Question 578837
We know that *[tex \LARGE P(head | B) = \frac{3}{4}]. We want to find *[tex \LARGE P(B |head)]. We use Bayes' theorem:


*[tex \LARGE P(B|head) = P(head|B)\frac{P(B)}{P(head)}].


We know that P(B) = 1/2 because coins A and B are equally likely to be picked. P(head) is also 1/2 because


*[tex \LARGE P(head) = P(A)P(head|A) + P(B)P(head|B)]


*[tex \LARGE = \frac{1}{2}(\frac{1}{4}) + \frac{1}{2}(\frac{3}{4}) = \frac{1}{2}]


Therefore,


*[tex \LARGE P(B|head) = \frac{3}{4}]


Additionally, P(A|tail) = 3/4. The answer is *not* 3/4 + 3/4 = 3/2, that's absurd. It is 3/4 because P(head) and P(tail) both have a 1/2 chance of occurring, and we need to multiply by 1/2 to compensate for each cases.