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.

Let the probability of winning the race is denoted by P(person)
P(P)=1/4,P(Q)=1/5,P(R)=1/6,P(S)=1/7
All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence,
Required probability:
= P(P) + P(Q) + P(R) + P(S)
=1/4 + 1/5 + 1/6 + 1/7 = 319/420