17. (1983 American High School Mathematics Examination) The probability of event $A$ occurring is $\frac{3}{4}$, and the probability of event $B$ occurring is $\frac{2}{3}$. 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. $\left[\frac{1}{12}, \frac{2}{3}\right]$
17. D. Reason: Let $P(E)$ denote the probability of event $E$ occurring.

From the formula $P(A \cup B)=P(A)+P(B)-P(A \cap B)$,
we have $p=P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$$
=\frac{3}{4}+\frac{2}{3}-P(A \cup B),
$$

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 $\frac{5}{12} \leqslant p \leqslant \frac{2}{3}$.