# 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 $$\text{P}(A' \wedge B') = 1 - \text{P}(A) - \text{P}(B) + \text{P}(A \wedge B) = 1 - \frac{3}{4} - \frac{2}{3} + p = p - \frac{5}{12},$$ 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 $\boxed{\textbf{(D)}}$.

## See Also

 1983 AHSME (Problems • Answer Key • Resources) Preceded byProblem 25 Followed byProblem 27 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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