Probability CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Probability CBQs

1. Question 1MCQ1 mark
   At a school fete, a lucky-dip box holds 60 tokens numbered 1 to 60. A visitor wins a prize only if the token drawn at random shows a number that is a multiple of both 4 and 6. The probability that a visitor wins a prize is
   • 1/4
   • 1/12
   • 1/6
   • 1/3
   Show step-wise solution

   1. [1 mark] Multiples of both 4 and 6 are multiples of LCM(4, 6) = 12: 12, 24, 36, 48, 60, i.e. 5 favourable tokens out of 60 equally likely, so P(prize) = 5/60 = 1/12.

   Answer: 1/12

2. Question 2MCQ1 mark
   A carton of 250 LED bulbs is known to contain 15 defective bulbs. Before packing, a shopkeeper takes out one bulb at random from the carton to test it. The probability that the bulb taken out is NOT defective is
   • 3/50
   • 3/47
   • 1/50
   • 47/50
   Show step-wise solution

   1. [1 mark] Non-defective bulbs = 250 - 15 = 235, so P(not defective) = 235/250 = 47/50.

   Answer: 47/50

3. Question 3Assertion-Reason1 mark
   At a funfair, a spinner has 10 equal sectors numbered 1 to 10, and a player wins if the arrow stops on a prime number. Assertion (A): The probability that the player wins is 2/5. Reason (R): Among 1 to 10 there are exactly 4 primes (2, 3, 5, 7), and the 10 outcomes are equally likely.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): 4 favourable out of 10 equally likely outcomes gives P = 4/10 = 2/5; R is true and is exactly the count that gives A.

   Answer: (A)

4. Question 4Short answer2 marks
   In a board game played at a school hobby club, a player whose token is 'locked' may unlock it only by rolling two fair dice together and getting a total of at least 10. Find the probability that a player unlocks the token on a single roll of the two dice.
   Show step-wise solution

   1. [1 mark] Total equally likely outcomes = 6 × 6 = 36. Totals of at least 10: total 10 → (4,6), (5,5), (6,4); total 11 → (5,6), (6,5); total 12 → (6,6); so 6 favourable outcomes.
   2. [1 mark] P(unlocking) = 6/36 = 1/6.

   Answer: 1/6

5. Question 5Case study4 marks
   Meera and her brother are playing a board game in which a player throws two ordinary dice together and moves the token forward by the SUM of the two numbers that come up. Meera's token is on square 88, and there is a snake whose mouth is on square 97. Based on the above information, answer the following questions: (i) Find the probability that Meera throws a doublet (the same number on both dice). [1 mark] (ii) Find the probability that the sum on the two dice is greater than 8. [1 mark] (iii) Find the probability that Meera's token lands on the snake's mouth on this throw. [2 marks]
   Show step-wise solution

   1. [1 mark] (i) There are 6 × 6 = 36 equally likely outcomes; the doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), i.e. 6 outcomes, so P(doublet) = 6/36 = 1/6.
   2. [1 mark] (ii) Sums 9, 10, 11, 12 occur in 4 + 3 + 2 + 1 = 10 outcomes, so P(sum > 8) = 10/36 = 5/18.
   3. [1 mark] (iii) To land on 97 from 88 she needs a sum of 97 − 88 = 9; favourable outcomes are (3,6), (4,5), (5,4), (6,3), i.e. 4 outcomes.
   4. [1 mark] (iii) P(lands on the snake) = 4/36 = 1/9.

   Answer: (i) 1/6 (ii) 5/18 (iii) 1/9

Study Probability

• Probability notes
• Probability Topic Hub
• All Class 10 CBQs