# Probability of selecting balls from urns

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?

• I think you need to think more about combinatorics before tackling this ...
– user451844
Sep 11, 2017 at 0:25
• The question is a little bit vague. Do you mean what is the probability of at least one of white ball in the remaining three, or the probability of drawing a white ball in the next draw? Sep 11, 2017 at 0:26
• Also your $P(W)$ and $P(B)$ are quite wrong, as well as the idea of how to obtain them Sep 11, 2017 at 0:27

You can directly calculate $P(W|B)$. To understand it better assume that you skip the second urn part and you directly draw 4 balls from the first urn one by one. As after the first draw we have a black ball, we're left with $5$ white and $3$ balls. The probability of drawing at least one white ball in the three attempts is "one minus" the probability of all of the three draws being black. Hence:
$$P(W|B) = 1 - \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{55}{56}$$