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 $P\left(\overline{A}\right)=\frac{1}{4},$ then events $A$ and $B$ are

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

Answer 1

The correct option is A Independent but not equally likely.

Given : $P\left(\overline{A\cup B}\right)=\frac{1}{6}$
$\Rightarrow 1-P(A\cup B)=\frac{1}{6}$
$\Rightarrow P(A\cup B)=\frac{5}{6}$
$\Rightarrow P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$P\left(B\right)=\frac{5}{6}+\frac{1}{4}-\frac{3}{4}=\frac{1}{3}$
$(\because P(\overline{A})=1/4)$
$\Rightarrow P\left(A\right)\ne P\left(B\right)$
$\Rightarrow P(A\cap B)=P\left(A\right)P\left(B\right)$
$\therefore A$ and $B$ are independent but not equally likely.