Class 10 Maths Chapter 15 Probability Practice Paper 4

PRACTICE PAPER 04 (2025-26) CHAPTER 14: PROBABILITY SUBJECT: MATHEMATICS STANDARD MAX. MARKS: 40 CLASS: X DURATION: 1½ hrs

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks each.

Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5 marks and

Section E comprises of 2 Case Study Based Questions of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted. SECTION – A Questions 1 to 10 carry 1 mark each.

1. In a bag containing 12 balls, 7 are white. What is the probability of drawing a ball which is not white?
(a) 5/12
(b) 7/12
(c) 1/2
(d) 2/3

2. A coin is tossed 3 times. What is the probability of getting exactly 1 head?
(a) 1/8
(b) 3/8
(c) 1/2
(d) 5/8

3. The probability of an event E is 0.05. What is the probability of 'not E'?
(a) 0.05
(b) 0.5
(c) 0.95
(d) 1.05

4. A number is selected from numbers 1 to 20. The probability that it is a multiple of 3 is:
(a) 1/5
(b) 3/10
(c) 2/5
(d) 1/2

5. If a card is selected at random from 52 cards, what is the probability that it is a heart or a club?
(a) 1/4
(b) 1/2
(c) 3/4
(d) 1

6. Which of the following cannot be the probability of an event?
(a) 0.3
(b) 3/5
(c) 15%
(d) 17/16

7. A letter is selected from the letters of the word 'TRIGONOMETRY'. What is the probability that it is 'T'?
(a) 1/12
(b) 1/6
(c) 1/4
(d) 1/3

8. Two dice are thrown. What is the probability of getting a sum of 6?
(a) 1/9
(b) 5/36
(c) 1/6
(d) 7/36 In the following questions 9 and 10, a statement of assertion
(a) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.

(d) Assertion
(a) is false but reason (R) is true.

9. Assertion
(a) : The probability of getting a prime number when a die is thrown is 1/2. Reason (R): Prime numbers on a die are 2, 3 and 5.

10. Assertion
(a) : If a bag contains 3 red and 2 blue balls, the probability of drawing a red ball is 3/5. Reason (R): Probability = (Favorable outcomes)/(Total outcomes) SECTION – B Questions 11 to 14 carry 2 marks each.

11. A bag contains tickets numbered 11 to 40. A ticket is drawn at random. Find the probability that the number on the ticket is: (i) a multiple of 5 (ii) a prime number greater than 20

12. All red face cards are removed from a pack of 52 cards. A card is drawn at random from the remaining cards. Find the probability of getting: (i) a black card (ii) a non-face card

13. Two different dice are thrown together. Find the probability that: (i) both show the same number (ii) the product of numbers is 12

14. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is randomly selected from the numbers 1, 4, 9 and 16. Find the probability that the product of x and y is less than 16. SECTION – C Questions 15 to 17 carry 3 marks each.

15. From a pack of 52 cards, two cards are drawn at random one after the other without replacement. Find the probability that both cards are kings.

16. A box contains 20 cards numbered 1 to 20. A card is drawn at random from the box. Find the probability that the number on the card is:
(a) a prime number
(b) divisible by 2 or 3
(c) a perfect square

17. In a Family of 2 children, find the probability that:
(a) both are boys
(b) at least one is a girl
(c) both are girls given that at least one is a girl SECTION – D Question 18 carries 5 marks.

18. A survey of 500 families was conducted to find out their monthly savings. The results are shown below: Monthly Savings (₹) Less than 5000 5000-10000 10000-15000 More than 15000 Number of Families 150 200 100 50 If a family is selected at random, find the probability that the family:
(a) saves less than ₹5000 per month
(b) saves at least ₹10000 per month
(c) saves between ₹5000 and ₹15000
(d) does not save more than ₹15000 (e) saves more than ₹5000 SECTION – E (CASE STUDY BASED QUESTIONS) Questions 19 to 20 carry 4 marks each.

19. Cricket Tournament In a school cricket tournament, the performance analysis of bowlers is done based on wickets taken. The data for 50 matches is given below: Wickets Taken 0-2 3-4 5 or more Number of Matches 20 18 12 Based on the above information, answer the following questions: (i) What is the probability that in a randomly selected match, 5 or more wickets were taken? (1) (ii) What is the probability that less than 3 wickets were taken in a match? (1) (iii)
(a) Find the probability that at least 3 wickets were taken. (1) OR
(b) Find the probability that at most 4 wickets were taken. (1) (iv) In which category do maximum matches fall? What is the probability for this category? (1)

20. Color Preference Survey A survey was conducted among 120 students to find their favorite color among Red, Blue, Green, and Yellow. The results showed: 30 students like Red 45 students like Blue 25 students like Green 20 students like Yellow Based on the above information, answer the following questions: (i)
(a) If a student is selected at random, what is the probability that the student likes Blue? (1) OR
(b) What is the probability that a randomly selected student likes Red or Yellow? (1) (ii) What is the probability that a student does not like Green? (1) (iii) If two students are selected at random, what is the probability that both like the same color? (Note:

Assume with replacement) (2) DETAILED ANSWER KEY

SECTION A - ANSWERS

1. Answer:
(a) 5/12 Total balls = 12, White balls = 7 Non-white balls = 12 - 7 = 5 P(not white) = 5/12

2. Answer:
(b) 3/8 Total outcomes = 2³ = 8 Exactly 1 head: HTT, THT, TTH = 3 outcomes P = 3/8

3. Answer:
(c) 0.95 P(not E) = 1 - P(E) = 1 - 0.05 = 0.95

4. Answer:
(b) 3/10 Multiples of 3 from 1-20: 3, 6, 9, 12, 15, 18 = 6 numbers P = 6/20 = 3/10

5. Answer:
(b) 1/2 Hearts = 13, Clubs = 13 Total = 26 cards P = 26/52 = 1/2

6. Answer:
(d) 17/16 Probability must be between 0 and 1 17/16 > 1, so it cannot be a probability

7. Answer:
(b) 1/6 TRIGONOMETRY has 12 letters Letter T appears 2 times P(T) = 2/12 = 1/6

8. Answer:
(b) 5/36 Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) = 5 outcomes P = 5/36

9. Answer:
(a) Both true and R explains A Prime numbers on die: 2, 3, 5 = 3 numbers P = 3/6 = 1/2 Both A and R are true, and R explains A

10. Answer:
(a) Both true and R explains A Total balls = 5, Red = 3 P(red) = 3/5 R correctly explains how to find probability

SECTION B - ANSWERS

11. Solution: Tickets: 11 to 40 = 30 tickets (i) Multiple of 5: 15, 20, 25, 30, 35, 40 = 6 tickets P = 6/30 = 1/5 (ii) Prime > 20: 23, 29, 31, 37 = 4 primes P = 4/30 = 2/15

12. Solution: Red face cards removed = 6 (3 hearts + 3 diamonds) Remaining = 52 - 6 = 46 cards (i) P(black): Black cards = 26 (all remain) P = 26/46 = 13/23 (ii) P(non-face): Total face cards originally = 12 Face cards removed = 6 Face cards remaining = 6 Non-face cards = 46 - 6 = 40 P = 40/46 = 20/23

13. Solution: Total outcomes = 36 (i) Same number (doublet): (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 P = 6/36 = 1/6 (ii) Product = 12: (2,6), (3,4), (4,3), (6,2) = 4 outcomes P = 4/36 = 1/9

14. Solution: x ∈ {1,2,3,4}, y ∈ {1,4,9,16} Total outcomes = 4 × 4 = 16 Product < 16: x=1: 1×1=1, 1×4=4, 1×9=9 (3 cases) x=2: 2×1=2, 2×4=8 (2 cases) x=3: 3×1=3, 3×4=12 (2 cases) x=4: 4×1=4 (1 case) Total favorable = 8 P = 8/16 = 1/2

SECTION C - ANSWERS

15. Solution: Total cards = 52, Kings = 4 P(first king) = 4/52 After one king, remaining cards = 51, remaining kings = 3 P(second king | first king) = 3/51 P(both kings) = (4/52) × (3/51) = 12/2652 = 1/221

16. Solution: Cards 1 to 20 = 20 cards
(a) Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 = 8 primes P = 8/20 = 2/5
(b) Divisible by 2 or 3: By 2: 2,4,6,8,10,12,14,16,18,20 = 10 By 3: 3,6,9,12,15,18 = 6 By both: 6,12,18 = 3 Total = 10 + 6 - 3 = 13 P = 13/20
(c) Perfect square: 1, 4, 9, 16 = 4 numbers P = 4/20 = 1/5

17. Solution: Sample space: {BB, BG, GB, GG} Total outcomes = 4
(a) Both boys: BB = 1 outcome P = 1/4
(b) At least one girl: BG, GB, GG = 3 outcomes P = 3/4
(c) Both girls given at least one girl: Given at least one girl: {BG, GB, GG} Both girls: {GG} = 1 outcome P(both girls | at least one girl) = 1/3

SECTION D - ANSWERS

18. Solution: Total families = 500
(a) P(< ₹5000): P = 150/500 = 3/10
(b) P(≥ ₹10000): Families = 100 + 50 = 150 P = 150/500 = 3/10
(c) P(between ₹5000-₹15000): Families = 200 + 100 = 300 P = 300/500 = 3/5
(d) P(not > ₹15000): Families = 500 - 50 = 450 P = 450/500 = 9/10 (e) P(> ₹5000): Families = 200 + 100 + 50 = 350 P = 350/500 = 7/10

SECTION E - ANSWERS

19. Cricket Tournament - Solutions: Total matches = 50 (i) P(5 or more wickets): P = 12/50 = 6/25 (ii) P(< 3 wickets): Matches with 0-2 wickets = 20 P = 20/50 = 2/5 (iii)
(a) P(at least 3 wickets): Matches = 18 + 12 = 30 P = 30/50 = 3/5 OR (iii)
(b) P(at most 4 wickets): Matches = 20 + 18 = 38 P = 38/50 = 19/25 (iv) Maximum category: 0-2 wickets (20 matches) P = 20/50 = 2/5

20. Color Preference - Solutions: Total students = 120 Red=30, Blue=45, Green=25, Yellow=20 (i)
(a) P(Blue): P = 45/120 = 3/8 OR (i)
(b) P(Red or Yellow): P = (30+20)/120 = 50/120 = 5/12 (ii) P(not Green): P = (120-25)/120 = 95/120 = 19/24 (iii) P(both like same color): P(both Red) = (30/120)² = 1/16 P(both Blue) = (45/120)² = 9/64 P(both Green) = (25/120)² = 25/576 P(both Yellow) = (20/120)² = 1/36 Total P = 1/16 + 9/64 + 25/576 + 1/36 = 36/576 + 81/576 + 25/576 + 16/576 = 158/576 = 79/288