Below I have inserted the question following a chapter on Bayes Theorem and Partition Theorem, and also my answer to see if anyone else can concur that my answer is correct. I will include also the method carried out
Exercise 1.52 Here are two routine problems about balls in urns. You are presented with two urns. Urn I contains 3 white and 4 black balls, and Urn II contains 2 White and 6 Black balls.
a) You pick a ball randomly from Urn I and place it in Urn II. Next you pick a ball randomly from Urn II. What is the Probability that the ball is black ?
My Answer
$P(1W)=\frac{3}{7}$ corresponds to the probability that a ball in Urn I picked at random is White.
$P(1B)=\frac{4}{7}$ Corresponds to the probability that a ball in Urn I picked at random is Black
$P(2W)$ Corresponds to the Probability of the 2nd ball drawn from Urn II being Black which is what we are trying to solve.
$P(2W) = P(2B|1W)*P(1W) + P(2B|1B)*P(1B)$
$=(\frac{6}{9})*(\frac{3}{7})+(\frac{7}{9})*(\frac{4}{7}) = 0.73$
b) This time you pick an urn at random, each of the two urns being picked with probability $\frac{1}{2}$, and you pick a ball at random from the chosen Urn. Given the ball is black, what is the probability you picked Urn I
Answer By a similar method to part a
$P(1U|B) = (\frac{P(B|1U)*P(1U)}{P(B|1U)*P(1U) + P(B|2U)*P(2U)})$
$ = \frac{(\frac{4}{7})*(\frac{1}{2})}{(\frac{4}{7})*(\frac{1}{2})+(\frac{6}{8})*(\frac{1}{2})} = 0.432$