Question

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

• A. $\frac{1}{2}\le P\left(B\right)\le \frac{3}{4}$
• 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. None of these

Answer 1

Answer: (B)

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

The odds against $A=\frac{P\left(A^c\right)}{P\left(A\right)}=\frac{2}{1}$

$\Rightarrow \frac{1-P\left(A\right)}{P\left(A\right)}=2$

$\Rightarrow \frac{1}{P\left(A\right)}-1=2$

$\Rightarrow \frac{1}{P\left(A\right)}=3$

$\Rightarrow P\left(A\right)=\frac{1}{3}$

Also, the odds in favour of $A\cup B=\frac{P(A\cup B)}{P{(A\cup B)}^c}=\frac{3}{1}$

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

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

$\Rightarrow \frac{1}{P(A\cup B)}=1+\frac{1}{3}=\frac{4}{3}$

$\Rightarrow P(A\cup B)=\frac{3}{4}$

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 result obtained above we get
$\frac{3}{4}=\frac{1}{3}+P\left(B\right)-P(A\cap B)$
On rearranging the terms we get,
$P\left(B\right)=P(A\cap B)+\frac{5}{12}$ from this equation we can see that$P(A\cap B)$ can never be equal to$P\left(B\right)$
$\Rightarrow \frac{5}{12}\le P\left(B\right)\le \frac{5}{12}+P\left(A\right)$ ( Here we have used the concept that $0\le P(A\cap B)\le P\left(A\right))$
$\Rightarrow \frac{5}{12}\le P\left(B\right)\le \frac{3}{4}$