# 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?

(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
• @Marc Yes. More precisely: $p_2=\frac{{11\choose 7}{988\choose 70}}{999\choose 77} = \frac{11!}{7!4!}\frac{988!}{70!918!}\frac{77!922!}{999!}$ – Graham Kemp Nov 11 '14 at 21:32