A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :

Solution

Maths sir · 3 steps

Weight each possible bag by how likely it is to give two blacks, then add up the bags with 5 or 6.

1. Set up hypotheses. Let the bag have k black balls, k = 0, 1, …, 6. With no other information, take all 7 cases as equally likely, each with prior 1/7.
2. Likelihood of two blacks. If the bag has k black balls, then P(both black) = C(k, 2)/C(6, 2) = C(k, 2)/15. The equal priors and the common 15 cancel in Bayes' theorem, so each case is weighted by C(k, 2) alone.

   Black balls k	Weight C(k, 2)
   0 or 1	0 (two blacks impossible)
   2	1
   3	3
   4	6
   5	10
   6	15

   P(k ≥ 5 | two black) = (10 + 15)/(1 + 3 + 6 + 10 + 15) = 25/35

3. So. The probability is 25/35 = 5/7, option (c).

The idea to keep. In Bayes' questions with equal priors, the posterior is just each likelihood divided by the sum of all likelihoods. List every case, weight it, and add the ones you want.

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