In an experiment, I am tossing $n$ balls at 10 boxes. The balls are equally likely (1/10) to land in any of the ten boxes. I have an expression for the probability that after throwing $n$ balls, at least one box contains at least 2 balls. I want to find $n$ such that this probability is exactly 1/2.
I would like to first say I am not great at solving problem containing factorials. Instead of find the probability of a box having at least 2 balls in it, I found the complement which is the probability of each box having less than 2 balls in it.
Sample Space |$S_n$| = $10^n$
Pr(A) = 1 - Pr($\overline A$)
and |$\overline A$| = ${\frac{10!}{(10-n)!}}$
Pr($\overline A$) = ${\frac{\frac{10!}{(10-n)!}}{10^n}}$
I am trying to solve Pr(A) = 1/2
Let's consider the complement event: $B=$all boxes have one or zero balls. If you toss $n=11$ balls into ten boxes, then by the pigeonhole principle, $B$ is impossible. On the other hand, if $n=1$, then $B$ is certain. As you increase $n$, the probability of $B$ should go down monotonically. From these facts, we know that the proability $P(B)=1/2$ occurs somewhere between $n=1$ and $n=11$.
Let's consider general $n$: we toss $n$ balls into ten boxes. $B$ occurs just when every ball goes into a different box. Disregarding the order in which we toss the balls, there are $10 \choose n$ ways to put $n$ indistinguishable balls into ten boxes that way. There are $10^n$ total ways to put balls into boxes, or $10^n/n!$ since again we're disregarding the order. Thus the probability of $B$ is:
$$P(B) = \frac{{10 \choose n}}{10^n/n!} = \frac{10!}{n! (10-n)!}\frac{n!}{10^n} = \frac{10!}{(10-n)! \;10^n}$$
The pattern is this:
$$P(B) = \frac{10}{10}\times \frac{9}{10} \times \frac{8}{10} \times \cdots \times \frac{10-[n-1]}{10}$$
which makes it easy to guess and check the answer, because $P(B_{n+1}) = P(B_{n}) \times \frac{10-n}{10}$.
If we plug in $n=4$, we find that
$$P(B) = \frac{10!}{6!} \frac{1}{10^4} = \frac{10\cdot 9\cdot 8\cdot 7}{10\cdot 10\cdot 10\cdot 10} = \frac{9\cdot 7}{5\cdot 5\cdot 5} = \frac{63}{125} = 0.504$$ which is an exact answer. Then $P(A) = 0.496$.