

### Question 4

*1 mark · MCQ*

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

- \frac{2}{3}
- -1.5
- 15%
- 0.7

**Solution**

1. The probability of any event E always lies between 0 and 1 inclusive, so $0 \le P(E) \le 1$.
2. Option (B) is $-1.5$, which is less than 0, so it cannot be a probability.

**Answer:** (B) -1.5

> Common mistake: Choosing a fraction greater than 1 by mistake.

### Question 5

*3 marks · Short answer*

If P(E) = 0.05, what is the probability of 'not E'?

**Solution**

1. Given: $P(E) = 0.05$.
2. Formula: $P(\bar{E}) = 1 - P(E)$.
3. Substitution: $P(\text{not } E) = 1 - 0.05$.
4. Result: $0.95$

**Answer:** 0.95

> Common mistake: Subtracting incorrectly from 1.

### Question 6

*3 marks · Short answer*

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?

**Part (i)**

1. The bag contains only lemon-flavored candies, so getting an orange-flavored candy is impossible.

Answer (i): 0

**Part (ii)**

1. Since all candies in the bag are lemon-flavored, drawing a lemon-flavored candy is a sure event.

Answer (ii): 1

**Answer:** (i) 0, (ii) 1

> Common mistake: Writing probabilities greater than 1 or less than 0.

### Question 7

*3 marks · Short answer*

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?

**Part (i)**

1. Let E be the event that the 2 students have the same birthday and $\overline{E}$ be the event that they do not have the same birthday.
2. We are given that $P(\overline{E}) = 0.992$.
3. We know that $P(E) + P(\overline{E}) = 1$, so $P(E) = 1 - P(\overline{E}) = 1 - 0.992 = 0.008$.

Answer (i): $0.008$

**Answer:** 0.008

> Common mistake: Subtracting incorrectly from 1.

### Question 8

*3 marks · Short answer*

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?

**Part (i)**

1. Total number of balls = $3 + 5 = 8$.
2. Number of favourable outcomes for a red ball = $3$.
3. Probability of drawing a red ball = $\frac{3}{8}$.

Answer (i): $\frac{3}{8}$

**Part (ii)**

1. Number of favourable outcomes for not getting a red ball (getting a black ball) = $5$.
2. Probability of not drawing a red ball = $\frac{5}{8}$.

Answer (ii): $\frac{5}{8}$

**Answer:** (i) $\frac{3}{8}$, (ii) $\frac{5}{8}$

> Common mistake: Confusing red and black ball counts.

### Question 9

*3 marks · Short answer*

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?

**Part (i)**

1. Total number of marbles = $5 + 8 + 4 = 17$.
2. Number of red marbles = $5$.
3. Probability of drawing a red marble = $\frac{5}{17}$.

Answer (i): $\frac{5}{17}$

**Part (ii)**

1. Number of white marbles = $8$.
2. Probability of drawing a white marble = $\frac{8}{17}$.

Answer (ii): $\frac{8}{17}$

**Part (iii)**

1. Number of marbles that are not green = $5 + 8 = 13$.
2. Probability of drawing a marble that is not green = $\frac{13}{17}$.

Answer (iii): $\frac{13}{17}$

**Answer:** (i) $\frac{5}{17}$, (ii) $\frac{8}{17}$, (iii) $\frac{13}{17}$

> Common mistake: Adding the total number of marbles incorrectly.

### Question 10

*3 marks · Short answer*

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?

**Part (i)**

1. Total number of coins = $100 + 50 + 20 + 10 = 180$.
2. Number of 50p coins = $100$.
3. Probability that the coin is a 50p coin = $\frac{100}{180} = \frac{5}{9}$.

Answer (i): $\frac{5}{9}$

**Part (ii)**

1. Number of Rs 5 coins = $10$.
2. Number of coins that are not Rs 5 coins = $180 - 10 = 170$.
3. Probability that the coin will not be a Rs 5 coin = $\frac{170}{180} = \frac{17}{18}$.

Answer (ii): $\frac{17}{18}$

**Answer:** (i) $\frac{5}{9}$, (ii) $\frac{17}{18}$

> Common mistake: Forgetting to sum all coin types for the total outcomes.

### Question 11

*3 marks · Short answer*

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 (see Fig. 14.4). What is the probability that the fish taken out is a male fish?

**Solution**

1. Total number of fish in the tank = $5 \text{ (male)} + 8 \text{ (female)} = 13$.
2. Number of male fish in the tank = $5$.
3. Probability that the fish taken out is a male fish = $\frac{5}{13}$.

**Answer:** $\frac{5}{13}$

> Common mistake: Using the number of female fish as the favorable outcome.

### Question 12

*3 marks · Short answer*

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 (see Fig. 14.5), 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?

**Part (i)**

1. Total possible outcomes = $8$.
2. Number of outcomes favourable to getting 8 = $1$.
3. Probability = $\frac{1}{8}$.

Answer (i): $\frac{1}{8}$

**Part (ii)**

1. Odd numbers on the spinner are $1, 3, 5, 7$ (total $4$).
2. Probability = $\frac{4}{8} = \frac{1}{2}$.

Answer (ii): $\frac{1}{2}$

**Part (iii)**

1. Numbers greater than 2 are $3, 4, 5, 6, 7, 8$ (total $6$).
2. Probability = $\frac{6}{8} = \frac{3}{4}$.

Answer (iii): $\frac{3}{4}$

**Part (iv)**

1. All numbers on the spinner are less than 9 (total $8$).
2. Probability = $\frac{8}{8} = 1$.

Answer (iv): $1$

**Answer:** (i) $\frac{1}{8}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{4}$, (iv) $1$

> Common mistake: Including 2 when counting numbers greater than 2.

### Question 13

*3 marks · Short answer*

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.

**Part (i)**

1. The total number of possible outcomes when a die is thrown once is $6$ ($1, 2, 3, 4, 5, 6$).
2. The prime numbers among these are $2, 3, 5$, so the number of favourable outcomes is $3$.
3. The probability of getting a prime number is $\frac{3}{6} = \frac{1}{2}$.

Answer (i): $\frac{1}{2}$

**Part (ii)**

1. The numbers lying between $2$ and $6$ are $3, 4, 5$, so the number of favourable outcomes is $3$.
2. The probability of getting a number lying between $2$ and $6$ is $\frac{3}{6} = \frac{1}{2}$.

Answer (ii): $\frac{1}{2}$

**Part (iii)**

1. The odd numbers among the possible outcomes are $1, 3, 5$, so the number of favourable outcomes is $3$.
2. The probability of getting an odd number is $\frac{3}{6} = \frac{1}{2}$.

Answer (iii): $\frac{1}{2}$

**Answer:** The probabilities are (i) $\frac{1}{2}$, (ii) $\frac{1}{2}$, (iii) $\frac{1}{2}$.

> Common mistake: Including $2$ or $6$ when counting numbers between $2$ and $6$.

### Question 14

*3 marks · Short answer*

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

**Part (i)**

1. The total number of possible outcomes is $52$.
2. There are $2$ kings of red colour (hearts and diamonds).
3. Therefore, the probability of getting a king of red colour is $\frac{2}{52} = \frac{1}{26}$.

Answer (i): \frac{1}{26}

**Part (ii)**

1. There are $12$ face cards in a deck ($4$ kings, $4$ queens, and $4$ jacks).
2. Therefore, the probability of getting a face card is $\frac{12}{52} = \frac{3}{13}$.

Answer (ii): \frac{3}{13}

**Part (iii)**

1. There are $6$ red face cards ($3$ in hearts and $3$ in diamonds).
2. Therefore, the probability of getting a red face card is $\frac{6}{52} = \frac{3}{26}$.

Answer (iii): \frac{3}{26}

**Part (iv)**

1. There is only $1$ jack of hearts in a deck.
2. Therefore, the probability of getting the jack of hearts is $\frac{1}{52}$.

Answer (iv): \frac{1}{52}

**Part (v)**

1. There are $13$ cards in the spade suit.
2. Therefore, the probability of getting a spade is $\frac{13}{52} = \frac{1}{4}$.

Answer (v): \frac{1}{4}

**Part (vi)**

1. There is only $1$ queen of diamonds in a deck.
2. Therefore, the probability of getting the queen of diamonds is $\frac{1}{52}$.

Answer (vi): \frac{1}{52}

**Answer:** Probabilities are (i) $\frac{1}{26}$ (ii) $\frac{3}{13}$ (iii) $\frac{3}{26}$ (iv) $\frac{1}{52}$ (v) $\frac{1}{4}$ (vi) $\frac{1}{52}$

> Common mistake: Confusing total face cards or red cards.

### Question 15

*3 marks · Short answer*

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?

**Part (i)**

1. The five cards are ten, jack, queen, king, and ace of diamonds, so the total number of possible outcomes is $5$.
2. There is only $1$ queen among these cards.
3. Therefore, the probability that the card is the queen is $\frac{1}{5}$.

Answer (i): \frac{1}{5}

**Part (ii)(a)**

1. If the queen is drawn and put aside, the remaining number of cards is $5 - 1 = 4$.
2. The remaining cards are ten, jack, king, and ace of diamonds.
3. Since there is $1$ ace among the remaining $4$ cards, the probability of getting an ace is $\frac{1}{4}$.

Answer (ii)(a): \frac{1}{4}

**Part (ii)(b)**

1. Since the queen has already been drawn and put aside, there are no queens left in the remaining $4$ cards.
2. The number of favourable outcomes for getting a queen is $0$.
3. Therefore, the probability of getting a second queen is $\frac{0}{4} = 0$.

Answer (ii)(b): 0

**Answer:** Probabilities are (i) $\frac{1}{5}$ (ii)(a) $\frac{1}{4}$ (ii)(b) $0$

> Common mistake: Not reducing the total number of cards to $4$ for the second draw.

### Question 16

*3 marks · Short answer*

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**

1. The number of defective pens is $12$ and the number of good pens is $132$.
2. The total number of pens in the lot is $12 + 132 = 144$, which are equally likely outcomes.
3. Let $E$ be the event that the pen taken out is a good one, so the number of outcomes favourable to $E$ is $132$.
4. The probability of getting a good pen is $P(E) = \frac{132}{144} = \frac{11}{12}$.

**Answer:** $\frac{11}{12}$

> Common mistake: Dividing by the number of good pens instead of the total number of pens.

### Question 17

*3 marks · Short answer*

(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 ?

**Part (i)**

1. Total number of bulbs in the lot = $20$, and number of defective bulbs = $4$.
2. Let E be the event 'the bulb drawn is defective'.
3. The probability that the bulb is defective is $\frac{4}{20} = \frac{1}{5}$.

Answer (i): \frac{1}{5}

**Part (ii)**

1. The bulb drawn is not defective and is not replaced, so the remaining total number of bulbs is $20 - 1 = 19$.
2. The number of remaining defective bulbs is still $4$, so the number of non-defective bulbs left is $19 - 4 = 15$.
3. The probability that the bulb drawn from the rest is not defective is $\frac{15}{19}$.

Answer (ii): \frac{15}{19}

**Answer:** Probabilities are (i) $\frac{1}{5}$ (ii) $\frac{15}{19}$

> Common mistake: Failing to decrease both the total number of bulbs and the target count for the second part.

### Question 18

*3 marks · Short answer*

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.

**Part (i)**

1. The total number of possible outcomes is $90$ (numbers from $1$ to $90$).
2. The numbers from $1$ to $9$ are single-digit numbers, so the number of two-digit numbers is $90 - 9 = 81$.
3. The probability that it bears a two-digit number is $\frac{81}{90} = \frac{9}{10}$.

Answer (i): \frac{9}{10}

**Part (ii)**

1. The perfect square numbers from $1$ to $90$ are $1, 4, 9, 16, 25, 36, 49, 64, 81$, which are $9$ in total.
2. The probability that it bears a perfect square number is $\frac{9}{90} = \frac{1}{10}$.

Answer (ii): \frac{1}{10}

**Part (iii)**

1. The numbers divisible by $5$ from $1$ to $90$ are $5, 10, 15, \dots, 90$, which are $\frac{90}{5} = 18$ in total.
2. The probability that it bears a number divisible by $5$ is $\frac{18}{90} = \frac{1}{5}$.

Answer (iii): \frac{1}{5}

**Answer:** Probabilities are (i) $\frac{9}{10}$ (ii) $\frac{1}{10}$ (iii) $\frac{1}{5}$

> Common mistake: Counting single-digit numbers instead of subtracting them to find two-digit numbers.

### Question 19

*3 marks · Short answer*

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

**Part (i)**

1. Total number of possible outcomes is 6 as the die has six faces.
2. The number of faces showing 'A' is 2.
3. Probability of getting A is $P(\text{A}) = \frac{2}{6} = \frac{1}{3}$.

Answer (i): \frac{1}{3}

**Part (ii)**

1. Total number of possible outcomes is 6.
2. The number of faces showing 'D' is 1.
3. Probability of getting D is $P(\text{D}) = \frac{1}{6}$.

Answer (ii): \frac{1}{6}

**Answer:** (i) $\frac{1}{3}$, (ii) $\frac{1}{6}$

> Common mistake: Counting the letter A as 1 instead of 2.

### Question 20

*3 marks · Short answer*

Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?

**Solution**

1. The rectangular region has dimensions $3\text{ m}$ by $2\text{ m}$, so its total area is $3 \times 2 = 6\text{ m}^2$.
2. The circle has a diameter of $1\text{ m}$, so its radius is $r = \frac{1}{2}\text{ m}$.
3. The area of the circular region is $\pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}\text{ m}^2$.
4. The probability that the die will land inside the circle is $\frac{\text{Area of the circle}}{\text{Area of the rectangle}} = \frac{\pi / 4}{6} = \frac{\pi}{24}$.

**Answer:** $\frac{\pi}{24}$

> Common mistake: Using diameter instead of radius when calculating the area of the circle.

### Question 21

*3 marks · Short answer*

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 ?

**Part (i)**

1. The total number of ball pens is $144$, out of which $20$ are defective, so the number of good pens is $144 - 20 = 124$.
2. Nuri will buy the pen if it is good, so the number of favourable outcomes for buying is $124$.
3. The probability that she will buy it is $\frac{124}{144} = \frac{31}{36}$.

Answer (i): $\frac{31}{36}$

**Part (ii)**

1. Nuri will not buy the pen if it is defective, so the number of favourable outcomes for not buying is $20$.
2. The probability that she will not buy it is $\frac{20}{144} = \frac{5}{36}$.

Answer (ii): $\frac{5}{36}$

**Answer:** The probabilities are (i) $\frac{31}{36}$, (ii) $\not from examination point of view}$.

> Common mistake: Using the number of defective pens for the first part instead of good pens.

### Question 22

*3 marks · Short answer*

Refer to Example 13. (i) Complete the following table:
Event : 'Sum on 2 dice' (2 through 12)
Probability (partial values given)
(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 \frac{1}{11}. Do you agree with this argument? Justify your answer.

**Part (i)**

1. The probabilities for sums 2 to 12 are $\frac{1}{36}$, $\frac{2}{36}$, $\frac{3}{36}$, $\frac{4}{36}$, $\frac{5}{36}$, $\frac{6}{36}$, $\frac{5}{36}$, $\frac{4}{36}$, $\frac{3}{36}$, $\frac{2}{36}$, and $\frac{1}{36}$ respectively.

Answer (i): \text{Table completed}

**Part (ii)**

1. The 11 sums are not equally likely because the number of favourable outcomes for each sum is different (e.g., sum 2 has only 1 outcome, while sum 7 has 6 outcomes).
2. Therefore, we do not agree with the student's argument.

Answer (ii): \text{No}

**Answer:** (i) Table completed with respective fractions, (ii) No, the outcomes are not equally likely.

> Common mistake: Assuming all sums of two dice have equal chances of occurring.

### Question 23

*3 marks · Short answer*

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**

1. When a one-rupee coin is tossed $3$ times, the total number of possible outcomes is $2^3 = 8$.
2. The complete set of possible outcomes is: $(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)$.
3. Hanif wins if he gets three heads $(H, H, H)$ or three tails $(T, T, T)$, which gives $2$ winning outcomes.
4. Hanif loses otherwise, so the number of outcomes favourable to Hanif losing is $8 - 2 = 6$.
5. The probability that Hanif will lose the game is $\frac{6}{8} = \frac{3}{4}$.

**Answer:** $\frac{3}{4}$

> Common mistake: Listing outcomes incorrectly or forgetting that losing means any outcome other than three heads or three tails.

### Question 24

*3 marks · Short answer*

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?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

**Part (i)**

1. Total number of outcomes when a die is thrown twice is $6 \times 6 = 36$.
2. The number of outcomes where 5 comes up at least once is 11.
3. The number of outcomes where 5 does not come up either time is $36 - 11 = 25$.
4. Probability that 5 will not come up either time is $\frac{25}{36}$.

Answer (i): \frac{25}{36}

**Part (ii)**

1. The outcomes where 5 comes up at least once are rows and columns containing 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5), making a total of 11 outcomes.
2. Probability that 5 will come up at least once is $\frac{11}{36}$.

Answer (ii): \frac{11}{36}

**Answer:** (i) $\frac{25}{36}$, (ii) $\frac{11}{36}$

> Common mistake: Double counting the outcome (5, 5) when finding outcomes where 5 appears at least once.

### Question 25

*3 marks · Short answer*

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 \frac{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 \frac{1}{2}.

**Part (i)**

1. The possible outcomes when two coins are tossed are $(H, H), (H, T), (T, H),$ and $(T, T)$, which means there are 4 possible outcomes.
2. The outcomes 'one head and one tail' consist of $(H, T)$ and $(T, H)$, so its probability is $\frac{2}{4} = \frac{1}{2}$, not $\frac{1}{3}$ because the outcomes are not equally likely.

Answer (i): Not correct, because the three outcomes are not equally likely.

**Part (ii)**

1. When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6, which are equally likely.
2. The odd numbers are 1, 3, and 5, so there are 3 favourable outcomes out of 6, giving a probability of $\frac{3}{6} = \frac{1}{2}$.

Answer (ii): Correct, because the two outcomes are equally likely.

**Answer:** Argument (i) is not correct, and argument (ii) is correct.

> Common mistake: Assuming that any set of listed outcomes in an experiment is automatically equally likely.

## Related pages

- [All Chapter 14 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/probability/ncert-solutions)

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