
permutation problem
The problem is from DeGroot 3rd edition, section 1.7,
problem 10: n=100 balls, r red balls. The balls
are chosen one at a time without replacement.
a) what is the pr that a red ball is chosen on the
first selection? b) what is the pr a red ball is
chosen on the 50th selection? c) what is the pr
a red ball is chosen on the 100th selection?
a) Pr(red ball 1st) = r/n, easy
c) Pr(red ball 100th)? There are 100!
arrangements of the n=100 balls because sampling
is w/o replacement. Of those arrangements, the
only way to guarantee a red ball is chosen last
is to arrange the rest of the r1 red balls in
the first 99 positions, which may be done in
P(99,r1) ways where P(n,k) is the permutation
function. Therefore, Pr(red ball 100th) =
P(99,r1) / 100!
b) Pr(red ball 50th)? Not so sure of this one.
I don't think I can use the above reasoning
because r might be > 50. There are a total of
100! permutations in 100 selections, 50! by the
50th selection. If r<=50 then there are P(50,r)
permutations of the red balls. To guarantee the
50th is a red ball, there are P(49,r1) ways to
arrange the red balls. If r=50, then Pr(50th
red)=1. If r>50, then Pr(50th red)=1. For r<50
Pr(50th red) should be P(49,r1)/50! But how to
combine the three cases for r to get one pr for
all r? I'm fairly sure I'm missing something
here.

Red Ball Selection
You might want to consider using a PR value from a
Hypergeometric Distribution.

I feel really dumb posting the solution becauses of all the hoops I jumped through. My professor said that the Pr(Red 1st) = Pr(Red 50th) = Pr(Red 100th) = r/100. Given no other information, the probability of a red ball on each selection is the same. Even sampling without replacement. Go figure. Trick question!