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

Answer 1

Answer: (C)

independent but not equally likely

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

and $P\left(\overline{A}\right)=\frac{1}{4}$

$=P(A\cup B)=5/6P\left(A\right)=3/4$

Also $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=>P\left(B\right)=\frac{5}{6}-\frac{3}{4}+\frac{1}{4}=\frac{1}{3}$

$\Rightarrow P\left(A\right)P\left(B\right)=\frac{3}{4}\times \frac{1}{3}=\frac{1}{4}=P(A\cap B)$

Hence $A$ and $B$ are independent but not equally likely.