The Geometric Distribution

Suppose we perform independent Bernoulli trials, each with success probability $p$ (and failure probability $1-p$). Let $X$ be the number of trials required to obtain the first success. Then $X$ follows the geometric distribution with parameter $p$, written $X \sim \text{Geom}(p).$

The probability mass function of $X$ is

$$P(X = k) = (1-p)^{k-1} \cdot p, \qquad k = 1, 2, 3, \ldots$$

The reasoning is direct: for the first success to occur on trial $k,$ the first $k-1$ trials must be failures (probability $(1-p)^{k-1}$) and the $k$th trial must be a success (probability $p$).

For example, suppose we flip a biased coin with $P(\text{heads}) = 0.4$ until we get heads. The probability that this takes exactly $3$ flips is

$$P(X = 3) = (0.6)^{2} \cdot (0.4) = 0.36 \cdot 0.4 = 0.144.$$