Bag B₁ contains 6 white and 4 blue balls, Bag B₂ contains 4 white and 6 blue balls, and Bag B₃ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability that the ball is drawn from Bag B₂ is:

Solution

Maths sir · 3 steps

Find every way a white ball can appear, then ask what share of that total came through B₂.

1. Paths to white. Each bag is chosen with probability 1/3. For each bag, multiply that by the chance of drawing white from it.

   Bag	P(bag)	P(W | bag)	P(bag) × P(W | bag)
   B₁	1/3	6/10	6/30
   B₂	1/3	4/10	4/30
   B₃	1/3	5/10	5/30

2. Total white. Add the three paths: P(W) = (6 + 4 + 5)/30 = 15/30 = 1/2.

   P(B₂ | W) = (P(B₂) P(W | B₂))/P(W) = (4/30)/(15/30) = 4/15

3. So. Given that the ball is white, the chance it came from B₂ is 4/15, which is option (b).

The idea to keep. In Bayes' problems, the denominator is the total probability of what you observed. The answer is the share of that total that came through the cause you are asked about.

Answer from the NTA final answer key. Solution by PrepWiser.