Question

There are 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.

Answer 1

$\frac{P\left(A^1\right)}{P\left(A\right)}=\frac{8}{3},P\left(A\right)=\frac{3}{11},P\left(A^1\right)=\frac{8}{11}$
$\frac{P\left(B^1\right)}{P\left(B\right)}=\frac{5}{2},P\left(B\right)=\frac{2}{7},P\left(B^1\right)=\frac{5}{7}$
Out of $A,B\&C$, one and only one can happen at a time.
$P\left(A\right)+P\left(B\right)+P\left(C\right)=1$
$P\left(C\right)=1-\frac{3}{11}-\frac{2}{7}$
$=\frac{34}{77}$
$P\left(C^1\right)=1-P\left(C\right)$
$=1-\frac{34}{77}$
$=\frac{43}{77}$
So odd against$C=\frac{P\left(C^1\right)}{P\left(C\right)}=\frac{43}{34}\Rightarrow 43:34$