Question

Let $A$ and $B$ be two events such that $P(A\cap B)=\frac{1}{3},P(A\cup B)=\frac{5}{6}$ and $P\left(\overline{A}\right)=\frac{1}{2}$. Then

• A. $A,B$ are independent
• B. $A,B$ are mutually exclusive
• C. $P\left(A\right)=P\left(B\right)$
• D. $P\left(B\right)\le P\left(A\right)$

Answer 1

The correct option is A $A,B$ are independent

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)alsoP\left(A\right)=1-P\left(\overline{A}\right)=\frac{1}{2}\Rightarrow \frac{5}{6}=\frac{1}{2}+P\left(B\right)-\frac{1}{3}\Rightarrow P\left(B\right)=\frac{2}{3}$
thus we see that $P\left(A\right)\ne P\left(B\right),P\left(B\right)>P\left(A\right)andP(A\cap B)=P\left(A\right)P\left(B\right)$