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A roulette has 38 slots (18 red, 18 black, 2 green). A customer bets on red until she has won 5 times. What is the probability that she made a total of 12 bets?

This is what I've done and I'm getting the wrong answer...

Wrong = ((20/38)^7)((18/38)^5)

Above, in my head, is the probabilities of her missing red 7 times and then hitting red 5 times. However, I'm guessing I'm suppose to put it in a conditional like this...

P(12 total bets | won 5 times) = P(12 total bets ^ won 5 times) / P(won 5 times) = ((20/38)^7)((18/38)^5) / ((18/38)^5)

Any help would be greatly appreciated.

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up vote 2 down vote accepted

In order for the fifth win to come on the $12$th trial, she has to have won exactly $4$ times in the first $11$ trials, and then won on the $12$th.

The probability of $4$ wins in $11$ trials is $\dbinom{11}{4}p^4(1-p)^7$, where $p=\frac{18}{38}$. Multiply by $p$ for the win on the $12$th, and we get that the probability is $$\binom{11}{4}p^5(1-p)^7,$$

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So, I was missing the (11 choose 4), figures! Thanks for taking the time to explain. – CODe Aug 19 '13 at 1:27
You are welcome. You were only counting one of the orders in which fifth win on the $12$th can happen. Definitely we need a W on the $12$th, but the other four W could have been anywhere. – André Nicolas Aug 19 '13 at 1:33

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