Question

If five balls are thrown at random into n boxes, and all throws are independent, what is the probability that no box contains more than two balls?

Short answer

The probability that no box contains more than two balls is \(\frac{{{n^4} - 10{n^2} + 15n - 6}}{{{n^4}}}\)

Step-by-step solution

Given information

Here, five balls are thrown randomly into n boxes.

All thrown are independent.

Finding the probability that no box contains more than two balls

Let, \({A_i}\) be the event that box i have at least three balls. Then,

\(\begin{aligned}{}\Pr \left( {{A_i}} \right) = \sum\limits_{j = 3}^5 {\Pr \left( {Box\,\,i\,\,has\,\,exactly\,\,j\,\,balls} \right)} \\ = \frac{{\left( {\begin{aligned}{{}{}}5\\3\end{aligned}} \right){{\left( {n - 1} \right)}^2}}}{{{n^5}}} + \frac{{\left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {n - 1} \right)}}{{{n^5}}} + \frac{1}{{{n^5}}}\\ = \frac{{10{{\left( {n - 1} \right)}^2}}}{{{n^5}}} + \frac{{5\left( {n - 1} \right)}}{{{n^5}}} + \frac{1}{{{n^5}}}\\ = \frac{{10{n^2} - 20n + 10 + 5n - 5 + 1}}{{{n^5}}}\\ = \frac{{10{n^2} - 15n + 6}}{{{n^5}}}\\ = p,\,\,say\end{aligned}\)

Since there are only five balls, two boxes can't have at least three balls simultaneously. Therefore, the events\({A_i}\)are disjoint, and the probability that at least one of the events,\({A_i}\)occurs is np.

Hence, the probability that no box contains more than two balls is:

\(\begin{aligned}{}1 - np = 1 - n \times \frac{{10{n^2} - 15n + 6}}{{{n^5}}}\\ = 1 - \frac{{10{n^2} - 15n + 6}}{{{n^4}}}\\ = \frac{{{n^4} - 10{n^2} + 15n - 6}}{{{n^4}}}\end{aligned}\).

That is \(\frac{{{n^4} - 10{n^2} + 15n - 6}}{{{n^4}}}\)