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 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

Answer: (D)

independent but not equally likely

$P\left(\overline{A\cup B}\right)=1-P(A\cup B)$
$=1-P\left(A\right)-P\left(B\right)+P(A\cap B)$
$\frac{1}{6}=1-\frac{3}{4}-P\left(B\right)+\frac{1}{4}$
$P\left(B\right)=\frac{1}{2}-\frac{1}{6}\Rightarrow P\left(B\right)=\frac{4}{12}=\frac{1}{3}$
now $P(A\cap B)=\frac{1}{4}=\frac{1}{3}\times \frac{3}{4}=P\left(A\right)P\left(B\right)$
so events are independent but not equally likely as $P\left(A\right)\ne P\left(B\right)$