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, the odds are $5$ to $2$ against $B$, find odds against $C$.

• A. $43:34$
• B. $45:34$
• C. $47:34$
• D. none of these

Answer 1

The correct option is A $43:34$

$P\left(A\right)=\frac{3}{11},P\left(B\right)=\frac{2}{7}$
Let us take $P\left(C\right)=x$
Since one must and only one can happen therefore A,B,C are mutually exclusive and exhaustive events. So,$P\left(A\right)+P\left(B\right)+P\left(C\right)=1\Rightarrow \frac{3}{11}+\frac{2}{7}+x=1\Rightarrow x=\frac{77-21-22}{77}=\frac{34}{77}$
$\therefore$ Odds against $C=(77-34):34$ i.e.,$43:34$