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1. There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

A. $\frac{6}{7}$

B. $\frac{1}{8}$

C. $\frac{3}{8}$

D. $\frac{5}{9}$

Solution:

Total cases of checking in the hotels =

4^{3}

ways.
Cases, when 3 men are checking in different hotels = 4 × 3 × 2 = 24 ways.
Required probability:

\begin{aligned} = \frac{24}{4^{3}} \\ = \frac{3}{8} \end{aligned}

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