Question

In a race, the odds in favor of four horses A, B, C, D are1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find the probability that one of them wins the race.
a) $\dfrac{{342}}{{420}}$
b) $\dfrac{{221}}{{417}}$
c) $\dfrac{{171}}{{420}}$
d) $\dfrac{{319}}{{420}}$

Answer 1

In this question we have to find the value of probability that one of the horse wins the race. For that, we are going to solve use the probability formula. And also we are going to calculate the probability using the given ratios that are given in a complete step-by-step solution.

Formulas used:
${\rm{P(A \, or \, B) = P(A) + P(B)}}$

Complete step by step answer:
It is given that Odds in favor of four horses ${\rm{A, B, C, D}}$ are ${\rm{1 : 3, 1 : 4, 1 : 5, 1 : 6}}$.
Here the odds are nothing but the ratio of their winning probabilities,
That is nothing but the odds in favor of horse A is $\dfrac{{P(A)}}{{P(\overline A )}} = \dfrac{1}{3}$
Also we know that $P(\overline A ) = 1 - P(A)$
Substituting in the above equation we get, $\dfrac{{P(A)}}{{P(\overline A )}} = \dfrac{{P(A)}}{{1 - P(A)}} = \dfrac{1}{3}$
Which on solving we get, $3P(A) = 1 - P(A)$
$\Rightarrow 4P(A) = 1$
Which in turn implies$P(A) = \dfrac{1}{4}$.
That is the probability of winning the horse ${\rm{A, P(A) = }}\dfrac{{\rm{1}}}{{\rm{4}}}$
The odds in favor of horse B is $\dfrac{{P(B)}}{{P(\overline B )}} = \dfrac{1}{4}$
Also we know that $P(\overline B ) = 1 - P(B)$
Substituting in the above equation we get, $\dfrac{{P(B)}}{{P(\overline B )}} = \dfrac{{P(B)}}{{1 - P(B)}} = \dfrac{1}{4}$
Which on solving we get, $4P(B) = 1 - P(B)$
$\Rightarrow 5P(B) = 1$
Which in turn implies $P(B) = \dfrac{1}{5}$.
The probability of winning the horse ${\rm{B, P}}\left( {\rm{B}} \right){\rm{ = }}\dfrac{{\rm{1}}}{{\rm{5}}}$
The odds in favor of horse C is $\dfrac{{P(C)}}{{P(\overline C )}} = \dfrac{1}{5}$
Also we know that $P(\overline C ) = 1 - P(C)$
Substituting in the above equation we get, $\dfrac{{P(C)}}{{P(\overline C )}} = \dfrac{{P(C)}}{{1 - P(C)}} = \dfrac{1}{5}$
Which on solving we get, $5P(C) = 1 - P(C)$
$\Rightarrow 6P(C) = 1$
Which in turn implies$P(C) = \dfrac{1}{6}$.
The probability of winning the horse ${\rm{C, P}}\left( {\rm{C}} \right){\rm{ = }}\dfrac{{\rm{1}}}{{\rm{6}}}$
The odds in favor of horse D is $\dfrac{{P(D)}}{{P(\overline D )}} = \dfrac{1}{6}$
Also we know that $P(\overline D ) = 1 - P(D)$
Substituting in the above equation we get, $\dfrac{{P(D)}}{{P(\overline D )}} = \dfrac{{P(D)}}{{1 - P(D)}} = \dfrac{1}{6}$
Which on solving we get, $6P(D) = 1 - P(D)$
$\Rightarrow 7P(D) = 1$
Which in turn implies $P(D) = \dfrac{1}{7}$.
The probability of winning the horse ${\rm{D, P}}\left( {\rm{D}} \right){\rm{ = }}\dfrac{{\rm{1}}}{{\rm{7}}}$
The probability that one of the horse winning the race ${\rm{ = }}\left( {{\rm{P}}\left( {\rm{A}} \right){\rm{ + P}}\left( {\rm{B}} \right){\rm{ + P}}\left( {\rm{C}} \right){\rm{ + P}}\left( {\rm{D}} \right)} \right)$
Let us now substitute the known values we get,
The probability of the one of the horse winning $= \left( {\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7}} \right)$
On solving the addition we get,
The probability of the one of the horse winning $= \dfrac{{\left( {210 + 168 + 140 + 120} \right)}}{{840}}$.

$\therefore$ The probability of the one of the horse winning $= \dfrac{{319}}{{420}}$.

Note:
The odds in favor of horse in ratio form which means the division of probability of success by the probability of failure that is $\dfrac{{P(A)}}{{P(\overline A )}}$ where the numerator is the probability of success and the denominator is the probability of failure. We should be careful with the odds of any events.