Question

Given two events $A$ and $B$ such that odds against $A$ are $2$ to $1$ and odds in favour of $A\cup B$ are $3$ to $1,$ then

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

Answer 1

Answer: (A), (C)

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

Given , odds against $A$ are $2$ to $1$ and odds in favour of $(A\cup B)$ are $3$ to $1,$
$\Rightarrow \frac{P\left(A^c\right)}{P\left(A\right)}=\frac{2}{1};P\left(A^c\right)=1-P\left(A\right)\Rightarrow P\left(A\right)=\frac{1}{3}$
and $\frac{P(A\cup B)}{P(A\cup B{)}^c}=\frac{3}{1}\Rightarrow P(A\cup B)=\frac{3}{4}=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow P\left(B\right)=P(A\cap B)+\frac{5}{12}\cdots \left(i\right)$
So, $\frac{5}{12}\le P\left(B\right)\le P\left(A\right)$
$[\because 0\le P(A\cap B)\le P(A\left)\right]$
$\Rightarrow \frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$