In a race,the odds in favour of horses A, B, | vedclass

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(D) Let $P(A), P(B), P(C),$ and $P(D)$ be the probabilities that horses $A, B, C,$ and $D$ win the race,respectively.Given the odds in favour of $A, B

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(D) Let $P(A), P(B), P(C),$ and $P(D)$ be the probabilities that horses $A, B, C,$ and $D$ win the race,respectively.Given the odds in favour of $A, B, C,$ and $D$ are $1:3, 1:4, 1:5,$ and $1:6,$ the probabilities are calculated as follows:$P(A) = \frac{1}{1+3} = \frac{1}{4}$$P(B) = \frac{1}{1+4} = \frac{1}{5}$$P(C) = \frac{1}{1+5} = \frac{1}{6}$$P(D) = \frac{1}{1+6} = \frac{1}{7}$Since these events are mutually exclusive (only one horse can win the race),the probability that one of them wins is the sum of their individual probabilities:$P(	ext{one of them wins}) = P(A) + P(B) + P(C) + P(D)$$= \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}$To add these fractions,find the least common multiple of $4, 5, 6,$ and $7$,which is $420$.$= \frac{105 + 84 + 70 + 60}{420} = \frac{319}{420}$Thus,the correct option is $D$.

In a race,the odds in favour of horses $A, B, C,$ and $D$ are $1:3, 1:4, 1:5,$ and $1:6,$ respectively. Find the probability that one of them wins the race.

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