Question

Given two events A and B, if the odds against Aare 2 to 1, and those in favour of A$\cup$B are 3 to1, then

• A. $\frac{1}{3}\le P\left(B\right)\le \frac{1}{2}$
• B. $\frac{1}{2}\le P\left(B\right)\le \frac{3}{4}$
• C. $\frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$
• D. $0\le P\left(B\right)\le 1$

Answer 1

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

$OddsagainstA=P\left(A^c\right)/P\left(A\right)=\frac{2}{1}\Rightarrow \{1-P(A\left)\right\}/P\left(A\right)=2\Rightarrow 1/P\left(A\right)-1=2\Rightarrow 1/P\left(A\right)=3\Rightarrow P\left(A\right)=1/3AlsoOddsinfavourofA\cup B=\frac{P(A\cup B)}{P{(A\cup B)}^c}=\frac{3}{1}\Rightarrow \frac{P(A\cup B)}{1-P(A\cup B)}=3\Rightarrow \frac{1-P(A\cup B)}{P(A\cup B)}=\frac{1}{3}\Rightarrow \frac{1}{P(A\cup B)}=1+\frac{1}{3}=\frac{4}{3}\Rightarrow P(A\cup B)=\frac{3}{4}$

since we know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

Hence putting the values of results obtained above we get $\Rightarrow \frac{3}{4}=\frac{1}{3}+P\left(B\right)-P(A\cap B)$

on rearranging the terms we get $\Rightarrow P\left(B\right)=P(A\cap B)+\frac{5}{12}$

$P(A\cap B)canneverbeequaltoP\left(B\right)\Rightarrow \frac{5}{12}\le P\left(B\right)\le \frac{5}{12}+P\left(A\right)$

$0\le P(A\cap B)\le P\left(A\right)\Rightarrow \frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$