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:
Total number of balls in bag1 = 7
Total number of balls in bag2 = 9
Let E1 denote the event that a red ball is transferred from bag1 to bag2
and Let E2 denote the event that a black ball is transferred from bag1 to bag2
So P(E1 ) = 3/7
P(E2 ) = 4/7
Let A denote the event that the ball drawn is red.
Now when a red ball is transferred from bag1 to bag2 P(A/E1 ) = 5/10 = 1/2
when a black ball is transferred from bag1 to bag2 P(A/E2 ) = 4/10 = 2/5
Probability that the transferred ball is black is denoted as P(E2 /A)
=> P(E2 /A) = {P(E2 )*P(A/E2 )}/{P(E1 )*P(A/E1 ) + P(E2 )*P(A/E2 )}
= {4/7 * 2/5}/{3/7 * 1/2 + 4/7 * 2/5}
= (8/5)/(3/2 + 8/5)
= (8/5)/{(15+16)/10}
= (8/5)/(31/10)
= (8*10)/5*31)
= (8*2)/31
= 16/31
So probability that the transferred ball is black is 16/31.
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