1. ## Nice Probability question

6 scavengers were analysed when flying to 20 bins in a line. The analyst found that 3 scavengers went to one bin, 2 scavengers another and 1 scavengers to a third bin. He was asked to work out the probability that this distribution occurred by chance i.e, a scavenger chooses a bin at random. (If this probability is small then it is likely that they are likely to congregate.) Explain your answer.

Well I said the $\displaystyle P(of this configuration)=_{20}C_3(\frac{1}{20})^6$. And that this probability is sufficiently low enough to be random. What do people think? Help greatly appreciated!

2. ## Re: Nice Probability question

Hello, MathJack!

6 scavengers were analysed when flying to 20 bins in a line.
The analyst found that 3 scavengers went to one bin, 2 scavengers another
and 1 scavenger to a third bin. He was asked to work out the probability
that this distribution occurred by chance i.e, a scavenger chooses a bin at random.

There are: $\;20^6$ possible outcomes.

Form an ordered partition of the 20 bins: $\:{20\choose3,2,1,14} \:=\:2,\!325,\!600$

But we want an unordered partition: $\:\dfrac{2,\!325,\!600}{4!} \:=\:96,\!900$

Therefore: $\:\text{Prob} \;=\;\dfrac{96,\!900}{20^6} \;=\;0.0015140625$