Class 10 Maths Chapter 15 Probability Practice Paper 1

PRACTICE PAPER 01 (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. A bag contains 5 red balls and 3 green balls. One ball is drawn at random. What is the probability that the ball drawn is NOT green?
(a) 3/8
(b) 5/8
(c) 1/2
(d) 2/3

2. Two coins are tossed simultaneously. What is the probability of getting at least one head?
(a) 1/4
(b) 1/2
(c) 3/4
(d) 1

3. A number is selected at random from numbers 1 to 30. What is the probability that the selected number is a prime number?
(a) 1/3
(b) 1/2
(c) 2/5
(d) 11/30

4. In a single throw of a die, what is the probability of getting a number greater than 4?
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3

5. A box contains cards numbered 11 to 60. A card is drawn at random. Find the probability that the number on the card is a perfect square.
(a) 1/5
(b) 1/10
(c) 3/25
(d) 2/25

6. An event is very likely to happen. Its probability can be:
(a) 0.09
(b) 0.9
(c) 0.009
(d) 0.0009

7. A letter is chosen at random from the word 'PROBABILITY'. What is the probability that it is a vowel?
(a) 3/11
(b) 4/11
(c) 5/11
(d) 2/11

8. From a well-shuffled pack of 52 cards, one card is drawn at random. What is the probability of getting a red king?
(a) 1/26
(b) 1/13
(c) 1/52
(d) 1/2 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 sum of probabilities of all elementary events of an experiment is 1. Reason (R): The probability of a sure event is 1.

10. Assertion
(a) : When a die is thrown, the probability of getting a number less than 7 is 1. Reason (R): The probability of an impossible event is 0. SECTION – B Questions 11 to 14 carry 2 marks each.

11. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: (i) white (ii) not black

12. A card is drawn at random from a well-shuffled pack of 52 playing cards. Find the probability of getting: (i) a red card (ii) a queen

13. Find the probability of getting 53 Mondays in a leap year.

14. Two dice are thrown simultaneously. Find the probability of getting: (i) the sum as 8 (ii) a doublet SECTION – C Questions 15 to 17 carry 3 marks each.

15. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.

16. Cards numbered from 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has: (i) an even number (ii) a square number (iii) a number which is a multiple of 10

17. Two different dice are thrown together. Find the probability of:
(a) getting a sum of 10
(b) getting a sum divisible by 3
(c) getting a sum less than or equal to 3 SECTION – D Question 18 carries 5 marks.

18. Red queens, black jacks and black kings are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards. Find the probability that the drawn card is:
(a) a king
(b) of red colour
(c) a face card
(d) a queen (e) of black colour SECTION – E (CASE STUDY BASED QUESTIONS) Questions 19 to 20 carry 4 marks each.

19. Blood Donation Camp A school organized a blood donation camp. Students from different classes participated enthusiastically. The data showing the blood groups of 80 students who donated blood is given below: Blood Group A B AB O Number of Students 24 28 8 20 Based on the above information, answer the following questions: (i) If one student is selected at random, find the probability that the student has blood group B. (1) (ii) If one student is selected at random, find the probability that the student has blood group AB or O. (1) (iii)
(a) Find the probability that the student selected has blood group A. (1) OR
(b) Find the probability that the student selected does not have blood group AB. (1) (iv) Which blood group has the maximum number of students? What is the probability of selecting a student with this blood group? (1)

20. Card Game at a Fair At a school fair, there is a game stall where a box contains 20 cards numbered 1 to 20. A player has to draw one card from the box. If the card shows an even number, the player wins a prize. If the card shows a prime number, the player gets another chance. If the card shows a number divisible by 5, the player gets a consolation prize. Based on the above information, answer the following questions: (i)
(a) What is the probability that a player wins a prize? (1) OR
(b) What is the probability that a player gets another chance? (1) (ii) What is the probability that a player gets a consolation prize? (1) (iii) What is the probability that a player neither wins a prize nor gets another chance? (2) DETAILED ANSWER KEY

SECTION A - ANSWERS

1. Answer:
(b) 5/8

Solution:

Total balls = 5 red + 3 green = 8 balls Number of balls that are NOT green = 5 red balls P(not green) = Number of red balls / Total balls = 5/8

2. Answer:
(c) 3/4

Solution:

When two coins are tossed, possible outcomes = {HH, HT, TH, TT} Total outcomes = 4 Favorable outcomes (at least one head) = {HH, HT, TH} = 3 outcomes P(at least one head) = 3/4

3. Answer:
(a) 1/3

Solution:

Numbers from 1 to 30 = 30 numbers Prime numbers from 1 to 30 = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} = 10 prime numbers P(prime number) = 10/30 = 1/3

4. Answer:
(b) 1/3

Solution:

When a die is thrown, possible outcomes = {1, 2, 3, 4, 5, 6} Total outcomes = 6 Numbers greater than 4 = {5, 6} = 2 outcomes P(number > 4) = 2/6 = 1/3

5. Answer:
(c) 3/25

Solution:

Cards numbered 11 to 60 = 50 cards Perfect squares from 11 to 60 = {16, 25, 36, 49} = 4 numbers Note: 4 = 2², 9 = 3², 16 = 4², 25 = 5², 36 = 6², 49 = 7², 64 = 8² So perfect squares in range are: 16, 25, 36, 49 However, we need to count cards: 16, 25, 36, 49 = 4 cards Wait, let me recount: From 11 to 60, we have 50 cards Perfect squares: 16, 25, 36, 49 = 4 cards P(perfect square) = 4/50 = 2/25 Correction: Answer should be
(d) 2/25

6. Answer:
(b) 0.9

Solution:

If an event is very likely to happen, its probability must be close to 1. Among the options, 0.9 is closest to 1. Therefore, the probability is 0.9

7. Answer:
(b) 4/11

Solution:

Word: PROBABILITY Total letters = 11 Vowels in PROBABILITY = {O, A, I, I} = 4 vowels P(vowel) = 4/11

8. Answer:
(a) 1/26

Solution:

Total cards = 52 Red kings = King of Hearts + King of Diamonds = 2 red kings P(red king) = 2/52 = 1/26

9. Answer:
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .

Solution:

Assertion
(a) is TRUE: The sum of probabilities of all elementary events equals 1. Reason (R) is TRUE: The probability of a sure event is 1. However, R does not explain A. They are both true statements but independent of each other.

10. Answer:
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .

Solution:

Assertion
(a) is TRUE: When a die is thrown, outcomes are {1,2,3,4,5,6}, all less than 7, so P = 6/6 = 1 Reason (R) is TRUE: The probability of an impossible event is 0. Wait, let me reconsider: R talks about impossible events (P=0), while A is about a sure event (P=1). They are opposite concepts, so R does NOT explain A. Correction: Answer should be
(c) Assertion
(a) is true but reason (R) is false. Actually, R is true (impossible events have P=0), but it doesn't explain A. Final Answer:
(b) Both true but R is not the correct explanation of A.

SECTION B - ANSWERS

11. Solution: Total balls = 4 red + 5 black + 6 white = 15 balls (i) P(white ball): Number of white balls = 6 P(white) = 6/15 = 2/5 (ii) P(not black): Number of balls that are not black = 4 red + 6 white = 10 balls P(not black) = 10/15 = 2/3

12. Solution: Total cards = 52 (i) P(red card): Red cards = Hearts + Diamonds = 26 cards P(red card) = 26/52 = 1/2 (ii) P(queen): Number of queens = 4 (one from each suit) P(queen) = 4/52 = 1/13

13. Solution: A leap year has 366 days = 52 weeks + 2 extra days 52 complete weeks will definitely have 52 Mondays For 53 Mondays, one of the 2 extra days must be Monday Possible combinations of 2 extra days: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun) Total combinations = 7 Favorable cases (Monday appears) = (Sun, Mon) and (Mon, Tue) = 2 cases P(53 Mondays) = 2/7

14. Solution: When two dice are thrown, total outcomes = 6 × 6 = 36 (i) P(sum = 8): Favorable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes P(sum = 8) = 5/36 (ii) P(doublet): Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes P(doublet) = 6/36 = 1/6

SECTION C - ANSWERS

15. Solution: Let the number of blue balls = x Total balls = 5 red + x blue = (5 + x) balls P(blue ball) = x/(5 + x) P(red ball) = 5/(5 + x) Given: P(blue) = 3 × P(red) x/(5 + x) = 3 × 5/(5 + x) x/(5 + x) = 15/(5 + x) Wait, this gives x = 15, let me recalculate: P(blue) = 3 × P(red) x/(5+x) = 3 × [5/(5+x)] x = 15 Therefore, number of blue balls = 15

16. Solution: Cards numbered 2 to 101 means 100 cards (2, 3, 4, ..., 101) (i) P(even number): Even numbers: 2, 4, 6, ..., 100 = 50 even numbers P(even) = 50/100 = 1/2 (ii) P(square number): Square numbers from 2 to 101: 4, 9, 16, 25, 36, 49, 64, 81, 100 = 9 square numbers P(square) = 9/100 (iii) P(multiple of 10): Multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 = 10 numbers P(multiple of 10) = 10/100 = 1/10

17. Solution: Total outcomes when two dice are thrown = 36
(a) P(sum = 10): Favorable outcomes: (4,6), (5,5), (6,4) = 3 outcomes P(sum = 10) = 3/36 = 1/12
(b) P(sum divisible by 3): Sums divisible by 3: 3, 6, 9, 12 Sum = 3: (1,2), (2,1) = 2 Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) = 5 Sum = 9: (3,6), (4,5), (5,4), (6,3) = 4 Sum = 12: (6,6) = 1 Total favorable = 2 + 5 + 4 + 1 = 12 P(sum divisible by 3) = 12/36 = 1/3
(c) P(sum ≤ 3): Sum = 2: (1,1) = 1 Sum = 3: (1,2), (2,1) = 2 Total favorable = 3 P(sum ≤ 3) = 3/36 = 1/12

SECTION D - ANSWERS

18. Solution: Cards removed: • Red queens = 2 (Queen of Hearts, Queen of Diamonds) • Black jacks = 2 (Jack of Spades, Jack of Clubs) • Black kings = 2 (King of Spades, King of Clubs) Total cards removed = 6 Remaining cards = 52 - 6 = 46 cards
(a) P(king): Kings removed = 2 black kings Remaining kings = 4 - 2 = 2 (both red kings) P(king) = 2/46 = 1/23
(b) P(red card): Red cards removed = 2 (red queens only) Red cards remaining = 26 - 2 = 24 P(red card) = 24/46 = 12/23
(c) P(face card): Face cards removed = 2 red queens + 2 black jacks + 2 black kings = 6 Total face cards = 12 Remaining face cards = 12 - 6 = 6 P(face card) = 6/46 = 3/23
(d) P(queen):

Queens removed = 2 (red queens) Remaining queens = 4 - 2 = 2 (both black queens) P(queen) = 2/46 = 1/23 (e) P(black card): Black cards removed = 2 jacks + 2 kings = 4 Black cards remaining = 26 - 4 = 22 P(black card) = 22/46 = 11/23

SECTION E - ANSWERS

19. Blood Donation Camp - Solutions: Total students = 80 Blood groups: A = 24, B = 28, AB = 8, O = 20 (i) P(blood group B): Number of students with blood group B = 28 P
(B) = 28/80 = 7/20 (ii) P(AB or O): Students with AB or O = 8 + 20 = 28 P(AB or O) = 28/80 = 7/20 (iii)
(a) P(blood group A): Number of students with blood group A = 24 P
(a) = 24/80 = 3/10 OR (iii)
(b) P(not AB): Students without AB = 24 + 28 + 20 = 72 P(not AB) = 72/80 = 9/10 (iv) Blood group with maximum students: Blood group B has maximum students = 28 P
(B) = 28/80 = 7/20

20. Card Game at Fair - Solutions: Cards numbered 1 to 20 = 20 cards Even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 = 10 cards Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 = 8 cards Numbers divisible by 5: 5, 10, 15, 20 = 4 cards (i)
(a) P(wins a prize - even number): P(even) = 10/20 = 1/2 OR (i)
(b) P(gets another chance - prime number): P(prime) = 8/20 = 2/5 (ii) P(consolation prize - divisible by 5): P(divisible by 5) = 4/20 = 1/5 (iii) P(neither wins nor gets another chance): This means: NOT even AND NOT prime Numbers that are neither even nor prime: 1, 9, 15 Wait, let me list all numbers from 1-20:

1: odd, not prime 9: odd, not prime (9 = 3²) 15: odd, not prime (15 = 3 × 5) 21 is not in range So favorable outcomes = 3 P(neither) = 3/20