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. $P\left(A\right)$ . $P\left(B\right)\ge \frac{1}{8}$

Answer 1

Answer: (A), (C)

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

We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=>P\left(A\right)+P\left(B\right)=P(A\cup B)+P(A\cap B)$
Now, $\frac{3}{4}\le P(A\cup B)\le 1and\frac{1}{8}\le P(A\cap B)\le \frac{3}{8}$
Thus, $(\frac{3}{4}+\frac{1}{8})\le P\left(A\right)+P\left(B\right)\le (1+\frac{3}{8})=>\frac{7}{8}\le P\left(A\right)+P\left(B\right)\le \frac{11}{8}$