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, the events $A$ 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

The correct option is B Independent, but not equally likely

Given, $P\left(\overline{A\cup B}\right)=\frac{1}{6}$
We know $1-P(A\cup B)=\frac{1}{6}$
$\Rightarrow 1-P\left(A\right)-P\left(B\right)+P(A\cap B)=\frac{1}{6}$
$\Rightarrow \frac{1}{4}-P\left(B\right)+\frac{1}{4}=\frac{1}{6}$
$\Rightarrow P\left(B\right)=\frac{1}{3}$
Therefore, $P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{1}{4}=\frac{3}{4}$