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Question: In a given race, the odds in favor of four horses \(A,B,C\& D\)are \(1:3,1:4,1:5\) and \(1:6\)respec...

In a given race, the odds in favor of four horses A,B,C&DA,B,C\& Dare 1:3,1:4,1:51:3,1:4,1:5 and 1:61:6respectively. Assuming that a dead heat is impossible, find the chance that one of them wins the race

(A)101420 (B)101210 (C)319420 (D)111420  (A) \dfrac{{101}}{{420}} \\\ (B) \dfrac{{101}}{{210}} \\\ (C) \dfrac{{319}}{{420}} \\\ (D) \dfrac{{111}}{{420}} \\\
Explanation

Solution

We will use the information of the ratio and by the help of the ratio we will determine the value of
P(A),P(B),P(C),P(D)P(A),P(B),P(C),P(D). So, by the given relation of the ratio we will be able to determine them and in the last we can add them to get the required solution.

Complete step-by-step answer:
In the given question it is given that the odds is in favor
Let us consider our first case that is for A
The ratio for A can be represented as follows
P(A)P(A)=13\dfrac{{P(A)}}{{P(\overline {A)} }} = \dfrac{1}{3}
Since we need to find the value of P(A)P(A) we need to know the relation of P(A)P(\overline {A)} in terms of P(A)P(A) that is 1P(A)1 - P(A) . So, substituting the value in the required place so that we can simplify for the specific value
P(A)1P(A)=13\dfrac{{P(A)}}{{1 - P(A)}} = \dfrac{1}{3}
Simplifying the above for the value of P(A)P(A) we get
3P(A)=1P(A)3P(A) = 1 - P(A)
Taking P(A)P(A) on one side
4P(A)=14P(A) = 1
Hence, we get
P(A)=14P(A) = \dfrac{1}{4}
So, similarly we can simplify for P(B)P(B)
P(B)P(B)=14\dfrac{{P(B)}}{{P(\overline {B)} }} = \dfrac{1}{4}
Since we need to find the value of P(B)P(B) we need to know the relation of P(B)P(\overline {B)} in terms of P(B)P(B) that is 1P(B)1 - P(B) . So, substituting the value in the required place so that we can simplify for the specific value
P(B)1P(B)=14\dfrac{{P(B)}}{{1 - P(B)}} = \dfrac{1}{4}
Simplifying the above for the value of P(B)P(B) we get
4P(B)=1P(B)4P(B) = 1 - P(B)
Taking P(B)P(B) on one side
5P(B)=15P(B) = 1
Hence, we get
P(B)=15P(B) = \dfrac{1}{5}
So, similarly we can simplify for P(C)P(C)
P(C)P(C)=15\dfrac{{P(C)}}{{P(\overline {C)} }} = \dfrac{1}{5}
Since we need to find the value of P(C)P(C) we need to know the relation of P(C)P(\overline {C)} in terms of P(C)P(C) that is 1P(C)1 - P(C) . So, substituting the value in the required place so that we can simplify for the specific value
P(C)1P(C)=15\dfrac{{P(C)}}{{1 - P(C)}} = \dfrac{1}{5}
Simplifying the above for the value of P(C)P(C) we get
5P(C)=1P(C)5P(C) = 1 - P(C)
Taking P(C)P(C) on one side
6P(C)=16P(C) = 1
Hence, we get
P(C)=16P(C) = \dfrac{1}{6}
So, similarly we can simplify for P(D)P(D)
P(D)P(D)=16\dfrac{{P(D)}}{{P(\overline {D)} }} = \dfrac{1}{6}
Since we need to find the value of P(D)P(D) we need to know the relation of P(D)P(\overline {D)} in terms of P(D)P(D) that is 1P(D)1 - P(D) . So, substituting the value in the required place so that we can simplify for the specific value
P(D)1P(D)=16\dfrac{{P(D)}}{{1 - P(D)}} = \dfrac{1}{6}
Simplifying the above for the value of P(D)P(D) we get
6P(D)=1P(D)6P(D) = 1 - P(D)
Taking P(D)P(D) on one side
7P(D)=17P(D) = 1
Hence, we get
P(D)=17P(D) = \dfrac{1}{7}
Hence, we get
P(A)=14P(A) = \dfrac{1}{4}
P(B)=15P(B) = \dfrac{1}{5}
P(C)=16P(C) = \dfrac{1}{6}
P(D)=17P(D) = \dfrac{1}{7}
The sum of the above
P(A)+P(B)+P(C)+P(D)=14+15+16+17P(A) + P(B) + P(C) + P(D) = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7}
So, On simplifying the above we get
P(A)+P(B)+P(C)+P(D)=319420P(A) + P(B) + P(C) + P(D) = \dfrac{{319}}{{420}}

Option C is the correct answer.

Note: Probability of any event is defined as the possible outcomes by the total number of outcomes.
Probability=Possible OutcomesTotal number of Outcomes{\text{Probability}}= \dfrac{\text{Possible Outcomes}}{\text{Total number of Outcomes}}
The probability of any event lies between 0 and 1.