There are three bags X, Y and Z. Bag X contains 5 one-rupee coins and 4 five-rupee coins; Bag Y contains 4 one-rupee coins and 5 five-rupee coins and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag Y, is :

Solution

Maths sir · 3 steps

Bags are equally likely, so the 1/3 cancels: just compare one-rupee counts, 4 out of 5 + 4 + 3.

1. Set up Bayes. Each bag is chosen with probability 1/3. Each bag holds 9 coins, so P(one | X) = 5/9, P(one | Y) = 4/9 and P(one | Z) = 3/9.

   Bag	P(bag)	P(one | bag)	Product
   X	1/3	5/9	5/27
   Y	1/3	4/9	4/27
   Z	1/3	3/9	3/27

   P(Y | one) = (4/27)/(5/27+4/27+3/27) = 4/12

2. Why it matches. The answer equals the prior 1/3 because bag Y has 4 one-rupee coins, exactly the average of 5, 4 and 3. Drawing a one-rupee coin tells you nothing new about bag Y.
3. So. P(Y | one) = 4/12 = 1/3, which is option (b).

The idea to keep. When every bag is equally likely and holds the same number of coins, the posterior is just that bag's count of the drawn type divided by the total count of that type across all bags.

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