Question

Given two events $A$ and $B.$ If odds against $A$ are as $2:1$ and those in favour of $A\cup B$ are as $3:1,$ then

• A. $\frac{1}{6}\le P\left(B\right)\le \frac{3}{4}$
• B. $\frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$
• C. $\frac{1}{4}\le P\left(B\right)\le \frac{3}{5}$
• D. $\frac{1}{8}\le P\left(B\right)\le \frac{3}{4}$

Answer 1

The correct option is B $\frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$

Odds against $A=\frac{2}{1}\Rightarrow P\left(A\right)=\frac{1}{3}$
Also, odds in favour of $A\cup B=\frac{3}{1}\Rightarrow P(A\cup B)=\frac{3}{4}$
Now, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\le P\left(A\right)+P\left(B\right)$
$\Rightarrow \frac{3}{4}\le \frac{1}{3}+P\left(B\right)$
$\Rightarrow \frac{5}{12}\le P\left(B\right)$
We have $B\subseteq A\cup B$
$\therefore P\left(B\right)\le P(A\cup B)=\frac{3}{4}$
Hence, $\frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$