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In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.

Correct Answer: 319/420

Explanation:

P ( P ) = 1 4 , P ( Q ) = 1 5 , P ( R ) = 1 6 , P ( S ) = 1 7
All the events are mutually exclusive hence,


 


Required probability = P(P)+P(Q)+P(R)+P(S)


 


1 4 + 1 5 + 1 6 + 1 7 = 319 420


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