Question:
There are \(4\) hotels in a town. If \(3\) men check into the hotels in a day, then the probability that each checks into a different hotel is
There are \(4\) hotels in a town. If \(3\) men check into the hotels in a day, then the probability that each checks into a different hotel is
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When people independently choose among several options, total outcomes are found using powers. For “all different” cases, use permutations of available choices.
Updated On: Jun 24, 2026
- \(\frac{6}{7}\)
- \(\frac{1}{8}\)
- \(\frac{3}{8}\)
- \(\frac{5}{9}\)
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The Correct Option is C
Solution and Explanation
Step 1: Find total possible ways.
Each of the \(3\) men can check into any one of the \(4\) hotels.
Therefore, total number of possible outcomes is \[ 4^3 \] \[ =64 \]
Step 2: Find favourable outcomes.
For all three men to stay in different hotels:
- First man can choose any of the \(4\) hotels.
- Second man can choose any of the remaining \(3\) hotels.
- Third man can choose any of the remaining \(2\) hotels.
Hence, favourable outcomes are \[ 4\times 3\times 2 \] \[ =24 \]
Step 3: Compute the probability.
\[ P=\frac{\text{Favourable outcomes}}{\text{Total outcomes}} \] \[ P=\frac{24}{64} \] \[ =\frac{3}{8} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{3}{8}} \]
Each of the \(3\) men can check into any one of the \(4\) hotels.
Therefore, total number of possible outcomes is \[ 4^3 \] \[ =64 \]
Step 2: Find favourable outcomes.
For all three men to stay in different hotels:
- First man can choose any of the \(4\) hotels.
- Second man can choose any of the remaining \(3\) hotels.
- Third man can choose any of the remaining \(2\) hotels.
Hence, favourable outcomes are \[ 4\times 3\times 2 \] \[ =24 \]
Step 3: Compute the probability.
\[ P=\frac{\text{Favourable outcomes}}{\text{Total outcomes}} \] \[ P=\frac{24}{64} \] \[ =\frac{3}{8} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{3}{8}} \]
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