Question

If $A,B$ 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)

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

Given that $P(A\cup B)\ge \frac{3}{4}$ and $\frac{1}{8}\le P(A\cap B)\ge \frac{3}{8}$

It is known that $P\left(A\right)+P\left(B\right)-P(A\cap B)=P(A\cup B)$

$\therefore P\left(A\right)+P\left(B\right)=P(A\cap B)+P(A\cup B)$

$\therefore P\left(A\right)+P\left(B\right)\ge \frac{1}{8}+\frac{3}{4}\ge \frac{7}{8}$

Also, $P(A\cup B)\le 1$ since it is a probability

$\therefore P\left(A\right)+P\left(B\right)\le \frac{3}{8}+1\le \frac{11}{8}$