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If r distinct balls are distributed at random into N (N ≤ r) boxes, what is the probability that box 1 will receive exactly j balls ( 0 ≤ ≤ r)?

my solution is [sample space] =$ N^r $

$$P=\frac{ 1}{N^r}\binom{r}{j}$$

I know there is something wrong. can you help me ?

leonbloy
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user311726
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2 Answers2

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You have the sample space correct. We calculate how many ways there are to place the balls so that $j$ are in box $1$.

  • Let's first pick $j$ balls from the $r$. There are ${{r}\choose{j}}$ ways to do that.

  • There is only $1$ way to place these $r$ balls since they must all go in box $1$.

  • Here is the part you are missing: For the remaining $r-j$ balls, we can place them each $N-1$ ways. We have $(N-1)^{(r-j)}$ ways to replace these remaining balls.

Now we multiply it all together. We have:

$${{r}\choose{j}}(N-1)^{(r-j)} $$

The probability is:

$$\frac{{{r}\choose{j}}(N-1)^{(r-j)}}{N^r}$$

CommonerG
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  • why we are calculating (r - j)? I mean why are writing it $(N - 1)^(r-j)$ – user311726 Feb 08 '16 at 23:33
  • $r-j$ is the number of balls not in box $1$. There are $N-1$ ways to place each of these balls. So, the number of ways to place all $r-j$ of these is $(N-1)^{(r-j)}$ – CommonerG Feb 08 '16 at 23:52
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CommonerG's answer is right. Another approach is considering it as a Binomial experiment: each ball is a try, when it falls into box 1 it's a success (prob=$1/N$), you are interested in getting $j$ successes. Then

$$ P = \binom{r}{j} \left(\frac{1}{N}\right)^j \left(1-\frac{1}{N}\right)^{r-j}$$

leonbloy
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