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?
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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 $\frac{1334961}{3628800}=\frac{16481}{44800}$.
robjohn
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Thank you! Is there any other solution to this problem without derangements,or none is easier then using derangements? – Sierra Res Mar 31 '17 at 18:35
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If we try to keep track of the various cases using the method outlined above in the first paragraph, the cases become very numerous and very hard to follow. Derangements is the simplest method I've seen. – robjohn Apr 01 '17 at 14:05
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This is very famous in probability under the name: Matching Problem
The probability density function of the number of matches is: $P(N_n=k)=\dfrac{1}{k!}\sum_{j=0}^{n-k}\dfrac{(-1)^j}{j!}$, $\forall k=0,...,n$
QFi
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