An urn contains five white and four black balls. Four balls are randomly selected and transferred to a second urn. A ball is then drawn from the second urn, and it happens to be black. Find the probability of drawing a white ball from the remaining three in the second urn
I am attempting this using conditional probability. To calculate $P(W | B)$, \begin{align*} P(W | B) = \frac{P(B|W)P(W)}{P(B)}. \end{align*} $P(W)$ is the probability of there being a white ball in the second urn, which is 5/9. $P(B)$ is the probability of a black ball being in the second urn, which is 4/9. $P(B | W)$ is the probability of there being a black ball in the urn assuming there's already a white one, which I think is 1/2 because out of the remaining 8 balls in urn 1 exactly 1/2 are black.
Putting this together we get $P(W|B) = (1/2)(5/9)/(4/9) = 5/8$.
How does this look?