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.
Solution
Correct Answer: Option A
\(P(P)=\frac14,P(Q)=\frac15,P(R)=\frac16,P(S)=\frac17\)
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
= \(\frac14+\frac15+\frac16+\frac17\)
= \(\frac{319}{420}\)
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