# Simple combinatorics and probability theory related question 5 apples are randomly distributed to 4

Simple combinatorics and probability theory related question
5 apples are randomly distributed to 4 boxes. We need to find probability that there are 2 boxes with 2 apples, 1 box with 1 apple and 1 empty box.
I'm getting the correct answer with $\frac{\frac{5!}{2!2!1!0!}\ast 4\ast 3}{{4}^{5}}=0.3515625$ (anyway, the answer is said to be 0.35, but I think it is a matter of rounding).
But I don't understand why there are ${4}^{5}$ elementary events in total. Firstly, I thought It should be $\left(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\left(\genfrac{}{}{0}{}{4}{5}\right)\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\right)$ - number of combinations with repetitions, but I couldn't get the proper answer.
Isn't approach with $\left(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\left(\genfrac{}{}{0}{}{4}{5}\right)\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\right)$ elementary events more correct?
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Rafael Dillon
Keep in mind that to use $Pr\left(A\right)=\frac{|A|}{|S|}$ where S is the size of the sample space, that the elements in the sample space must be equally likely to occur.
It is heavily implied that the apples are being randomly distributed in a "natural" way. The most natural interpretation of how the apples are distributed would be that we take a first apple, pick a box uniformly at random, and put the apple in it. We then take the next apple and independently choose the box for the next apple uniformly at random, etc...
As such, we see that it far more likely that the first box has two apples while the remaining boxes have one apple each than it is for the first box to have all five apples.
It follows then that the sample space of size $\left(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\left(\genfrac{}{}{0}{}{4}{5}\right)\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\right)$ is not an equiprobable sample space and so should not be used if we want to use counting techniques to calculate probabilities. The sample space of size ${4}^{5}$ which corresponds to the order in which boxes were selected to receive the next apples on the other hand will be equiprobable by our assumption on how the apples are being distributed.