CA Foundation P3 · Chapter 15

Probability MCQs with Answers

9 multiple-choice questions on Probability for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   Two fair dice are thrown. The probability that the sum of the numbers is 8 is:

   • A) 1/6
   • B) 1/9
   • C) 5/36
   • D) 5/12
   Show answer & explanation

   Answer: C) 5/36

   Favourable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2), i.e. 5 out of 36 equally likely outcomes. Probability = 5/36. Missing (4,4) or double counting gives 4/36 = 1/9 or 6/36 = 1/6.

2. Question 2

   A bag contains 5 red and 7 blue balls. Two balls are drawn at random without replacement. The probability that both are red is:

   • A) 25/144
   • B) 5/33
   • C) 5/12
   • D) 10/33
   Show answer & explanation

   Answer: B) 5/33

   P = ⁵C₂/¹²C₂ = 10/66 = 5/33. Equivalently (5/12) x (4/11) = 20/132 = 5/33. Drawing with replacement would give (5/12)² = 25/144.

3. Question 3

   If P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, then P(A ∪ B) is:

   • A) 0.7
   • B) 0.9
   • C) 0.2
   • D) 0.5
   Show answer & explanation

   Answer: A) 0.7

   By the addition theorem, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.4 − 0.2 = 0.7. Simply adding gives 0.9, which double counts the overlap.

4. Question 4

   The odds in favour of an event are 3 : 5. The probability of the event is:

   • A) 3/5
   • B) 5/8
   • C) 2/5
   • D) 3/8
   Show answer & explanation

   Answer: D) 3/8

   Odds in favour a : b means P = a/(a + b) = 3/(3 + 5) = 3/8. 5/8 is the probability that the event does not occur, and 3/5 confuses odds with probability.

5. Question 5

   If P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8, then P(A | B) is:

   • A) 0.5
   • B) 0.3
   • C) 0.6
   • D) 0.375
   Show answer & explanation

   Answer: C) 0.6

   P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.5 − 0.8 = 0.3. Then P(A | B) = P(A ∩ B)/P(B) = 0.3/0.5 = 0.6. Dividing by P(A ∪ B) gives 0.375.

6. Question 6

   In a game, a player wins ₹50 with probability 0.2, wins ₹20 with probability 0.3, and loses ₹10 with probability 0.5. The expected gain per game is:

   • A) ₹16
   • B) ₹20
   • C) ₹21
   • D) ₹11
   Show answer & explanation

   Answer: D) ₹11

   E(X) = Σ x·p = 50(0.2) + 20(0.3) + (−10)(0.5) = 10 + 6 − 5 = ₹11. Treating the loss as a gain gives 10 + 6 + 5 = ₹21, and ignoring the loss gives ₹16.

7. Question 7

   Three fair coins are tossed together. The probability of getting at least one head is:

   • A) 1/8
   • B) 7/8
   • C) 3/8
   • D) 1/2
   Show answer & explanation

   Answer: B) 7/8

   P(at least one head) = 1 − P(no head) = 1 − (1/2)³ = 1 − 1/8 = 7/8. 1/8 is the probability of no head (all tails), and 3/8 is the probability of exactly one head.

8. Question 8

   Machines A and B produce 60% and 40% of a factory's output. 2% of A's output and 5% of B's output is defective. An item chosen at random is found defective. The probability that it was produced by machine B is:

   • A) 0.625
   • B) 0.400
   • C) 0.375
   • D) 0.032
   Show answer & explanation

   Answer: A) 0.625

   By Bayes' theorem: P(D|A)P(A) = 0.02 x 0.6 = 0.012 and P(D|B)P(B) = 0.05 x 0.4 = 0.020. Total P(D) = 0.032. P(B|D) = 0.020/0.032 = 0.625. 0.375 is P(A|D), and 0.032 is the total probability of a defective item.

9. Question 9

   A card is drawn at random from a well-shuffled pack of 52 cards. The probability that it is a king or a heart is:

   • A) 17/52
   • B) 1/52
   • C) 1/4
   • D) 4/13
   Show answer & explanation

   Answer: D) 4/13

   P(King) = 4/52, P(Heart) = 13/52, P(King of hearts) = 1/52. P(King or Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. Not subtracting the king of hearts gives 17/52.