Question

Let $A$ and $B$ be two events such that $P(A\cap B)=\frac{1}{6},P(A\cup B)=\frac{31}{45}$ and $P\left(\overline{B}\right)=\frac{7}{10}$ then

• A. $A$ and $B$ are independent
• B. $A$ and $B$ are mutually exclusive
• C. $P\left(\frac{A}{B}\right)<\frac{1}{6}$
• D. $P\left(\frac{B}{A}\right)<\frac{1}{6}$

Answer 1

The correct option is A $A$ and $B$ are independent

$P\left(B\right)=1-P\left(\overline{B}\right)=1-\frac{7}{10}=\frac{3}{10}$
We know that,
$P\left(A\right)=P(A\cup B)-P\left(B\right)+P(A\cap B)\Rightarrow P\left(A\right)=\frac{5}{9}$
So,
$P\left(A\right)\times P\left(B\right)=\frac{5}{9}\times \frac{3}{10}\Rightarrow P\left(A\right)\times P\left(B\right)=\frac{1}{6}=P(A\cap B)$

Henc they are independent events.