If \(A\) is an event in the finite sample space \(\mathcal{S}\), then \(P(A)\) equals the sum of the probabilities of the individual outcomes comprising \(A\).

**Proof:**

To prove this, let \(E_{1}, E_{2}, \ldots, E_{n}\) be the \(n\) outcomes comprising event \(A\), so that we can write \(A=E_{1} \cup E_{2} \cup \cdots \cup E_{n}\). Since the \(E\) 's are individual outcomes, they are mutually exclusive, and by Theorem \(3.4\) we have

\[

\begin{aligned}

P(A) &=P\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right) \\

&=P\left(E_{1}\right)+P\left(E_{2}\right)+\cdots+P\left(E_{n}\right)

\end{aligned}

\]

which completes the proof.