# randomly putting 77 balls in 999 numbered boxes

77 balls are distributed randomly to 999 numbered boxes. I want to know the probability that 7 balls total are in the boxes 1-11 in two cases: (1) each box may contain arbitrary numbers of balls and (2) each box may contain only one ball.

How do I tackle this problem?

Hints:

(1) The probability of any given ball going into boxes $1$ through $11$ is $\frac{11}{999}$. This is a binomial probability situation.

(2) Choose $77$ of the $999$ boxes to receive balls. You're interested in cases where $7$ of the $11$ boxes from $1$ to $11$ are among the chosen boxes (and so $70$ of the remaining boxes are chosen).

• Ah, thanks. So in (1) the probability that a certain ball is not in box 1-11 is (1-p) and the probability for exactly 7 balls to be in 1-11 is $\binom{n}{k} p^7 (1-p)^{70}$. Correct?
– Marc
Nov 11 '14 at 13:48
• Yes, that looks right to me. Nov 11 '14 at 13:49
• Nice. I still feel a bit uncomfortable about (2). I have m = 999 998 ... (999-77) possibilities to choose 77 of 999 boxes. For my probability, I suspect p = [11 10 ... (11-7)] * [70 69 ... (988-70)] / m. All these factors are permutations without repetition. Is this correct?
– Marc
Nov 11 '14 at 14:11
• @Marc Yes. More precisely: $p_2=\frac{{11\choose 7}{988\choose 70}}{999\choose 77} = \frac{11!}{7!4!}\frac{988!}{70!918!}\frac{77!922!}{999!}$ Nov 11 '14 at 21:32
• Hmm, but this isn't the same as what I have written. I think we need permutations (without repetition) because the order of the boxes matters. Your formula divides each of my factors by k! and therefore counts combinations not permutations, doesn't it?
– Marc
Nov 11 '14 at 22:04