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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?

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    $\begingroup$ I think you need to think more about combinatorics before tackling this ... $\endgroup$ – user451844 Sep 11 '17 at 0:25
  • $\begingroup$ 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? $\endgroup$ – Stefan4024 Sep 11 '17 at 0:26
  • $\begingroup$ Also your $P(W)$ and $P(B)$ are quite wrong, as well as the idea of how to obtain them $\endgroup$ – Stefan4024 Sep 11 '17 at 0:27
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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:

$$P(W|B) = 1 - \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{55}{56}$$

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  • $\begingroup$ So is the bit about two urns meant to confuse you? $\endgroup$ – Joao Noch Sep 11 '17 at 0:55
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    $\begingroup$ OP, yes, the second urn is irrelevant in the analysis, but I would say it's not "meant to confuse", but to train you to remove what is unnecessary and simplify the problem. But the way you have posted the question also has some ambiguity. After the first draw of the black ball, is there just one draw (and your question is what is the probability of picking white) OR are there 3 draws and the question is about getting at least one white in those 3 draws? $\endgroup$ – Mathemagical Sep 11 '17 at 2:46

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