Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Solution

Maths sir · 2 steps

All urns are equally likely and hold 12 balls each, so just compare the black counts: 5 out of 18.

1. Set up Bayes. Each urn is picked with probability 1/3. Given the ball is black, find how much of the total chance of black comes from urn A.

   Urn	P(urn)	P(black | urn)	P(urn) × P(black | urn)
   A	1/3	5/12	5/36
   B	1/3	7/12	7/36
   C	1/3	6/12	6/36

   P(A | black) = (5/36)/(5/36+7/36+6/36) = 5/18

2. So. The common factor 1/36 cancels, leaving 5/(5+7+6) = 5/18. That is option (c).

The idea to keep. In Bayes' theorem, when every source is equally likely and has the same total, the answer is just that source's favourable count over all the favourable counts.

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