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1. 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.

A. $\frac{9}{17}$

B. $\frac{114}{121}$

C. $\frac{319}{420}$

D. $\frac{27}{111}$

Solution:

Let the probability of winning the race is denoted by P(person)

\begin{aligned} P (P) = \frac{1}{4}, \\ P (Q) = \frac{1}{5}, \\ P (R) = \frac{1}{6}, \\ P (S) = \frac{1}{7} \end{aligned}

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

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