Question

Let $A$ and $B$ be two events such that $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{7}{12}$ and $P$(not $A$ or not $B$)$=\frac{1}{4},$ then which of the following is/are true ?

• A. $A$ and $B$ are mutually exclusive events.
• B. $A$ and $B$ are not mutually exclusive events.
• C. $P(A\cap B)=\frac{3}{4}$
• D. $P(A\cup B)=\frac{1}{3}$

Answer 1

Answer: (B), (C), (D)

$P(A\cup B)=\frac{1}{3}$

Given : $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{7}{12}$
$P$(not $A$ or not $B$)$=\frac{1}{4}$
$\Rightarrow P(\overline{A}\cup \overline{B})=\frac{1}{4}$
$\Rightarrow P(\overline{A}\cup \overline{B})=P\left(\overline{A}\right)+P\left(\overline{B}\right)-P(\overline{A}\cap \overline{B})$
$\Rightarrow P(\overline{A}\cup \overline{B})=1-P\left(A\right)+1-P\left(B\right)-P(\overline{A}\cap \overline{B})$
$\frac{1}{4}=\frac{1}{2}+\frac{5}{12}-P(\overline{A}\cap \overline{B})$
$\Rightarrow P(\overline{A}\cap \overline{B})=\frac{2}{3}$
$\Rightarrow 1-P(A\cup B)=\frac{2}{3}$
$\therefore P(A\cup B)=\frac{1}{3}$
$\Rightarrow P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$P(A\cap B)=\frac{1}{2}+\frac{7}{12}-\frac{1}{3}$
$\therefore P(A\cap B)=\frac{3}{4}\ne 0$
$\Rightarrow A$ and $B$ are not mutually exclusive events