Expectation of number of trails till $r$ successes.

Question

Let $X$ be the number of Bernoulli trails till $r$ successes with probability $p$ (including the last one). Find $\mathbb{E}(X)$.

My attempt:

$$P(X<r) = 0$$ $$P(X=r) = p^r$$ $$P(X=r+s) = p^r(1-p)^s$$ $$\mathbb{E}(X) = p^r\sum_{n=r}^{\infty}n(p-1)^{n-r}$$

I think I can simplify this using the geometric series but does this make sense as an answer?

Answer 1

The correct expression for $\Pr(X=n)$ has been dealt with in comments. From that expression one can, by manipulation, find $E(X)$. But there are easier ways.

Let $X_1$ be the number of trials until the first success. Let $X_2$ be the number of trials from (but not including) the first success to (and including) the second success. Similarly, define $X_3,\dots,X_r$.

Then $X=X_1+X_2+\cdots +X_r$, so by the linearity of expectation we have $E(X)=E(X_1)+E(X_2)+\cdots+E(X_r)$.

Each of the $X_i$ has geometric distribution with parameter $p$, so $E(X_i)=\frac{1}{p}$ for all $i$. It follows that $E(X)=\frac{r}{p}$.