# Independent Bernoulli trials - help with question!

• Jan 31st 2011, 02:49 PM
Champ83
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...
• Feb 1st 2011, 02:12 AM
emakarov
Suppose n and x are fixed. Then \$\displaystyle P(X=x)={n\choose x}p^x(1-p)^{n-x}\$ (binomial distribution). Also, \$\displaystyle 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.
• Feb 1st 2011, 02:31 PM
Champ83
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.