>[EMAIL PROTECTED] (Xiao Li) wrote in message 
>news:<[EMAIL PROTECTED]>...
>
>> A theater has a set of seats consisting of four rows and twelve
>> columns for a total of 48 seats.  10 people walk into the theater and
>> each person takes a seat at random.  Let X be the number of people
>> with someone sitting directly in front of them.  What is the expected
>> value of X?

In article <[EMAIL PROTECTED]>,
Ray Koopman <[EMAIL PROTECTED]> wrote:

>E(X) = 12*E(Y), where Y is the number of people in any given column
>that have someone sitting directly in front of them...

Another, simpler, way is to note that E(X) = sum_i E(Y_i), where Y_i
is the number of people sitting directly in front of person i (in 
some arbitrary labelling).  Note that Y_i is either 0 or 1, so 
E(Y_i) = P(Y_i=1).  Note also that all the Y_i are the same (nothing
distinguishes one person from another a priori).  So E(X) = 10 * E(Y_1).

If person i sits in the front row, no one will be in front of them.
Otherwise, as happens with probability 3/4, the probability that someone 
will be directly in front of them is just the probability that one of the 
other 9 people will sit in the chair in front of them.  These other 9
people will occupy 9 of the 47 other chairs.  The probability that the 
chair in front is one of the occupied chairs is therefore 9/47.

The final answer is therefore E(X) = 10 * (3/4) * (9/47) = 1.43617.
This matches the answer obtained by Ray Koopman.

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Radford M. Neal                                       [EMAIL PROTECTED]
Dept. of Statistics and Dept. of Computer Science [EMAIL PROTECTED]
University of Toronto                     http://www.cs.utoronto.ca/~radford
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