Five men are trying to find their hats in a closet. The closet contains their hats as well as 5 additional hats. (total 10 hats) What is the probability that at least one man retrieves his own hat?
Thank you for your help.
If there are $\displaystyle n$ men and $\displaystyle n$ hats. The probability that $\displaystyle k$ men select there own hat is:
$\displaystyle \frac{(n-k)!}{n!}\times P_{n-k} \times {{n}\choose{k}}$ Where, $\displaystyle P_n = \sum^n \frac{(-1)^i}{i!}$ is the probability that no matches occur.
For your case, extend the number of hats to $\displaystyle 15,$ and let $\displaystyle k $ to be $\displaystyle 0,$ then to be $\displaystyle 1.$ Now take $\displaystyle 1$ minus the sum of your values for $\displaystyle k=0$ and $\displaystyle k=1.$
Hope this helps.
Anonymous
I derived the formula above through recursion. I didn't type it out because I'm lazy. But, I just realized if you wanna be slick about it.
$\displaystyle Pr($k men select there own hat$\displaystyle )$ ~ $\displaystyle Hypergeometric(5, k, 15)$
Booyah-Kasha .
Anonymous
I do not follow the first response. It may work.
Here is another.
Let $\displaystyle ^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, the number of permutations of N things taken k at a time.
Use the inclusion/exclusion principle.
The number of ways for at least one man to get his own hat is $\displaystyle \sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)} $
Divide that number by $\displaystyle ^{10}\mathcal{P}_5$