What is a time of waiting for 5th success in bernoulli's sequence with p - probability.

Question

What is a time of waiting for 5th success in bernoulli's sequence with p - probability.

Hum, what exactly should I found? Should I use Newton distribution for r=5, but what is my k?

Answer 1

Let $f_k(n;p)$ be the probability of the $k$-th success occuring at step $n$ for a success probability $p$; you are looking for an expression for $f_5(n;p)$. (I will sometimes omit the $;p$ for notational brevity.)

For the $k$-th success to occur at step $n$, there must be precisely $k-1$ successes in the previous $n-1$ steps, and the last step must be a success. Thus (using $(n-1)-(k-1)=n-k$) $$ f_k(n) = p^{k-1}(1-p)^{n-k} \binom{n-1}{k-1} p = p^{k}(1-p)^{n-k}\binom{n-1}{k-1} $$

For $k= 5$ your answer is $$ f_5(n;p)=p^{5}(1-p)^{n-5}\binom{n-1}{4} $$

If you want the expectation value, that is much easier: It is $k$ times the waiting time for the first success, thus in your case $$ E(f_5(n;p)) = \frac{5}{p} $$