There are three events $\mathrm{A}, \mathrm{B}, \mathrm{C}$, one of which must and only one can happen. The…
There are three events $\mathrm{A}, \mathrm{B}, \mathrm{C}$, one of which must and only one can happen. The odds are 8:3 against $\mathrm{A}, 5: 2$ against B and the odds against C is $43: 17 \mathrm{k}$, then value of $k$ is
$\frac{1}{2}$
2
$\frac{1}{3}$
$\frac{1}{4}$
Solution
The odds against A are $8: 3$.
$\therefore \quad \mathrm{P}(\mathrm{~A})=\frac{3}{11}$
Odds against B are $5: 2$
$\begin{array}{ll}
\therefore & P(B)=\frac{2}{7} \\
\therefore & P(A)+P(B)+P(C)=1 \\
& \Rightarrow \frac{3}{11}+\frac{2}{7}+P(C)=1 \\
\therefore & P(C)=1-\frac{2}{7}-\frac{3}{11} \\
\therefore & P(C)=\frac{34}{77}
\end{array}$
$\therefore \quad$ odds against $\mathrm{P}(\mathrm{C})=\frac{77-34}{34}=\frac{43}{34}$
But odds against $\mathrm{C}=\frac{43}{17 \mathrm{k}}$
$\begin{aligned}
& \therefore \quad \frac{43}{17 k}=\frac{43}{34} \\
& \therefore \quad k=2
\end{aligned}$