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Hey guys I was struggling with this question and was looking for some help:

An urn contains 10 red and 10 white balls. The balls are drawn from the urn at random, one at a time. Find the probabilities that the fourth white ball is the fourth, fifth, sixth, or seventh ball drawn if the sampling is done with replacement?

I know that the probability that it's the fourth ball drawn to be $(\frac{1}{2})^4 $

But the book says that on the fifth ball drawn the probability would be $(\frac{1}{8})$ and I'm not exactly sure why. Any help would be appreciated!

Thanks!

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    $\begingroup$ Hint: to get it on the fifth trial you need to have drawn one R and three W (in some order) in the first four. First compute the probability, $p$, of that. It follows that the probability of getting the fourth W on trial five is $\frac p2$. $\endgroup$ – lulu Sep 28 '15 at 19:47
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Hint: $RWWWW, WRWWW, WWRWW, WWWRW$

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  • $\begingroup$ I am an idiot. Sometimes just seeing it makes it click in the good ol' noggin. Thank you so much for your help! $\endgroup$ – WhatsAGuitar Sep 28 '15 at 19:56
  • $\begingroup$ Now when the problem is extended to the fourth white ball occurring on the sixth ball drawn, is there an easier way to go about this problem rather than writing out all possible scenarios and summing the probabilities? $\endgroup$ – WhatsAGuitar Sep 28 '15 at 19:59
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    $\begingroup$ Treat it as a form of NegBin probability $\endgroup$ – Alex Sep 28 '15 at 20:32

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