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(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. Odds in favour of $A$ are $4:6$,so $P(A) = \fr

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$7:3$

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(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. Odds in favour of $A$ are $4:6$,so $P(A) = \frac{4}{4+6} = \frac{4}{10} = \frac{2}{5}$. Odds against $B$ are $7:3$,which means odds in favour of $B$ are $3:7$,so $P(B) = \frac{3}{3+7} = \frac{3}{10}$. Substituting these into the equation: $\frac{2}{5} + \frac{3}{10} + P(C) = 1$. $\frac{4}{10} + \frac{3}{10} + P(C) = 1$ $\Rightarrow \frac{7}{10} + P(C) = 1$ $\Rightarrow P(C) = 1 - \frac{7}{10} = \frac{3}{10}$. Odds against $C$ are given by $\frac{1 - P(C)}{P(C)} = \frac{1 - \frac{3}{10}}{\frac{3}{10}} = \frac{\frac{7}{10}}{\frac{3}{10}} = \frac{7}{3}$. Thus,the odds against $C$ are $7:3$.

$A, B, C$ are three events,one of which must and only one can happen. The odds in favour of $A$ are $4:6$,and the odds against $B$ are $7:3$. Then,the odds against $C$ are:

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