A,B,and C are three events,one of which must | vedclass

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

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$7 : 3$

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(A) Given that $A$,$B$,and $C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. Odds in favor of $A$ are $4 : 6$,so $P(A) = \frac{4}{4+6} = \frac{4}{10}$. Odds against $B$ are $7 : 3$,which means odds in favor of $B$ are $3 : 7$,so $P(B) = \frac{3}{3+7} = \frac{3}{10}$. Substituting these into the sum: $\frac{4}{10} + \frac{3}{10} + P(C) = 1$. $\frac{7}{10} + P(C) = 1 \implies P(C) = 1 - \frac{7}{10} = \frac{3}{10}$. The probability of the complement event $C'$ is $P(C') = 1 - P(C) = 1 - \frac{3}{10} = \frac{7}{10}$. The odds against $C$ are defined as $P(C') : P(C) = \frac{7}{10} : \frac{3}{10} = 7 : 3$.

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

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