\begin{aligned}  Expected number of trials until first success is; Therefore, expected number of failures until first success is; Hence, we expect failures before the rth success. × (½)4× (½)1= 5/32 P(x = 5) = 5C5 p… Suppose we flip a coin repeatedly and count the number of heads (successes). \begin{aligned} A geometric distribution is a special case of a negative binomial distribution with $$r=1$$. V(X) &= \frac{rq}{p^2}\\ Toss a fair coin until get 8 heads. & \quad\quad \qquad 0