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.

$P\left(P\right)=\frac{1}{4},P\left(Q\right)=\frac{1}{5},P\left(R\right)=\frac{1}{6},P\left(S\right)=\frac{1}{7}$
All the events are mutually exclusive hence,

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

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