# Randomly distribute r balls in n boxes. Find the probability that the rst box is empty [closed]

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!

• 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.
– whuber
Commented Sep 21, 2021 at 21:29
• Could you be more specific? Commented Sep 21, 2021 at 21:50
• 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? Commented Sep 21, 2021 at 22:07

$$P(1\text{st box empty})=1-P(1\text{st not empty})$$
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: $$P(1\text{st box empty})=1-\dfrac{\binom{n+r-2}{r-1}}{\binom{n+r-1}r}$$
• Isn't the answer simply $(1 - 1/n)^r$? Commented Sep 25, 2021 at 18:21