SOLUTION TO PROBLEM 17 OF CHAPTER 1
We will prove by induction on nthe following: For any choice of positive integers, r
and b, if at any given time the urn contains rred balls and bblack balls, then the
probability of drawing a red ball at the nth trial (i.e., ndraws later) equals r/(r+b).
Clearly the statement is true for n= 1. Assuming its truth for n−1, we must
establish its truth for n. To begin, for each positive integer k,let
R
k= event that a red ball is drawn at the kth trial
and
Bk= event that a black ball is drawn at the kth trial.
We have by the stratified sampling theorem,
P(Rn)=P(R
1
)P(R
n|R
1
)+P(B
1
)P(R
n|B
1
).(1)
Now, given that event R1occurs, the urn contains r+cred balls and bblack
balls after the first draw. Therefore, by the induction assumption, the probability
of drawing a red ball n−1 draws later equals (r+c)/(r+c+b); in other words,
P(Rn|R1)=(r+c)/(r+c+b) . Similarly, P(Rn|B1)=r/(r+b+c) . Substi-
tuting into (1), we conclude that
P(Rn)= r
r+b·r+c
r+c+b+b
r+b·r
r+b+c=r
r+b
,
as required.