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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.
Solution
Bag I contains 3 red and 4 black balls.
Bag II contains 4 red and 5 black balls.
Let E1 and E2 be the events of drawing a red ball and a black ball from bag I, then
∴ P(E1) = `3/7`, P(E2) = `4/7`
Event A: Drawing a red ball
A red ball is taken out from bag I and placed in bag II. Thus bag II has 5 red and 5 black balls.
∴ `P(A/E_1) = 5/10`
A black ball is taken out from bag I and placed in bag II. Thus bag II contains 4 red and 6 black balls.
∴ `P(A/E_2) = 4/10`
By Bayes' theorem,
`P(E_2/E) = (P(E_2) xx P(E/E_2))/(P(E_1) xx P(E/E_1) + P(E_2) xx P(E/E_2))`
= `(4/7 xx 4/10)/(3/7 xx 5/10 + 4/7 xx 4/10)`
= `16/(15 + 16)`
= `16/31`
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