# Math Help - permutation problem

1. ## 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 r-1 red balls in
the first 99 positions, which may be done in
P(99,r-1) ways where P(n,k) is the permutation
function. Therefore, Pr(red ball 100th) =
P(99,r-1) / 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,r-1) 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,r-1)/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.

2. ## Red Ball Selection

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

3. 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!