Question

If A and B are two events such that P(A) = $\frac{1}{2}$ and P(B) = $\frac{2}{3}$, then

• A.
• B.
• C.
• D. All of the above

Answer 1

The correct option is D All of the above

We have $P(A\cup B)\ge$ max. $\left\{P\right(A),P(B\left)\right\}=\frac{2}{3}$
$P(A\cap B)\le$ min. $\left\{P\right(A),P(B\left)\right\}=\frac{1}{2}$
and $P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)\ge P\left(A\right)+P\left(B\right)-1=\frac{1}{6}$
$\Rightarrow \frac{1}{6}\le P(A\cap B)\le \frac{1}{2}$
$P(A^{\prime }\cap B)=P\left(B\right)-P(A\cap B)$
Hence $\frac{2}{3}-\frac{1}{2}\le P(A^{\prime }\cap B)\le \frac{2}{3}-\frac{1}{6}$
$\Rightarrow \frac{1}{6}\le P(A^{\prime }\cap B)\le \frac{1}{2}$