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 the complement of the event A. Then, the events A and B are ?

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

Answer 1

The correct option is A Independent but not equally likely

Given, $P\left(\overline{A\cup B}\right)=\frac{1}{6};P(A\cap B)=\frac{1}{4};P\left(\overline{A}\right)=\frac{1}{4}$
$\therefore P(A\cup B)=1-P\left(\overline{A\cup B}\right)=1-\frac{1}{6}=\frac{5}{6}$
And $P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{1}{4}=\frac{3}{4}$
$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow \frac{5}{6}=\frac{3}{4}+P\left(B\right)-\frac{1}{4}$
$\Rightarrow P\left(B\right)=\frac{1}{3}\Rightarrow$ A and B are not equally likely $P(A\cap B)=P\left(A\right).P\left(B\right)=\frac{1}{4}$
So, events $A\text{and}B$ are independent.