---
title: "Probability: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/probability
---

# Probability: CBSE Class 10 Maths previous year questions

24 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 18

*1 mark · MCQ*

A die is thrown once. Probability of getting a number other than 3 is :

- \frac{1}{6}
- \frac{3}{6}
- \frac{5}{6}
- 1

**Solution**

1. When a die is thrown once, the total number of possible outcomes is 6.
2. The outcomes other than 3 are 1, 2, 4, 5, and 6, which are 5 outcomes in total.
3. The required probability is $\frac{5}{6}$.

**Answer:** (c) \frac{5}{6}

> Common mistake: Writing the probability of getting the number 3 instead of a number other than 3.

### Question 19

*1 mark · Assertion and reason*

Assertion (A) : The probability that a leap year has 53 Mondays is $\frac{2}{7}$.
Reason (R) : The probability that a non-leap year has 53 Mondays is $\frac{5}{7}$.

- Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.

**Solution**

1. A leap year has 366 days, which consists of 52 complete weeks and 2 extra days.
2. The 2 extra days can be {Sun, Mon}, {Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, or {Sat, Sun}.
3. Out of these 7 possible outcomes, 2 outcomes contain Monday, so the probability of 53 Mondays in a leap year is $\frac{2}{7}$, making Assertion (A) true.
4. A non-leap year has 365 days, which consists of 52 complete weeks and 1 extra day.
5. The 1 extra day can be any of the 7 days of the week, so the probability of 53 Mondays in a non-leap year is $\frac{1}{7}$, making Reason (R) false.

**Answer:** Assertion (A) is true, but Reason (R) is false.

> Common mistake: Confusing the number of extra days in a non-leap year and incorrectly calculating the probability as 5/7.

### Question 31

*3 marks · Short answer*

Two dice of different colours are thrown at the same time. Write down all the possible outcomes. What is the probability that :
(i) same number appears on both the dice ?
(ii) different number appears on both the dice ?

**Part (i)**

1. Total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
2. Outcomes where the same number appears on both dice are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$, which are 6 in number.
3. Probability = $\frac{6}{36} = \frac{1}{6}$.

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

**Part (ii)**

1. Outcomes where different numbers appear on both dice are the remaining outcomes, or $36 - 6 = 30$.
2. Probability = $\frac{30}{36} = \frac{5}{6}$.

Answer (ii): \frac{5}{6}

**Answer:** Total possible outcomes are 36.

> Common mistake: Listing outcomes incorrectly or miscalculating the number of favorable outcomes for different numbers.

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 18

*1 mark · MCQ*

In a random experiment of throwing a die, which of the following is a sure event?

- Getting a number between 1 and 6
- Getting an odd number < 7
- Getting an even number < 7
- Getting a natural number < 7

**Solution**

1. When throwing a standard die, the possible outcomes are {1, 2, 3, 4, 5, 6}.
2. Every outcome is a natural number less than 7, making it a sure event with probability 1.

**Answer:** (d) Getting a natural number < 7

> Common mistake: Choosing an event like getting a number between 1 and 6, which excludes 1 and 6.

### Question 31

*3 marks · Short answer*

A lot consists of 200 pens of which 180 are good and the rest are defective. A customer will buy a pen if it is not defective. The shopkeeper draws a pen at random and gives it to the customer. What is the probability that the customer will not buy it? Another lot of 100 pens containing 80 good pens is mixed with the previous lot of 200 pens. The shopkeeper now draws one pen at random from the entire lot and gives it to the customer. What is the probability that the customer will buy the pen?

**Solution**

1. Total number of pens in the first lot = $200$, number of good pens = $180$, so defective pens = $200 - 180 = 20$.
2. The customer will not buy the pen if it is defective. Probability that the customer will not buy = $\frac{\text{Number of defective pens}}{\text{Total pens}} = \frac{20}{200} = \frac{1}{10}$.
3. Another lot of $100$ pens containing $80$ good pens is mixed with the first lot.
4. New total number of pens = $200 + 100 = 300$.
5. New total number of good pens = $180 + 80 = 260$.
6. The customer will buy the pen if it is good. Probability that the customer will buy the pen = $\frac{\text{Total good pens}}{\text{New total pens}} = \frac{260}{300} = \frac{13}{15}$.

**Answer:** Probability that the customer will not buy the first pen is $\frac{1}{10}$, and the probability that the customer will buy a pen from the mixed lot is $\frac{13}{15}$.

> Common mistake: Confusing the number of good pens with defective pens, or failing to update the total number of pens and good pens correctly after mixing.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 13

*1 mark · MCQ*

A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is :

- 3/13
- 2/13
- 1/2
- 3/26

**Solution**

1. Total number of cards in a deck is 52.
2. Number of red face cards is 6 (3 in hearts and 3 in diamonds).
3. Probability = $\frac{6}{52} = \frac{3}{26}$.

**Answer:** (d) 3/26

> Common mistake: Taking total face cards as 12 instead of only red face cards.

### Question 19

*1 mark · Assertion and reason*

Assertion (A) : The probability of selecting a number at random from the numbers 1 to 20 is 1. Reason (R): For any event E, if P(E) = 1, then E is called a sure event.

- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.

**Solution**

1. The total number of favorable outcomes for selecting a number from 1 to 20 is 20, and the total possible outcomes are 20, so the probability is $20/20 = 1$.
2. Therefore, Assertion (A) is true, and Reason (R) correctly defines a sure event where $P(E) = 1$.

**Answer:** Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).

> Common mistake: Confusing the probability of selecting a number from 1 to 20 with selecting a specific subset.

### Question 31

*3 marks · Short answer*

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.

**Solution**

1. When two dice are thrown simultaneously, the total number of possible outcomes is $6 \times 6 = 36$.
2. Let $E$ be the event that the difference of the numbers on the two dice is $2$.
3. The favorable outcomes for this event are $(1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), \text{ and } (6, 4)$.
4. The total number of favorable outcomes is $8$.
5. The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes, which is $\frac{8}{36} = \frac{2}{9}$.

**Answer:** $\frac{2}{9}$

> Common mistake: Counting pairs like (1,3) and (3,1) as a single outcome or missing some valid pairs of differences.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 7

*1 mark · MCQ*

If probability of winning a game is $p$, then probability of losing the game is :

- $1 + p$
- $-p$
- $p - 1$
- $1 - p$

**Solution**

1. Let $E$ be the event of winning the game and $\bar{E}$ be the event of losing the game.
2. We know that $P(E) + P(\bar{E}) = 1$, so $P(\bar{E}) = 1 - P(E) = 1 - p$.

**Answer:** (d) $1 - p$

> Common mistake: Confusing the probability of the complement with $p - 1$ or $1 + p$.

### Question 10

*1 mark · MCQ*

A card is drawn from a well shuffled deck of 52 playing cards. The probability that drawn card is a red queen, is :

- $\frac{1}{13}$
- $\frac{2}{13}$
- $\frac{1}{52}$
- $\frac{1}{26}$

**Solution**

1. Total number of possible outcomes = 52. Number of red queens in a deck is 2 (Queen of hearts and Queen of diamonds).
2. Probability of drawing a red queen = $\frac{2}{52} = \frac{1}{26}$.

**Answer:** (d) $\frac{1}{26}$

> Common mistake: Taking 4 queens instead of 2 red queens.

### Question 18

*1 mark · MCQ*

Two dice are rolled together. The probability of getting a doublet is :

- $\frac{2}{36}$
- $\frac{1}{36}$
- $\frac{1}{6}$
- $\frac{5}{6}$

**Solution**

1. When two dice are rolled, the total number of possible outcomes is $6 \times 6 = 36$.
2. The doublets are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$, which are $6$ outcomes.
3. The probability of getting a doublet is $\frac{6}{36} = \frac{1}{6}$.

**Answer:** (c) $\frac{1}{6}$

> Common mistake: Forgetting to simplify the fraction $\frac{6}{36}$ to $\frac{1}{6}$.

### Question 25

*2 marks · Short answer*

A bag contains 4 red, 5 white and some yellow balls. If probability of drawing a red ball at random is $\frac{1}{5}$, then find the probability of drawing a yellow ball at random.

**Solution**

1. Let the number of yellow balls be $x$.
2. Total number of balls in the bag $= 4 + 5 + x = 9 + x$.
3. Number of red balls $= 4$.
4. Given that the probability of drawing a red ball is $\frac{1}{5}$, so $\frac{4}{9 + x} = \frac{1}{5}$.
5. Cross-multiplying gives $9 + x = 20$, which means $x = 11$.
6. Total number of balls $= 9 + 11 = 20$, and the number of yellow balls is $11$.
7. The probability of drawing a yellow ball at random is $\frac{11}{20}$.

**Answer:** $\frac{11}{20}$

> Common mistake: Forgetting to add the number of red and white balls to form the correct total number of outcomes.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 2

*1 mark · MCQ*

If the probability of a player winning a game is $0.79$, then the probability of his losing the same game is :

- $1.79$
- $0.31$
- $0.21\%$
- $0.21$

**Solution**

1. Let $E$ be the event of winning the game, so $P(E) = 0.79$.
2. The probability of losing the game is $P(\bar{E}) = 1 - P(E) = 1 - 0.79 = 0.21$.

**Answer:** (d) $0.21$

> Common mistake: Subtracting incorrectly or confusing percentage with decimal.

### Question 8

*1 mark · MCQ*

From the data $1, 4, 7, 9, 16, 21, 25$, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :

- $\frac{2}{5}$
- $\frac{1}{5}$
- $\frac{1}{7}$
- $\frac{2}{7}$

**Solution**

1. The given data is $1, 4, 7, 9, 16, 21, 25$. Removing all even numbers ($4, 16$), the remaining numbers are $1, 7, 9, 21, 25$.
2. The total number of remaining outcomes is $5$. Among these, the prime number is only $7$, so the number of favorable outcomes is $1$.
3. The probability is $\frac{1}{5}$.

**Answer:** (b) $\frac{1}{5}$

> Common mistake: Students often consider $1$ as a prime number.

### Question 16

*1 mark · MCQ*

Two dice are rolled together. The probability of getting sum of numbers on the two dice as $2, 3$ or $5$, is :

- $\frac{7}{36}$
- $\frac{11}{36}$
- $\frac{5}{36}$
- $\frac{4}{9}$

**Solution**

1. Total number of possible outcomes when two dice are rolled is $36$.
2. Outcomes with sum $2$: $(1, 1)$ -> $1$ outcome.
3. Outcomes with sum $3$: $(1, 2), (2, 1)$ -> $2$ outcomes.
4. Outcomes with sum $5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ -> $4$ outcomes.
5. Total favourable outcomes = $1 + 2 + 4 = 7$, so the probability is $\frac{7}{36}$.

**Answer:** (a) $\frac{7}{36}$

> Common mistake: Forgetting to include some pairs or counting pairs twice.

### Question 22

*2 marks · Very short answer*

In a pack of 52 playing cards one card is lost. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is queen of heart, if the lost card is a black card.

**Solution**

1. Total number of cards in a standard deck is $52$.
2. Given that one lost card is a black card, the remaining deck still contains all $4$ cards of hearts, including the queen of hearts.
3. The total number of remaining cards is $52 - 1 = 51$.
4. The probability of drawing a queen of heart from the remaining cards is $\frac{1}{51}$.

**Answer:** $$\frac{1}{51}$$

> Common mistake: Taking the total remaining cards as $52$ or assuming the lost card could be the queen of hearts.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 3

*1 mark · MCQ*

A card is drawn at random from a well-shuffled deck of $52$ cards. The probability of getting a red card is :

- $\frac{1}{26}$
- $\frac{1}{13}$
- $\frac{1}{4}$
- $\frac{1}{2}$

**Solution**

1. Total number of cards in a deck is $52$, and the number of red cards is $26$.
2. Probability of getting a red card = $\frac{26}{52} = \frac{1}{2}$.

**Answer:** (d) $\frac{1}{2}$

> Common mistake: Forgetting the total number of red cards in a standard deck.

### Question 11

*1 mark · MCQ*

A die is thrown once. Find the probability of getting a number less than 7.

- $\frac{5}{6}$
- $1$
- $\frac{1}{6}$
- $0$

**Solution**

1. When a fair die is thrown once, the possible outcomes are $1, 2, 3, 4, 5, 6$, all of which are less than 7.
2. The number of favorable outcomes is 6, and the total number of possible outcomes is 6, so the probability is $\frac{6}{6} = 1$.

**Answer:** (b) $1$

> Common mistake: Treating numbers greater than 7 incorrectly or confusing probability with zero.

### Question 38

*4 marks · Case-based*

Blood group describes the type of blood a person has. It is a classification of blood based on the presence or absence of inherited antigenic substances on the surface of red blood cells. Blood types predict whether a serious reaction will occur in a blood transfusion.
In a sample of $50$ people, $21$ had type O blood, $22$ had type A, $5$ had type B and rest had type AB blood group.
Based on the above, answer the following questions :
(i) What is the probability that a person chosen at random had type O blood ?
(ii) What is the probability that a person chosen at random had type AB blood group ?
(iii) What is the probability that a person chosen at random had neither type A nor type B blood group ?

**Part (i)**

1. Total number of people in the sample = $50$.
2. Number of people with type O blood = $21$.
3. Probability of choosing a person with type O blood = $\frac{21}{50}$.

Answer (i): $\frac{21}{50}$

**Part (ii)**

1. Number of people with type A blood = $22$, type O = $21$, type B = $5$.
2. Number of people with type AB blood = Total - (Type O + Type A + Type B) = $50 - (21 + 22 + 5) = 50 - 48 = 2$.
3. Probability of choosing a person with type AB blood = $\frac{2}{50} = \frac{1}{25}$.

Answer (ii): $\frac{1}{25}$

**Part (iii)**

1. People having neither type A nor type B blood group will have either type O or type AB blood group.
2. Number of people with type O or type AB blood = $21 + 2 = 23$. Alternatively, subtract people with type A and B from total: $50 - (22 + 5) = 50 - 27 = 23$. Wait, type O is 21 and AB is 2, so $21 + 2 = 23$. Let us recheck: total is 50, type A is 22, type B is 5. Neither A nor B means O or AB, which is $21 + 2 = 23$. Wait, rest had AB, so rest = $50 - (21+22+5) = 2$. Type O = 21, AB = 2. Total neither A nor B = $21 + 2 = 23$. Let us check $50 - (22 + 5) = 23$.
3. Probability = $\frac{23}{50}$.

Answer (iii): $\frac{23}{50}$

**Answer:** Probabilities are (i) $\frac{21}{50}$, (ii) $\frac{1}{25}$, (iii) $\frac{27}{50}$.

> Common mistake: Miscalculating the number of people with AB blood group or misunderstanding 'neither type A nor type B'.

### Question 38 (OR)

*4 marks · Case-based*

(iii) What is the probability that person chosen at random had either type A or type B or type O blood group ?

**Part (iii)**

1. Total number of people in the sample $n(S) = 50$.
2. Number of people having either type A, type B, or type O blood group $= 22 + 5 + 21 = 48$.
3. Probability $= \frac{48}{50} = \frac{24}{25}$.

Answer (iii): $\frac{24}{25}$

**Answer:** Probability: $48/50$

> Common mistake: Including type AB people when finding the sum for A, B, and O.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 16

*1 mark · MCQ*

Probability of happening of an event is denoted by $p$ and probability of non-happening of the event is denoted by $q$. Relation between $p$ and $q$ is

- $p + q = 1$
- $p = 1, q = 1$
- $p = q - 1$
- $p + q + 1 = 0$

**Solution**

1. Let $E$ be an event, with $p$ being the probability of happening of $E$, so $p = P(E)$.
2. The probability of non-happening of the event is $q = P(\text{not } E) = P(\bar{E})$.
3. We know that $P(E) + P(\text{not } E) = 1$, therefore $p + q = 1$.

**Answer:** (a) $p + q = 1$

> Common mistake: Confusing complementary probabilities with independent events.

### Question 17

*1 mark · MCQ*

A girl calculates that the probability of her winning the first prize in a lottery is $0.08$. If $6000$ tickets are sold, how many tickets has she bought?

- 40
- 240
- 480
- 750

**Solution**

1. Let the number of tickets she bought be $x$.
2. The probability of winning is given by $\frac{\text{Number of tickets bought}}{\text{Total number of tickets sold}} = 0.08$.
3. Therefore, $\frac{x}{6000} = 0.08$.
4. Solving for $x$, we get $x = 0.08 \times 6000 = 480$.

**Answer:** (c) 480

> Common mistake: Making decimal multiplication errors while calculating $0.08 \times 6000$.

### Question 18

*1 mark · MCQ*

In a group of 20 people, 5 can't swim. If one person is selected at random, then the probability that he/she can swim, is

- $\frac{3}{4}$
- $\frac{1}{3}$
- 1
- $\frac{1}{4}$

**Solution**

1. Total number of people in the group is 20, and the number of people who can't swim is 5.
2. Number of people who can swim = $20 - 5 = 15$.
3. Probability that the selected person can swim = $\frac{15}{20} = \frac{3}{4}$.

**Answer:** (a) $\frac{3}{4}$

> Common mistake: Using the number of people who can't swim (5) instead of the number of people who can swim (15) in the numerator.

### Question 25

*2 marks · Short answer*

A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random from the bag. Find the probability that drawn ball is (i) red (ii) yellow.

**Part (i)**

1. Total number of balls = $4 + 3 + 2 = 9$.
2. Number of red balls = $4$.
3. Probability of drawing a red ball = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{9}$.

Answer (i): \frac{4}{9}

**Part (ii)**

1. Total number of balls = $9$.
2. Number of yellow balls = $2$.
3. Probability of drawing a yellow ball = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{2}{9}$.

Answer (ii): \frac{2}{9}

**Answer:** (i) $\frac{4}{9}$, (ii) $\frac{2}{9}$

> Common mistake: Adding the number of balls incorrectly while finding the total number of possible outcomes.

## Related pages

- [Probability: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/probability/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student