Question

5 persons entered a lift cabin in the cellar of a 7-floor building apart from cellar. If each of the independently and with equal probability can leave the cabin at any floor out of the 7 floors beginning with the first, then the probability of all the 5 persons leaving the cabin at different floors is

• A. \(\frac{360}{2401}\)
• B. \(\frac{5}{54}\)
• C. \(\frac{51}{71}\)
• D. \(\frac{5}{18}\)

Answer 1

The Correct Option is A

To find the probability that five persons exit a lift cabin at different floors, we proceed as follows:

1. Defining the Sample Space:
Let $F_i$ be the floor number where the $i$-th person exits the lift cabin, for $i=1,2,3,4,5$. Each person can exit at any floor from 1 to 7. Since each of the 5 persons independently chooses one of the 7 floors, the total number of possible outcomes is:

$ 7^5 $
Calculating $7^5$:

$ 7^5 = 7 \times 7 \times 7 \times 7 \times 7 = 16807 $

2. Defining the Favorable Outcomes:
We want the probability that all 5 persons leave the cabin at different floors, meaning $F_i \neq F_j$ for $i \neq j$. To count the number of favorable outcomes, consider the choices for each person:

- The first person can choose any of the 7 floors.
- The second person can choose any floor except the one chosen by the first person, so they have 6 choices.
- The third person can choose any floor except those chosen by the first and second persons, so they have 5 choices.
- The fourth person has 4 choices.
- The fifth person has 3 choices.

Thus, the number of ways for all 5 persons to exit at different floors is:

$ 7 \times 6 \times 5 \times 4 \times 3 $
Calculating:

$ 7 \times 6 = 42 $
$ 42 \times 5 = 210 $
$ 210 \times 4 = 840 $
$ 840 \times 3 = 2520 $
So, there are 2520 favorable outcomes.

3. Alternative Counting Method:
Alternatively, we can choose 5 floors out of the 7 for the persons to exit, which can be done in $\binom{7}{5}$ ways, and then assign each of the 5 persons to one of those floors, which can be done in $5!$ ways. Thus:

$ \binom{7}{5} \cdot 5! $
Compute $\binom{7}{5}$:

$ \binom{7}{5} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21 $
Compute $5!$:

$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $
So:

$ 21 \times 120 = 2520 $
This confirms the number of favorable outcomes is 2520.

4. Calculating the Probability:
The probability that all 5 persons leave the cabin at different floors is the number of favorable outcomes divided by the total number of possible outcomes:

$ P(\text{all 5 persons leave at different floors}) = \frac{7 \times 6 \times 5 \times 4 \times 3}{7^5} = \frac{2520}{16807} $ = \(\frac{360}{2401}\)

Final Answer:
The probability that all 5 persons exit at different floors is $ \frac{360}{2401} $.