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

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(B) Given that events $A, B, C$ are mutually exclusive and exhaustive,we have $P(A) + P(B) + P(C) = 1$. Odds against $A$ are $8:3$,so $P(A) = \frac{3}

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$2$

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(B) Given that events $A, B, C$ are mutually exclusive and exhaustive,we have $P(A) + P(B) + P(C) = 1$. Odds against $A$ are $8:3$,so $P(A) = \frac{3}{8+3} = \frac{3}{11}$. Odds against $B$ are $5:2$,so $P(B) = \frac{2}{5+2} = \frac{2}{7}$. Since $P(A) + P(B) + P(C) = 1$,we have $P(C) = 1 - (\frac{3}{11} + \frac{2}{7}) = 1 - (\frac{21+22}{77}) = 1 - \frac{43}{77} = \frac{34}{77}$. The odds against $C$ are given by $\frac{P(C^c)}{P(C)} = \frac{1 - P(C)}{P(C)} = \frac{1 - 34/77}{34/77} = \frac{43/77}{34/77} = \frac{43}{34}$. Given the odds against $C$ are $43:17k$,we equate: $\frac{43}{17k} = \frac{43}{34}$. Thus,$17k = 34$,which implies $k = 2$.

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$0.5$	$0.35$	$.........$	$0.7$

There are three events $A, B, C$,one of which must and only one can happen. The odds are $8:3$ against $A$,$5:2$ against $B$,and the odds against $C$ are $43:17k$. Then the value of $k$ is:

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