 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Put $q=1-p$. Since $X$ takes positive integer values, $\mathbb P(X>0)=1=q^0$. Taking differences of the given tail <probabilities> therefore yields, also for $k=1$,
$$
\boxed{\mathbb P(X=k)=\mathbb P(X>k-1)-\mathbb P(X>k)=pq^{k-1},\qquad k\ge1.}
$$
This is the <probability mass function> of the <geometric distribution> on $\{1,2,\ldots\}$.

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