A bag contains 9 white and 5 yellow balls, and another bag contains 6 white and 8 yellow balls.
Qualcomm technical mcq question, verified with a worked answer. Free to practise - no sign-up.
A bag contains 9 white and 5 yellow balls, and another bag contains 6 white and 8 yellow balls. If one of the bags is selected at random and two balls are drawn at random from the bag, then the probability that both the balls are white is:
Show answer & explanation
Using the law of total probability, we calculate the probability of drawing two white balls from either bag. Bag 1 has probability 1/2 × C(9,2)/C(14,2) = 1/2 × 36/91 = 18/91. Bag 2 has probability 1/2 × C(6,2)/C(14,2) = 1/2 × 15/91 = 15/182. Total probability = 18/91 + 15/182 = 36/182 + 15/182 = 51/182.
Step-by-step Derivation:
Step 1: Identify the two bags.
- Bag 1: 9 white, 5 yellow (total 14 balls)
- Bag 2: 6 white, 8 yellow (total 14 balls)
Step 2: Probability of selecting each bag = 1/2
Step 3: Calculate P(both white | Bag 1)
- Number of ways to select 2 white from 9: C(9,2) = 9×8/2 = 36
- Total ways to select 2 from 14: C(14,2) = 14×13/2 = 91
- P(both white | Bag 1) = 36/91
Step 4: Calculate P(both white | Bag 2)
- Number of ways to select 2 white from 6: C(6,2) = 6×5/2 = 15
- Total ways to select 2 from 14: C(14,2) = 91
- P(both white | Bag 2) = 15/91
Step 5: Apply law of total probability
- P(both white) = P(Bag 1) × P(both white | Bag 1) + P(Bag 2) × P(both white | Bag 2)
- P(both white) = 1/2 × 36/91 + 1/2 × 15/91
- P(both white) = 18/91 + 15/182
- Convert to common denominator (182): 36/182 + 15/182 = 51/182