Question

If $A$ and $B$ are two event 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. $P(\overline{A}\cap B)\le \frac{1}{4}$
• C. $\frac{3}{8}\le P(A\cap B)\le \frac{5}{8}$
• D. All of these

Answer 1

The correct option is D All of these

$P(A\cup B)\ge max\left\{P\right(A),P(B\left)\right\}$

Here $P\left(A\right)=\frac{3}{4},P\left(B\right)=\frac{5}{8}$

$\Rightarrow P(A\cup B)\ge \frac{3}{4}$

So, option A is correct.

$P(A\cap B)\le min\left\{P\right(A),P(B\left)\right\}\Rightarrow P(A\cap B)\le \frac{5}{8}$ ......(i)

We know $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$\Rightarrow P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)\Rightarrow P(A\cap B)\ge P\left(A\right)+P\left(B\right)-1\Rightarrow P(A\cap B)\ge \frac{3}{4}+\frac{5}{8}-1\Rightarrow P(A\cap B)\ge \frac{3}{8}$ .......(ii)

From (i) and (ii), we have

$\frac{3}{8}\le P(A\cap B)\le \frac{5}{8}$

So, option C is correct.

$\frac{3}{8}\le P(A\cap B)\le \frac{5}{8}\Rightarrow -\frac{5}{8}\le -P(A\cap B)\le -\frac{3}{8}$

$\Rightarrow P\left(B\right)-\frac{5}{8}\le P\left(B\right)-P(A\cap B)\le P\left(B\right)-\frac{3}{8}\Rightarrow \frac{5}{8}-\frac{5}{8}\le P(\stackrel{\_}{A}\cap B)\le \frac{5}{8}-\frac{3}{8}\Rightarrow 0\le P(\stackrel{\_}{A}\cap B)\le \frac{1}{4}\Rightarrow P(\stackrel{\_}{A}\cap B)\le \frac{1}{4}$