In a party there are 10 people with same umbrellas,at the end of the party they get one umbrella randomly.What is the probability that none gets its own umbrella?

I was thinking on solving this in the followin way:

The probability of the first person picking a wrong umbrella is $\frac{9}{10} $, then for the second guy it will be $\frac{8}{9}$, for the third $\frac {7}{8}$ and so on,so the probability of event A to happen would be:

$$p(A) = \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdots \frac{1}{2} \cdot 1$$

Is this solution correct,does the possibility of the first person getting the second person hats affect this solution?

# [Math] Out of 10 people whats the probability that none gets his own umbrella…

combinatoricsprobability

## Best Answer

The solution you give is not correct, and your comment at the end is the key to why. The first person does indeed have a $\frac9{10}$ chance of getting the wrong umbrella. However, the second person has a $\frac11$ probablity of getting the wrong umbrella if the first person picked his umbrella. Otherwise, if the first person picked someone else's umbrella, there would be an $\frac89$ chance that the second person would pick the wrong umbrella.

The proper way to look at this kind of question is to look at Derangements. The correct probability is $\frac{1334961}{3628800}=\frac{16481}{44800}$.