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Question

If the odds in favour of winning a race by three horses are 1:4,1:5 and 1:6, find the probability that exactly one of these horses will win.

A
3760
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B
39105
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C
41105
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D
51105
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Solution

The correct option is A 3760
Let P(A), P(B), P(C) be the responsibilities of winning of the horses A, B, C respectively. Then
P(A)=14,P(B)=15,P(C)=16
Since the above events are mutually exclusive, the chance that one of them wins
=P(ABC)=P(A)+P(B)+P(C)=14+15+16=30+24+20120=74120=3760
is the required probability.

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