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I am stuck on this problem where I have been asked to find the distribution of stops for the scenario where there are $5$ levels and $4$ people.

For example, when stops $(S)$ is $= 1$, $P(S=1)$. This means that the lift stops at exactly one level. I need to find the probability that all people exit at exactly one floor. For this, there are five levels to choose where the lift stops. Hence, $P(S=1) = \frac{5}{5^4}$.

For $P(S=2)$, the lift stops at exactly two floors. This could happen when $3$ students exit at one floor and the other exits at a different floor OR, $2$ people exit at one floor and other $2$ exit at a different floor. For Case $1$, $3$ people have $5$ options, these $3$ people could be chosen $4$ ways and there are $4$ floors remaining so I have concluded that case $1$ $=$ $5 \cdot 4 \cdot 4$. For the second scenario I am not sure.

Then further, I am having issues resolving $P(S=3)$ and $P(S=4)$.

Assistance is greatly appreciated.

TShiong
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ZordXXL
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1 Answers1

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Let $m$ be the number of levels and $n$ be the number of people. The number of ways to distribute $n$ people into $k$ non-empty levels is given by $k!\left\{ {n \atop k}\right\}$, where $\left\{ {n \atop k}\right\}$ is the notation for the stirling number of the second kind. Since the number of ways we can choose the non-empty levels are $\binom{m}{k}$, then $$ P(S = k) = \frac{k!}{m^n} \left\{ {n \atop k}\right\} \binom{m}{k}.$$

For the case $m=5$ and $n=4$, we have \begin{align*} P(S = 1) &= \frac{1}{125},\\ P(S = 2) &= \frac{28}{125},\\ P(S = 3) &= \frac{72}{125},\\ P(S = 4) &= \frac{24}{125},\\ P(S = 5) &= 0,\\ \end{align*} which indeed sum up to $1$.

P.S. Dester
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