Question

The odds in favour of winning a race for three horses A, B and C respectively are 1:2,1:3 and 1:4. Find the probability for winning any one of them.
[a] $\dfrac{1}{60}$
[b] $\dfrac{25}{60}$
[c] $\dfrac{33}{60}$
[d] $\dfrac{47}{60}$

Answer 1

Hint: Use the fact that if the odds in favour of an event A are a:b, then $P\left( A \right)=\dfrac{a}{a+b}$ and $P\left( A' \right)=\dfrac{b}{a+b}$. Use the fact that if A, B and C are disjoint events, i.e. occurrence of any one of the events automatically disqualifies the occurrence of other events, then \[P\left( A\bigcup B\bigcup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)\].

Complete step-by-step answer:
Let X be the event: Horse A wins the race
Let Y be the event: Horse B wins the race
and Let Z be the event : Horse C wins the race.
Since the winning of one horse automatically disqualifies the winning of the other two horses, we have
X, Y and Z are disjoint events.
Now we know that if the odds in favour of an event A are a:b, then $P\left( A \right)=\dfrac{a}{a+b}$
Hence P(X) $=\dfrac{1}{1+2}=\dfrac{1}{3}$, P(Y) $=\dfrac{1}{1+3}=\dfrac{1}{4}$ and P(Z) $=\dfrac{1}{1+4}=\dfrac{1}{5}$
Hence $P\left( X\bigcup Y\bigcup Z \right)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{20+15+12}{60}=\dfrac{47}{60}$
Hence the probability for winning any one of them is $\dfrac{47}{60}$.
Hence, option [d] is correct.

Note: Odds in favour of an event are defined as the ratio of the probability of that event to the probability of the complement of that event.
Hence if a:b are the odds in favour of an event A, the we have a : b :: P(A) : P(A’)
Now, we know that if a : b :: c : d, then a : a+b :: c : c+d.
Using, we get
a : a+b :: P(A) : P(A)+P(A’)
Now, we know that P(A) + P(A’) = 1.
Hence, we have
a : a+b :: P(A):1
Hence P(A) $=\dfrac{a}{a+b}$ .