Question

Let $A$ and $B$ be two events such that $P\left(\overline{A\cup B}\right)=\frac{1}{6},$ $P(A\cap B)=\frac{1}{4}$ and $P\left(\overline{A}\right)=\frac{1}{4}$, where, $\overline{A}$ stands for complement of event $A$. Then, event $A$ and $B$ are

• A. Mutually exclusive and independent
• B. Independent but not equally likely
• C. Equally likely but not independent
• D. Equally likely and mutually exclusive

Answer 1

Answer: (B)

Independent but not equally likely

Given that, $P\left(\overline{A\cup B}\right)=\frac{1}{6},$ $P(A\cap B)=\frac{1}{4},$ $P\left(\overline{A}\right)=\frac{1}{4}$

$P\left(\overline{A\cup B}\right)=\frac{1}{6}$
$\Rightarrow 1-P(A\cup B)=\frac{1}{6}$
$\Rightarrow [1-P\left(A\right)]-P\left(B\right)+P(A\cap B)=\frac{1}{6}$

$\Rightarrow P\left(\overline{A}\right)-P\left(B\right)+P(A\cap B)=\frac{1}{6}$
$\Rightarrow \frac{1}{4}-P\left(B\right)+\frac{1}{4}=\frac{1}{6}$
$\Rightarrow P\left(B\right)=\frac{2}{4}-\frac{1}{6}$

$\Rightarrow P\left(B\right)=\frac{1}{3}$
and $P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{1}{4}=\frac{3}{4}$

Since, $P(A\cap B)=P\left(A\right)\cdot P\left(B\right)$, so the events $A$ and $B$ are independent events but not equal likely.