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An urn $U_1$ has $3$ white balls and $5$ black balls. $4$ balls were drawn from this urn and put into another empty urn $U_2$. Now $2$ balls were drawn from the urn $U_2$ and were found to be white. If a third ball is drawn from urn $U_2$ then what is the probability that it is white?

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  • $\begingroup$ Welcome to MSE. Please include your thoughts and efforts (work in progress) in this and future posts. You are more likely to receive positive/constructive feedback that way. Formatting your post helps too. Formatting tips here. $\endgroup$ – Em. Mar 22 '16 at 20:45
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Urn 2 is irrelevant. We can imagine we are looking at the balls as they come out of Urn 1. If the first two are white, there is one white left, so the probability the third is white is $1/6$.

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With care, you can also solve this without that insight:

After moving $4$ balls to urn $2$, the probability distribution for the number of white balls is $\{ 0 : \frac{5}{70}, 1: \frac{30}{70}, 2: \frac{30}{70}, 3: \frac{5}{70}\}$. (For example, we can choose 2 white balls in $\binom32 \binom52 = 30$ ways.)

So the chances of pulling $2$ white balls out of urn $2$ are $$ \frac{5}{70}\frac34\frac23 + \frac{30}{70}\frac12\frac13= \frac5{140}+\frac{10}{140} = \frac{30}{280} $$ And the chances of pulling $2$ white balls out of urn $2$ and then pulling a third white ball out are $$ \frac{5}{70}\frac34\frac23\frac12 = \frac5{280} $$ So the probability of that second event given that the first event happened is $$\frac{5}{30} = \frac16 $$

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  • $\begingroup$ Nicely explained. I was hoping someone would give the conditional probability argument. $\endgroup$ – André Nicolas Mar 22 '16 at 22:43

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