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Bernoulli Binomial Distribution
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[QUOTE="RUber, post: 5293736, member: 524408"] If p is the probability of a win, then p^k is the probability of winning k times in a row. If you have n trials and only win k times, then you lose the rest (n-k) of te trials. So the probability of winning the first k and then losing the rest would be ##p^k(1-p)^{n-k}##. Now comes the combination part. As I said, the probability of winning the first k and losing the rest is that piece of the formula. But the probability would be the same if you lost the first n-k and won the last k. In all, you have to add up all the possible ways to win k times out of n. Each way has the same probability, so the total probability of winning k times is the probability of one of the ways times the number of combinations of k wins and n-k losses. [/QUOTE]
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