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 drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
$\frac{16}{31}$
The transferred ball is red with probability \(3/7\) (Bag I has 3 red out of 7) and black with probability \(4/7\).
If the transferred ball is red, Bag II has 5 red and 5 black balls (10 total), so \(P(\text{draw red}\mid\text{transferred red}) = 5/10=1/2\).
If the transferred ball is black, Bag II has 4 red and 6 black balls (10 total), so \(P(\text{draw red}\mid\text{transferred black}) = 4/10=2/5\).
By the law of total probability, \(P(\text{draw red}) = \left(\frac{3}{7}\right)\left(\frac{1}{2}\right)+\left(\frac{4}{7}\right)\left(\frac{2}{5}\right) = \frac{3}{14}+\frac{8}{35} = \frac{15}{70}+\frac{16}{70}=\frac{31}{70}\).
By Bayes' theorem, \(P(\text{transferred black}\mid\text{draw red}) = \dfrac{(4/7)(2/5)}{31/70} = \dfrac{8/35}{31/70} = \dfrac{16}{31}\).
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