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. $\frac{37}{60}$
• B. $\frac{39}{105}$
• C. $\frac{41}{105}$
• D. $\frac{51}{105}$

Answer 1

The correct option is A $\frac{37}{60}$

Let P(A), P(B), P(C) be the responsibilities of winning of the horses A, B, C respectively. Then

$P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{5},P\left(C\right)=\frac{1}{6}$

Since the above events are mutually exclusive, the chance that one of them wins

$=P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)⟹=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{30+24+20}{120}⟹=\frac{74}{120}=\frac{37}{60}$

is the required probability.