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

Answer 1

Answer: (A), (C)

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

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow \frac{3}{4}\le P\left(A\right)+P\left(B\right)-P(A\cap B)\le 1\Rightarrow \frac{3}{4}+P(A\cap B)\le P\left(A\right)+P\left(B\right)\le 1+P(A\cap B)\Rightarrow \frac{3}{4}+\frac{1}{8}\le P\left(A\right)+P\left(B\right)\le 1+\frac{3}{8}$ since $\frac{1}{8}\le P(A\cap B)\le \frac{3}{8}$
$\Rightarrow \frac{7}{8}\le P\left(A\right)+P\left(B\right)\le \frac{11}{8}$
Hence option A and C are correct.
Since A and B are not given to be independent hence $P(A\cap B)\ne P\left(A\right)P\left(B\right)$ hence it will be wrong to say $P\left(A\right)P\left(B\right)\le \frac{3}{8}$