Question

$A$ and $B$ are two events such that $P(A\cup B)=\frac{5}{6}$ and $P(A\cap B)=\frac{1}{3}$, if $P\left(B^c\right)=\frac{1}{2}$, then events are

• A. dependent
• B. Independent
• C. M.E.
• D. None of these

Answer 1

Answer: (B)

Independent

Given $P\left(B^c\right)=\frac{1}{2}$ $\therefore P\left(B\right)=1-1/2=1/2$
Also $P(A\cup B)=\frac{5}{6}$
$\Rightarrow P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{5}{6}$
$\therefore P\left(A\right)=\frac{5}{6}+\frac{1}{3}-\frac{1}{2}$
$P\left(A\right)=\frac{2}{3}$ so $P\left(A^c\right)=1/3$
Now $P\left(A\right)\times P\left(B\right)=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}=P(A\cap B)$
So events are independent.