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 the complement of the event $A$. Then the events $A$ and $B$ are

• A. mutually exclusive and independent
• B. equally likely but not independent
• C. independent but not equally likely
• D. independent and eqaully likely

Answer 1

The correct option is C independent but not equally likely

Given: $P\left(\overline{A\cup B}\right)=\frac{1}{6},P(A\cap B)=\frac{1}{4}\text{and}P\left(\overline{A}\right)=\frac{1}{4}$
Since, $P(A\cup B)=1-P\left(\overline{A\cup B}\right)=1-\frac{1}{6}=\frac{5}{6}$
and, $P\left(A\right)=1-P\left(\overline{A}\right)=\frac{3}{4}$
Now, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow \frac{5}{6}=\frac{3}{4}+P\left(B\right)-\frac{1}{4}$
$\Rightarrow P\left(B\right)=\frac{1}{3}$
$P\left(A\right)\ne P\left(B\right)$, So not equally likely
Now, $P\left(A\right)\cdot P\left(B\right)=\frac{1}{3}\cdot \frac{3}{4}$
$\Rightarrow P\left(A\right)\cdot P\left(B\right)=\frac{1}{4}=P(A\cap B)$
Hence, events $A$ and $B$ are independent but not equally likely