A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is :

Solution

Maths sir · 3 steps

Each possible bag is a hypothesis. Weigh it by how easily it gives 2 white and 2 black, then compare.

1. Set up hypotheses. Let the bag hold k white balls, with k = 0, 1, …, 8. With no other information, all 9 compositions are equally likely. Bags with k = 0, 1, 7, 8 cannot give 2 white and 2 black, so only k = 2 to 6 matter.

   P(E | k) = C(k, 2)C(8-k, 2)/C(8, 4)

   White balls k	C(k, 2)C(8-k, 2)
   2	1 × 15 = 15
   3	3 × 10 = 30
   4	6 × 6 = 36
   5	10 × 3 = 30
   6	15 × 1 = 15

2. Apply Bayes. The equal priors 1/9 and the common denominator C(8, 4) = 70 cancel. So the required probability is the k = 4 count divided by the total count: 36/(15 + 30 + 36 + 30 + 15) = 36/126.
3. So. 36/126 = 2/7, which is option (b).

The idea to keep. In Bayes problems, each hypothesis counts in proportion to prior times likelihood. When the priors are equal they cancel, and you just compare the likelihoods.

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