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Question: Randomly distribute r balls in n boxes. Find the probability that the first box is empty.

I think I should make the question into 3 cases, namely, r=n, r<n, and r>n.

CASE r=n:

${r \choose 2}$ because there are two balls of r balls that will be put into the same box.

${n-1 \choose 1}$ because we choose one box that will put the two balls.

$(r-2)!$ put the rest r-2 balls into the rest n-2 boxes.

$$\frac{{r \choose 2} {n-1 \choose 1} (r-2)! }{r^{n}}$$

CASE r>n: $$(\frac{n-1}{n})^r$$

CASE r<n: $$\frac{{(n-1)Pr}}{nPr}$$

Am I on the right track? Please give me some advice, thank you!

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    $\begingroup$ If by "randomly" you mean independently and uniformly, then there's no need to break into cases: apply the definition of independence to compute the chance that all balls are in boxes 2, 3, ..., n. $\endgroup$
    – whuber
    Commented Sep 21, 2021 at 21:29
  • $\begingroup$ Could you be more specific? $\endgroup$
    – anonyx2
    Commented Sep 21, 2021 at 21:50
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    $\begingroup$ What is the probability that the first ball ends up in box 2, ..., n? What is the probability that the second ball ends up in box 2, ..., n? $\endgroup$ Commented Sep 21, 2021 at 22:07

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Just because the first box is empty doesn't mean all the other boxes aren't.

$$ P(1\text{st box empty})=1-P(1\text{st not empty}) $$

If you "give" the first box 1 ball, you are left with r-1 balls and n boxes, you can express it as: $$ x_1+x_2+\ldots+x_n=r-1 $$

The sample space similarly is: $$ x_1+x_2+\ldots+x_n=r $$

Using stars and bars method: $$ P(1\text{st box empty})=1-\dfrac{\binom{n+r-2}{r-1}}{\binom{n+r-1}r} $$

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    $\begingroup$ Isn't the answer simply $(1 - 1/n)^r$? $\endgroup$ Commented Sep 25, 2021 at 18:21

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