13. Let $A, B, C$ be pairwise independent and equally probable events, and $A \cap B \cap C=\varnothing$. Find the maximum possible value for the probability $\mathrm{P}(A)$.
Solution. Let $\mathrm{P}(A)=p$. By the condition, $\mathrm{P}(B)=\mathrm{P}(C)=p$, $\mathrm{P}(A B)=\mathrm{P}(B C)=\mathrm{P}(A C)=p^{2}$.

Consider the probability

$$
\begin{aligned}
& \mathrm{P}(A \backslash(B \cup C))=\mathrm{P}(A \backslash(A(B \cup C)))= \\
& =\mathrm{P}(A)-\mathrm{P}(A(B \cup C))=\mathrm{P}(A)-\mathrm{P}(A B \cup A C)= \\
& \quad=\mathrm{P}(A)-\mathrm{P}(A B)-\mathrm{P}(A C)+\mathrm{P}(A B C)=p-2 p^{2}
\end{aligned}
$$

Since the probability of any event is non-negative, $p \geqslant 0$ and $p-$ $-2 p^{2} \geqslant 0$. Solving the quadratic inequality, we get that $0 \leqslant p \leqslant 1 / 2$.

Let's provide an example that satisfies the conditions of the problem with $\mathrm{P}(A)=1 / 2$. Consider the Bernoulli trials scheme with $n=2$ and $p=1 / 2$. Let

$$
\begin{aligned}
& A=\{\text { success on the first trial }\} \\
& B=\{\text { success on the second trial }\}, \\
& C=\{\text { exactly one success }\} .
\end{aligned}
$$

Then $\mathrm{P}(A)=\mathrm{P}(B)=\mathrm{P}(C)=1 / 2, \mathrm{P}(A B)=\mathrm{P}(B C)=\mathrm{P}(A C)=1 / 4$ and $A B C=\varnothing$.

Thus, it is shown that $\mathrm{P}(A) \leqslant 1 / 2$, and an example is provided where $\mathrm{P}(A)=$ $=1 / 2$, meaning that $1 / 2$ is the maximum possible value for the probability $\mathrm{P}(A)$.