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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 $\frac{\frac{5!}{2!2!1!0!} * 4 * 3}{4^5} = 0.3515625$ (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 $(\!\!\binom{4}{5}\!\!)$ - number of combinations with repetitions, but I couldn't get the proper answer.

Isn't approach with $(\!\!\binom{4}{5}\!\!)$ elementary events more correct? Apples don't seem distinct to me - that's the reason. Or, may be, we can solve this problem with sample space in which apples are not distinct?

m73
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  • I've never seen the symbol you use. What does it mean? In any event, the denominator is the number of ways to distribute $5$ distinct objects into $4$ distinct boxes. Why should it be anything else? – saulspatz Aug 05 '19 at 19:16
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    @saulspatz it is one of the ways in which people denote the multichoose function, the solution to the stars-and-bars problem of distributing $n$ balls into $k$ bins where bins are distinct and balls are identical. You would know it as $\binom{n+k-1}{k-1}$ instead. Some authors will sadly use $\left(!\binom{n}{r}!\right)$ while others might use $\left(!\binom{r}{n}!\right)$ – JMoravitz Aug 05 '19 at 19:19
  • @saulspatz "why should it be anything else?" Because if the apples are considered identical, there are the number of outcomes he refers to... but as alluded to in my answer below, those outcomes are not equally likely to occur, hence the error. This is an acceptable choice of sample space, but not if you want to use counting techniques to calculate with. – JMoravitz Aug 05 '19 at 19:20
  • @JMoravitz Then you aren't distributing apples; you're choosing a distribution of apples. It doesn't seem like a possible interpretation of the question at all. – saulspatz Aug 05 '19 at 19:23
  • y is the number of possible ways to distribute 5 apples to 4 boxes. x is the number of possible ways to put 2 apples in one box, another two apples in another box, one apple in a box and leave a fourth empty box. The probability we are looking for is x/y. – Ahmed Hossam Aug 05 '19 at 19:24
  • @saulspatz sample spaces don't need to be equiprobable to be a sample space. It is perfectly acceptable to choose a distribution of apples. It is also perfectly acceptable (in some problems) to use the sample space ${0,1,2,3,4,5}$ where the number describes the number of apples in the first box. So long as the event you are wishing to describe is a subset of the sample space, that is all that is necessary. However, as already mentioned, to use counting techniques to calculate probabilities, you should tend towards using equiprobable sample spaces. – JMoravitz Aug 05 '19 at 19:26
  • If you want to distribute five apples to four boxes, then you take the first apple and count the possible number of ways of putting this first apple in one of the four boxes. Then you repeat this process for each of the five apples. For each apple you have four possible ways, because each time you can choose one of the four boxes seperately. Using this simple counting rule https://en.m.wikipedia.org/wiki/Rule_of_product you get $y=4\cdot 4\cdot 4\cdot 4\cdot 4=4^5$ for the number of possible ways of putting five apples in four boxes. – Ahmed Hossam Aug 05 '19 at 19:33

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Keep in mind that to use $Pr(A) = \dfrac{|A|}{|S|}$ 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 $\left(\!\!\binom{4}{5}\!\!\right)$ 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.

JMoravitz
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