[EMAIL PROTECTED] (Xiao Li) wrote in message news:<[EMAIL PROTECTED]>...
> This problem was on my Probability final a month ago... and to this
> day, I still haven't figured it out yet.  Here it is:
> 
> 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?
> 
> Note: 
> 
> In the following case, Sally is sitting directly in front of Mike:
> Empty Seat
> Sally
> Mike
> 
> In the following case, Fred is NOT sitting directly in front of
> George:
> Fred
> Empty Seat
> George

People can sit in the columns in the groupings (4, 4, 2), (4, 3, 3)
and so on.  If 4 people sit in a column, exactly 3 people have someone
sitting in front of them.  If 1 person sits in a column, then nobody
is in front of that person.  For 3 persons in a column there are 2
possibilities, 2 persons with somebody in front and 1 person with
somebody in front, and for 2 persons there are also 2 possibilities,
namely 1 person with somebody in front and 0 persons with somebody in
front.
Consider the grouping (3, 3, 3, 1).  The number of ways in which this
can be achieved is C(10, 3)�C(7,3)�C(4,3)�C(1,1)�4!/(3!1!)=67200. 
This value has to be found for each of the other groupings and the
total found.  Let's say that this total is T.  The probability of the
grouping (3, 3, 3, 1) is 67200/T.
There are 8 permutations for the columns with 3 people, each with
probability 1/8.  The numbers of such persons in these permutations
are 6, 5, 5, 4, 5, 4, 4, 3 respectively.
For the grouping (3, 3, 3, 1) the contribution to the expected value
is (6+5+5+4+5+4+4+3)/8�67200/T.  This has to be done for all the other
groupings and the total taken to give the expected value.
.
.
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