Question

If $A$ and $B$ are two events such that$P(A\cup B)\ge \frac{3}{4}$ and $\frac{1}{8}\le P(A\cap B)\le \frac{3}{8}$, then

• A. $P\left(A\right)+P\left(B\right)\ge \frac{7}{8}$
• B. $P\left(A\right)+P\left(B\right)\le 11/8$
• C. $P\left(A\right)P\left(B\right)\le \frac{3}{8}$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A $P\left(A\right)+P\left(B\right)\ge \frac{7}{8}$
B $P\left(A\right)+P\left(B\right)\le 11/8$

$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\therefore 1\ge P\left(A\right)+P\left(B\right)-P(A\cap B)\ge 3/4.$
But minimum of $P(A\cap B)=18$
$\therefore P\left(A\right)+P\left(B\right)-\frac{1}{8}\ge 3/4$
$\Rightarrow P\left(A\right)+P\left(B\right)\ge \frac{7}{8}$
As the maximum of $P(A\cap B)=\frac{3}{8}$
$\therefore 1\ge 1P\left(A\right)+P\left(B\right)-3/8$
$\Rightarrow P\left(A\right)+P\left(B\right)\le 11/8$