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)
$$\eqalign{ & P(P) = \frac{1}{4}, \cr & P(Q) = \frac{1}{5}, \cr & P(R) = \frac{1}{6}, \cr & P(S) = \frac{1}{7} \cr} $$
All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence,
Required probability:
$$ = P(P) + P(Q) + P(R)$$     $$ + P(S)$$
$$\eqalign{ & = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \cr & = \frac{{319}}{{420}} \cr} $$