# Thread: Independent Bernoulli trials - help with question!

1. ## Independent Bernoulli trials - help with question!

Suppose that independent Bernoulli trials with parameter p are performed successively. Let N be the number of trials needed to get x successes, and X be the number of successes in the first n trials. Show that P(N=n) = (x/n) P(X=x)

I have no idea how to show it...

2. Suppose n and x are fixed. Then $P(X=x)={n\choose x}p^x(1-p)^{n-x}$ (binomial distribution). Also, $P(N=n)={n-1\choose x-1}p^x(1-p)^{n-x}$ (negative binomial distribution, see "Alternative formulations"). The idea for the second formula is that the last trial must be a success (n is the minimum number of trials to get x successes), so we need to squeeze x - 1 successes into n - 1 trials.

Now it is easy to prove the required equality if you express the binomial coefficients through factorials.

3. Thank you so much! I'm actually surprised that I couldn't figure it out myself. Maybe I took it as a more difficult question then it was.