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.
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 $$\binom{11}{4}p^5(1-p)^7,$$