1983 AHSME Problems/Problem 26

Problem

The probability that event $A$ occurs is $\frac{3}{4}$ ; the probability that event B occurs is $\frac{2}{3}$ . Let $p$ be the probability that both $A$ and $B$ occur. The smallest interval necessarily containing $p$ is the interval

$\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big]$

Solution

Firstly note that $p \leq \frac{3}{4}$ and $p \leq \frac{2}{3}$ , as clearly the probability that both $A$ and $B$ occur cannot be more than the probability that $A$ or $B$ alone occurs. The more restrictive condition is $p \leq \frac{2}{3}$ , since $\frac{2}{3} < \frac{3}{4}$ . Now, by the Inclusion-Exclusion Principle, we also have $\mathrm{P}(A' \wedge B') = 1 - \mathrm{P}(A) - \mathrm{P}(B) + \mathrm{P}(A \wedge B) = 1 - \frac{3}{4} - \frac{2}{3} + p = p - \frac{5}{12}$ , and as a probability must be non-negative, $p - \frac{5}{12} \geq 0$ , so $p \geq \frac{5}{12}$ . Therefore, combining our inequalities gives $\frac{5}{12} \leq p \leq \frac{2}{3}$ , which is answer choice $\boxed{\textbf{D}}$ .

Page

Toolbox

Search

1983 AHSME Problems/Problem 26

Problem

Solution