Suppose n people throw their car keys in a hat and then each picks one key at random. SO what is the expected value of X , the number of people who gets back their own key.


Lets denote for i th person, a random variable Xi.

If that person goes with his own key then Xi=1 and Xi=0 otherwise.

Here there are N people.

So P(Xi=1)= $ \frac{1}{n}\! $

and so that P(Xi=0)= 1-(1/n)

so E[Xi]=$ 1*\frac{1}{n}\ + 0*(1-\frac{1}{n}\!) $

         =$ \frac{1}{n}\! $

Now we have X= X1+X2+X3+.....+Xn

So E[X]=E[X1]+E[X2]+E[X3]+.......+E[Xn]

      =$ n*\frac{1}{n}\! $

Back to ECE302 Fall 2008 Prof. Sanghavi

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn