Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

Solution

Maths sir · 3 steps

Split by what was transferred, weight each case, then ask what share of all black draws came from red.

1. Set up cases. Bag I has 10 balls, so the transferred ball is red, black or white with probabilities 3/10, 4/10, 3/10. After the transfer, Bag II always has 9 + 1 = 10 balls.

   Transferred ball	Prior	Black in Bag II now	Prior × P(black)
   Red	3/10	5 of 10	15/100
   Black	4/10	6 of 10	24/100
   White	3/10	5 of 10	15/100

2. Total probability. The chance of drawing black at all is the sum of the last column: (15 + 24 + 15)/100 = 54/100.

   P(Red | Black) = (15/100)/(54/100) = 15/54 = 5/18

3. So. The probability that the transferred ball was red is 5/18, option (b).

The idea to keep. In Bayes questions, list every way the observed event can happen, multiply each prior by its likelihood, and divide the case you want by the total.

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