Question

Box A contains 4 red and 5 yellow balls while box B contains 6 red and 5 yellow balls. A box is chosen at random and a ball is selected from it. The probability of choosing box A is 58%. Find the the probability that the ball came from box B if the ball was red.

Answer 1

P(A|B) = P(A intersection B) / P(B)

P(B) = "being red" = (4/9)*0.58 + (6/11)*0.42

P(A intersection B) = ?
now you do the rest of the work ;)

Answer 2

Perhaps a more straightforward way to approach this problem is with a tree.



So, the total probability of drawing a red ball is the sum of the probabilities for each path leading to a red ball, as I circumscribed in the above diagram. In this case, there are only two such paths: start->A->red and start->B->red. Therefore, the sum of these probabilities is: [probablity of choosing box A * probability of choosing a red ball from box A] + [probability of choosing box B * probability of choosing a red ball from box B].

Now, we're given that a red ball was chosen, so what's the probability that this red ball came from box B? Well, it's just the total the probability of going down the path, start->B->red, divided by the total probability of choosing a red ball.

You'll get the same result if you use Bayes' Theorem, which in this case is: P(B|r) = P(r|B) / P(r) = P(r|B) / (P(r|B)*P(B) + P(r|A)*P(A))