Question

There are only three events $A,B,C$ one of which must and only one can happen; the odds are $8$ to $3$ against $A,5$ to $2$ against $B$; find the odds against $C$

• A. $12:5$
• B. $5:12$
• C. $34:43$
• D. $43:34$

Answer 1

Answer: (D)

$43:34$

Given odds against $A$ are $8:3$,
So, $P\left(A\right)=\frac{3}{8+3}=\frac{3}{11}$
Given odds against $B$ are $5:2$,
$P\left(B\right)=\frac{2}{5+2}=\frac{2}{7}$
Since the events $A,B,C$ are mutually exclusive and totally exhaustive,
So, $P\left(A\right)+P\left(B\right)+P\left(C\right)=1$
$\frac{3}{11}+\frac{2}{7}+P\left(C\right)=1$
$\Rightarrow P\left(C\right)=\frac{34}{77}$
Odds in favour of $C$ are $34:77$
Hence, the odds against $C$ are as $(77-34):34$ i.e, $43:34$