Question

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black 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 red in colour. Find the probability that the transferred ball is black.

Answer 1

Let E 1 be the event that a red ball is transferred from bag I to bag II and E 2 be the event that a black ball is transferred from bag I to bag II.

Total number of balls in bag I is 7 and total number of balls in bag II is 9.

Probability of occurrence of event E 1 is,

P( E 1 )= Number of red balls in bag I Total number of balls in bag I = 3 7 Probability of occurrence of event E 2 is,

P( E 2 )= Number of black balls in bag I Total number of balls in bag I = 4 7 Let A be the event that a red ball is drawn.

P( A|E 1 ) denotes the event of drawing a red ball from bag II, when a red ball is already transferred to bag II.

P( A|E 1 )= Number of red balls in bag II Total number of balls in bag II = 5 10 = 1 2

Similarly, if a black ball is transferred from bag I to bag II, then probability of drawing of red ball is,

P( A|E 2 )= Number of red balls in bag II Total number of balls in bag II = 4 10 = 2 5 Now, the probability that transferred ball is black is,

P( E 2 |A )= P( E 2 )P( A| E 2 ) P( E 1 )P( A| E 1 )+P( E 2 )P( A| E 2 ) = 4 7 × 2 5 3 7 × 1 2 + 4 7 × 2 5 = 16 31

Thus, the probability that the ball transferred from bag I to bag II was black is 16 31 .