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.

Answer: Option C

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