Question

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

• A. $P\left(A\right)+P\left(B\right)\le \frac{11}{8}$
• B. $P\left(A\right)\cdot P\left(B\right)\le \frac{3}{8}$
• 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 $P\left(A\right)+P\left(B\right)\ge \frac{7}{8}$

We know that
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow P\left(A\right)+P\left(B\right)=P(A\cup B)+P(A\cap B)$
We know that
$1\ge P(A\cup B)\ge \frac{3}{4}\text{and}\frac{3}{8}\ge P(A\cap B)\ge \frac{1}{8}$
Addding both we get
$\frac{7}{8}\le P(A\cup B)+P(A\cap B)\le \frac{11}{8}$
So,
$\frac{11}{8}\ge P\left(A\right)+P\left(B\right)\ge \frac{7}{8}$