Question

Suppose that$n$balls are randomly distributed into $N$compartments. Find the probability that $m$balls will fall into the first compartment. Assume that all $N^n$arrangements are equally likely.

Short answer

$\left(\begin{array}{c}n\\ m\end{array}\right){(N-1)}^{n-m}$

Step-by-step solution

Given Information.

Suppose that $n$balls are randomly distributed into $N$compartments.

Explanation.

$n$balls.

$N$compartments.

Each of the$N^n$outcomes is equally likely

localid="1649038659363" $ProbabilityofA=exactlymballsfallintothefirstcompartment$

The outcome space $S$is a set of $n$valued vectors where each element is from $\{1,2,\dots ,N\}$and describes in which compartment did that ball went.

As these events are equally likely -

$A\subseteq S\to P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

There are$\left(\begin{array}{l}n\\ m\end{array}\right)$choices for the balls that fall into the first compartment. And for each choice of these balls, the rest of the balls $(n-m$of them) have to be in some of the$N-1$remaining compartments.

$\text{So}\left|A\right|=\left(\begin{array}{c}n\\ m\end{array}\right)(N-1{)}^{n-m}$

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}=\frac{\left(\begin{array}{c}n\\ m\end{array}\right)(N-1{)}^{n-m}}{N^n}$