Question

$A$ and $B$ are two events. Odds against $A$ and $2:1$. Odds in favor of $A\cup B$ are $3:1$. If $x\le P\left(B\right)\le y$, then the ordered pair $(x,y)$ is

• A. $(\frac{5}{12},\frac{3}{4})$
• B. $(\frac{2}{3},\frac{3}{4})$
• C. $(\frac{1}{3},\frac{3}{4})$
• D. None of these

Answer 1

The correct option is A $(\frac{5}{12},\frac{3}{4})$

We have $P\left(A\right)=\frac{1}{3}$ and $P(A\cap B)=\frac{3}{4}$
$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow \frac{3}{4}=\frac{1}{3}+P\left(B\right)-P(A\cap B)\Rightarrow \frac{5}{12}=P\left(B\right)-P(A\cap B)$
$\Rightarrow P\left(B\right)=\frac{5}{12}+P(A\cap B)\Rightarrow P\left(B\right)\ge \frac{5}{12}$ ...(1)
Again $\Rightarrow P\left(B\right)=\frac{5}{12}+P(A\cap B)\Rightarrow P\left(B\right)\le \frac{5}{12}+P\left(A\right)=\frac{5}{12}+\frac{1}{3}=\frac{3}{4}$ ...(2)
$[\because P(A\cap B)\le P\left(A\right)]$
From (1) and (2), we obtain $\frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$.
Hence $x=\frac{5}{12}$ and $y=\frac{3}{4}$