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1. In a race where 12 cars are running, the chance that car X will win is $\frac{1}{6},$ that Y will win is $\frac{1}{10}$ and that Z will win is $\frac{1}{8}$ . Assuming that a dead heat is impossible. Find the chance that one of them will win.

A. $\frac{47}{120}$

B. $\frac{1}{480}$

C. $\frac{1}{160}$

D. $\frac{1}{240}$

Solution:

Required probability = P(X) + P(Y) + P(Z) (all the events are mutually exclusive)

\begin{aligned} = \frac{1}{6} + \frac{1}{10} + \frac{1}{8} \\ = \frac{47}{120} \end{aligned}

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