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A bag contains 5 red balls and 6 green balls. If two balls are taken out successively, without replacement, what is the probability that green is picked on the first pick given that at least one red is chosen?

I know the answer is 3/8, however I have no idea how to get there.

I've tried taking it as 6/11 first pick red, 5/10 second green, wrong, along with several other variations in that vein, but it generally boils down to the fact that I don't understand what to do with "at least one green is chosen." Am I supposed to take that as 5/10 chance knowing the green is second? Treating the second event as 100% chance of being green doesn't work either?

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  • $\begingroup$ You need to first work with the definition of $P[A|B]$ for the events $A, B$ relevant to your problem. $\endgroup$ – Michael Apr 23 '18 at 4:53
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If at least one red is chosen, it is either $RR, RG, GR$ (that is we just have to exclude $GG$) and you are interested in $GR$.

Hence the probability is

\begin{align}\frac{P(GR)}{1-P(GG)}&=\frac{\frac{6}{11}\cdot \frac{5}{10}}{1-\frac{6}{11}\cdot \frac{5}{10}} \\ &=\frac{\frac{3}{11}}{\frac{8}{11}}\\ &= \frac{3}{8}\end{align}

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$A$ be the event that green is chosen on the first try. $B$ be the event that atleast one red is chosen $$P(A|B)=\frac{P(B|A).P(A)}{P(B|A).P(A)+P(B|\bar{A})P(\bar{A})} \\ =\frac{(5/10)(6/11)}{(5/10)(6/11)+1.(5/11)}=\frac38$$

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