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, finds the odds against C.

Answer 1

$P\left(\overline{A}\right):P\left(B\right)=8:3$

$\Rightarrow \frac{1-P\left(A\right)}{P\left(A\right)}=\frac{8}{3}$

$\Rightarrow P\left(A\right)=\frac{3}{11}$

$P\left(\overline{B}\right):P\left(B\right)=5:2$

$\Rightarrow \frac{1-P\left(B\right)}{P\left(B\right)}=\frac{5}{2}$

$\Rightarrow \frac{1}{P\left(B\right)}=\frac{5}{2}+1=\frac{7}{2}$

$\Rightarrow P\left(B\right)=\frac{2}{7}$

$\because$ A,B, and C are mutually exhaustive

$\therefore A\cup B\cup C=S$

$\Rightarrow P(A\cup B\cup C)=P\left(S\right)$

$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

$P\left(C\right)=1-\left\{P\right(A)+P(B\left)\right\}$

= $1-(\frac{3}{11}+\frac{2}{7})$

= $1-\frac{43}{77}$

$\Rightarrow P\left(\overline{C}\right)=1-P\left(C\right)$

= $1-\frac{34}{77}=\frac{43}{77}$

$\therefore$ Odds against C is

$P\left(\overline{C}\right):P\left(C\right)=\frac{43}{77}:\frac{34}{77}$

= 43 : 34