Question

Let $A$ and $B$ be two events such that $P(\overline{A\cup B})=\frac{1}{6},P(A\cap B)=\frac{1}{4}$ and $P(\overline{A})=\frac{1}{4}$, 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

Given, $P\left(\overline{A\cup B}\right)=\frac{1}{6}$
$P(A\cap B)=\frac{1}{4},P\left(\overline{A}\right)=\frac{1}{4}\because P\left(\overline{A\cup B}\right)=\frac{1}{6}$

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

$\Rightarrow P(A\cup B)=1-\frac{1}{6}=\frac{5}{6}$

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

$\Rightarrow P\left(B\right)=\frac{5}{6}-\frac{3}{4}+\frac{1}{4}\because P\left(\overline{A}\right)=\frac{1}{4}\Rightarrow P\left(A\right)=\frac{3}{4}$

$\Rightarrow P\left(B\right)=\frac{1}{3}$

Hence, $P(A\cap B)=P\left(A\right)P\left(B\right)$ but $P\left(A\right)\ne P\left(B\right)$