There are three events A, B, C one of which m | vedclass

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(NONE OF THE ABOVE (CALCULATED: 43:34)) Since the odds against event $A$ are $8:3$,the probability of the occurrence of event $A$ is $P(A) = \frac{3}{

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$43:44$

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(NONE OF THE ABOVE (CALCULATED: 43:34)) Since the odds against event $A$ are $8:3$,the probability of the occurrence of event $A$ is $P(A) = \frac{3}{8+3} = \frac{3}{11}$.Similarly,the odds against event $B$ are $5:2$,so the probability of the occurrence of event $B$ is $P(B) = \frac{2}{5+2} = \frac{2}{7}$.Since events $A, B, C$ are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.Therefore,$P(A) + P(B) + P(C) = 1$.Substituting the values: $\frac{3}{11} + \frac{2}{7} + P(C) = 1$.Calculating the sum: $\frac{21 + 22}{77} + P(C) = 1 \Rightarrow \frac{43}{77} + P(C) = 1$.Thus,$P(C) = 1 - \frac{43}{77} = \frac{34}{77}$.The probability of the non-occurrence of $C$ is $P(\bar{C}) = 1 - P(C) = 1 - \frac{34}{77} = \frac{43}{77}$.The odds against $C$ are given by $P(\bar{C}) : P(C) = \frac{43}{77} : \frac{34}{77} = 43:34$.

There are three events $A, B, C$ one of which must and only one can happen. The odds are $8$ to $3$ against $A$ and $5$ to $2$ against $B$. Find the odds against $C$.

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