$14 \cdot 36$ event $A$ occurs with a probability of $\frac{3}{4}$, and event $B$ occurs with a probability of $\frac{2}{3}$. Let $p$ be the probability that both $A$ and $B$ occur. Then the interval containing $p$ is
(A) $\left[\frac{1}{12}, \frac{1}{2}\right]$.
(B) $\left[\frac{5}{12}, \frac{1}{2}\right]$.
(C) $\left[\frac{1}{2}, \frac{2}{3}\right]$.
(D) $\left[\frac{5}{12}, \frac{2}{3}\right]$.
(E) $\left[\frac{1}{12}, \frac{2}{3}\right]$.
(34th American High School Mathematics Examination, 1983)
[Solution] Let $P(E)$ denote the probability of event $E$ occurring.
By the formula $P(A \cup B)=P(A)+P(B)-P(A \cap B)$,
$$
\begin{aligned}
\therefore \quad p & =P(A \cap B)=P(A)+P(B)-P(A \cup B) \\
& =\frac{3}{4}+\frac{2}{3}-P(A \cup B),
\end{aligned}
$$

where $\quad 1 \geqslant P(A \cup B) \geqslant \max \{P(A), P(B)\}=\frac{3}{4}$.
Therefore, $\frac{3}{4}+\frac{2}{3}-1 \leqslant p \leqslant \frac{3}{4}+\frac{2}{3}-\frac{3}{4}$,

which means $\square$
$$
\frac{5}{12} \leqslant p \leqslant \frac{2}{3} \text {. }
$$

Thus, the answer is $(D)$.