Question

If $A$ and $B$ are two events such that $P\left(A\right)=\frac{3}{4}$ and $P\left(B\right)=\frac{5}{8}$, then

• A. $P(A\cup B)\ge \frac{3}{4}$
• B. $P(A^{\prime }\cup B)\ge \frac{1}{4}$
• C. $\frac{3}{8}\le P(A\cup B)\le \frac{5}{8}$
• D. $\frac{3}{8}\le P(A\cap B)\le \frac{5}{8}$

Answer 1

Answer: (A), (D)

The correct options are
A $P(A\cup B)\ge \frac{3}{4}$
D $\frac{3}{8}\le P(A\cap B)\le \frac{5}{8}$

$A\subseteq A\cup B\Rightarrow P\left(A\right)\le P(A\cup B)\Rightarrow P(A\cup B)\ge \frac{3}{4}$
Also,
$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)$
$\ge P\left(A\right)+P\left(B\right)-1$
$=\frac{3}{4}+\frac{5}{8}-1=\frac{3}{8}$
Now, $A\cap B\subseteq B\Rightarrow P(A\cap B)\le P\left(B\right)=\frac{5}{8}\therefore \frac{3}{8}\le P(A\cap B)\le \frac{5}{8}\cdots \left(1\right)$
and
$P(A\cap B^{\prime })=P\left(A\right)-P(A\cap B)\Rightarrow \frac{3}{4}-\frac{5}{8}\le P(A\cap B^{\prime })\le \frac{3}{4}-\frac{3}{8}\therefore P(A\cap B)=P\left(B\right)-P(A^{\prime }\cap B)\left[\text{Using}\right(1\left)\right]\Rightarrow 0\le P(A^{\prime }\cap B)\le \frac{1}{4}$