Given, odds in favour of A is \(\frac{P(A)}{P(\bar A)}\) = \(\frac{1}{3}\)

⇒ \(\frac{P(A)}{1-P(A)} = \frac{1}{3}\)

 ⇒ 1 – P(A) = 3P(A) 

⇒ 4P(A) = 1 

⇒ P(A) = \(\frac{1}{4}\)

Odds in favour of horse B is  \(\frac{P(B)}{P(\bar B)}\) = \(\frac{1}{4}\) 

\(\frac{P(B)}{1-P(B)} = \frac{1}{4}\) 

⇒ 1 – P(B) = 4P(B) 

⇒ 5P(B) = 1 

⇒ P(B) = \(\frac{1}{5}\) 

Odds in favour of horse C is \(\frac{P(C)}{P(\bar C)}\) = \(\frac{1}{5}\) 

⇒ \(\frac{P(C)}{1-P(C)} = \frac{1}{5}\) 

⇒ 1 – P(C) = 5P(C) 

⇒ 6P(C) = 1 

⇒ P(C) = \(\frac{1}{6}\) 

Odds in favour of horse D is \( \frac{P(D)}{P(\bar D)}\) = \(\frac{1}{6}\) 

⇒ \(\frac{P(D)}{1-P(D)} = \frac{1}{6}\)  

⇒ 1 – P(D) = 6P(D) 

⇒ 7P(D) = 1 

⇒ P(D) = \(\frac{1}{7}\) 

We have to find the probability that one of the horses win the race. 

∵ only one horse can win the race ⇒ A ,B,C and D are mutually exclusive events. 

We need to find P(A∪B∪C∪D). 

∵ A ,B,C and D are mutually exclusive events. 

∴ P(A ∪ B ∪ C ∪ D) = P(A) + P(B) + P(C) + P(D)

= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\) = \(\frac{319}{420}\) 

Hence, 

probability that one of the horses win the race = \(\frac{319}{420}\)