Question

There are three events $A,B$ and $C$ one of which must, and only one can happen, the odds are $8$ to $3$ against $A,5$ to $2$ against $B.$ If the odds against $C$ is $\frac{p}{q},$ then the value of $p-q$ is

Answer 1

Given, the odds are $8$ to $3$ against $A$
$\therefore P\left(A\right)=\frac{3}{11}$
The odds are $5$ to $2$ against $B$
$\therefore P\left(B\right)=\frac{2}{7}$

As out of three event one of which must and only one can happen.
$\therefore$ All events will form exhaustive set of system.
$\therefore P\left(A\right)+P\left(B\right)+P\left(C\right)=1$
$\therefore P\left(C\right)=\frac{34}{77}$
$\Rightarrow$ odds against $C=\frac{43}{34}$
$\therefore p-q=9$