Question

If $A$ and $B$ are two events such that $P(A\cup B{)}^{\prime }=\frac{1}{6}$, $P(A\cap B)=\frac{1}{4}$ and $P\left(A^{\prime }\right)=\frac{1}{4}$, then events $A$ and $B$ are

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

Answer 1

The correct option is D

independent but not equally likely

$P(A\cup B{)}^{\prime }=\frac{1}{6}=1-P(A\cup B)$

$\Rightarrow P(A\cup B)=\frac{5}{6}$

$\Rightarrow P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{5}{6}$

$\Rightarrow P\left(A\right)+P\left(B\right)=\frac{5}{6}+\frac{1}{4}=\frac{13}{12}$

$\Rightarrow \frac{3}{4}+P\left(B\right)=\frac{13}{12}$ [$\because P\left(A\right)=1-P\left(A^{\prime }\right)$]

$\Rightarrow P\left(B\right)=\frac{13}{12}-\frac{3}{4}=\frac{1}{3}$

We have $P\left(A\right).P\left(B\right)=\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)=\frac{1}{4}=P(A\cap B)$

Thus, events $A$ and $B$ are independent but not equally likely.