Conference rusure::math

     Choose one of the twenty.  With probability 7/20 it is a
     boy, and then there is probability 13/19 that choosing the
     next person results in selecting a girl.  With probability
     13/20 the first person chosen is a girl, and then there is
     probability 7/19 that the second person chosen is a boy. 
     Thus there is a probability of
     
          (7/20)(13/19) + (13/20)(7/19) = 91/190
     
     that choosing two of the people (without replacement) at
     random results in a boy and girl standing next to each
     other.  Since there are 19 person-person spots in the row
     of twenty people, one expects
     
          (91/190)(19) = 91/10 = 9.1
     
     boy-girl or vice versa links.  So the answer to select is
     answer (A), closest to 9.
     
     Dan                      

	I was surprised no one challenged .1.  The motivating
	factors behind it were:

		lay out all of the possible rows end to end,
		to get a huge string of B's and G's, and it
		might look like a random sequence with the
		probabilities given in .1

		the answer computed this way matches the answer
		given from a computer assisted count. :-)

	Dan

    Re .1:
    
    I thought about it.  Clearly the probabilities of what the neighbors
    will be are not independent.  But it is also intuitive that the
    interdependence of the probabilities will not change the mean.
    
    
    				-- edp

    Given that this was only 1 of 30 problems that had to be solved
    within a mere 90 minutes, I figured that it had to have a much more
    simplistic approach.  After all, the kids wouldn't have access to
    a computer, and I'm not sure about how much deep thought and
    calculation time would be available to work out all of the
    probabilities.  Anyway, try this out for size:
    
    Estimating the "average" value for S could be a matter of knowing
    what the MINIMUM, MAXIMUM values for S are, and whether there is
    any reason to assume that all the integer values in between are
    also possible.  The average, then, would involve adding up all the
    different choices and dividing by the number of choices.  Simple
    arithmetic, right?
    
    Here goes:
              
    
     (minimum S) + ... all different values between ... + (maximum value of S)  
     -------------------------------------------------------------------------
                  number of different values S can be
    
    
    Minimum value of S = 1    (Like:  BBBBBBB GGGGGGGGGGGGG)
                                             ^
    Maximum value of S = 15   (Like:  G BG BG BG BG BG BG BG GGGGG)
                                       ^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ 
                               It doesn't seem to matter where the extra
                               girls stand.  It wouldn't make S any
                               bigger.
    Number of different
    values of S = 15          (Every combo in between minimum and maximum)
    
    (1+2+3+4+...+15)/15 = 120/15 = 8     (It even makes students use
                                         one of the "classic" formulas
                                         they learn in high school,
                                         that of adding up a series
                                         of consecutive integers.
    
    Answer:  8 is closest to 9 or choice (A).  And it takes only about
    30 seconds to get to the answer by this method.
    
    Michele
    
    That's very simple arithmetic and yields the "same" answer for the
    test.

    Let #(s,n) be the number of arrangements of n boys & girls in a line
    that have exactly s places where a boy and a girl are adjacent.
    
    We have:
    
    	#(0,1) = 2,
    	#(s+1,n+1) = #(s,n) + #(s+1,n),
    	#(s,n) = 0 if s >= n.
    
    So we have:
    
          |           s
    #(s,n)| 0   1   2   3   4   5
    ------+----------------------
        1 | 2
        2 | 2   2
        3 | 2   4   2
     n  4 | 2   6   6   2
        5 | 2   8  12   8   2
        6 | 2  10  20  20  10   2
    ...
    
    But this is just a table of binomial coefficients, times two.
    
    	#(s,n) = 2 C(n-1,s).
    
    To get the average ...
    
    	               n-1
                     2 sum s * C(n-1,s)
                       s=0
    	average(n) = ------------------
    	                      n
    	                     2

     re .5
     
     Using a recurrence to solve the problem is a good idea, but
     how does dividing an integer by 2^n get to the result of .0,
     which is 9.1?  You need at least one factor of five in the
     denominator to reproduce that result.
     
     Dan

Following are two solutions from the Solutions Pamphlet.

My opinion is that the solution in .1 is much clearer then either of these, 
although their second solution parallels it.  In both I think a statement needs
to be made that the sum of expectations of random variables equals the
expectation of the sum of the random variables even when the random variables
aren't independently distributed.  Perhaps, for high school folk you might even
include a proof (almost any probability or statistics book would have one).

Anyway here are the solutions:


Notation:	x[i] = x
			i

		C(n,r) = number of combinations of n things r at a time

Suppose that John and Carol are two of the people.  For i=1,2,...,19, let J[i]
and C[i] be the numbers of orderings (out of all 20!) in which the i'th and
(i+1)'st persons are John and Carol, or Carol and John, respectively.  Then
J[i] = C[i] = 18! is the number of orderings of the remaining persons.

For i = 1,2,...,19 let N[i] be the number of times a boy-girl or girl-boy pair
occupies positions i and i + 1.  Since there are 7 boys and 13 girls, 
N[i] = 7*13*(J[i] + C[i]).  Thus the average value of S is

	N[1]+N[2]+...+N[19]   19{7*13*(18!+18!)}   91
        ------------------- = ------------------ = --
	        20!                  20!           10

				OR

In general, suppose there are k boys and n-k girls.  For i=1,2,...,n-1 let A[i]
be the probability that there is a boy-girl pair in positions (i,i+1) in the
line.  Since there is either 0 or 1 pair in (i,i+1), A[i] is also the expected
number of pairs in these positions.  By symmetry, all A[i]'s are the same (or
note that the argument below is independent of i).  Thus the answer is
(n-1)A[i].  We may consider the boys indistinguishable and likewise the girls. 
(Why?)  Then an order is just a sequence of k B's and n-k G's.  To have a pair
at (i,i+1) we must have BG or GB in those positions, and the remaining n-2
positions must have k-1 boys and n-k-1 girls.  Thus there are 2*C(n-2,k-1)
sequences with a pair at (i,i+1).  Since there are C(n,k) sequences,

			     (n-1)*2*C(n-2,k-1)   2*k*(n-k)
	answer = (n-1)A[i] = ------------------ = ---------
				   C(n,k)	      n

In our case, the answer is (2*7*13)/20 = 91/10.

re .6, .5:

	Yes, .5 doesn't help solve the problem in .0.  Note .5 considers
	the average number of 'couples' over all 2^20 possible boy/girl
	combinations, while note .0 is specifically interested in 7 boys
	and 13 girls.