Question

If $A$ and $B$ are two events such that $P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{2}{3}$, then which of the following is/are correct?

• A. $P(A\cup B)\ge \frac{2}{3}$
• B. $P_{max}(A\cap B^{\prime })=\frac{1}{3}$
• C. $\frac{1}{6}\le P(A\cap B)\le \frac{1}{2}$
• D. $P_{max}(A^{\prime }\cap B)=\frac{1}{2}$

Answer 1

Answer: (A), (B), (C), (D)

$P_{max}(A^{\prime }\cap B)=\frac{1}{2}$

Given
$P\left(A\right)=\frac{1}{2}\text{and}P\left(B\right)=\frac{2}{3}\Rightarrow P(A\cup B)\ge max\left\{P\right(A),P(B\left)\right\}\Rightarrow P(A\cup B)\ge \frac{2}{3}$
Also,
$P(A\cap B)\le min\left\{P\right(A),P(B\left)\right\}\Rightarrow P(A\cap B)\le \frac{1}{2}$
$P(A\cup B)\le 1\Rightarrow P(A\cap B)\ge P\left(A\right)+P\left(B\right)-1\Rightarrow P(A\cap B)\ge \frac{1}{6}\Rightarrow \frac{1}{6}\le P(A\cap B)\le \frac{1}{2}$
$P(A\cap B^{\prime })=P\left(A\right)-P(A\cap B)\le (\frac{1}{2}-\frac{1}{6})=\frac{1}{3}\therefore P_{max}(A\cap B^{\prime })=\frac{1}{3}P(A^{\prime }\cap B)=P\left(B\right)-P(A\cap B)\le (\frac{2}{3}-\frac{1}{6})=\frac{1}{2}{P}_{max}(A^{\prime }\cap B)=\frac{1}{2}$