Math 5010-1, Spring 2005
Assignment 3
Problems
#2, p. 104. First note that P (first = 6 , sum = i) = P (first = 6 , second = 6 − i). By independence, this is (1/36) if i = 7, . . . , 12, and zero otherwise. On the other hand, by the
Bayes formula,
P {sum = i} =
i−1
X
P ( sum = i | first = j) P {first = j} =
j=1
i−1
.
36
I have used the fact that P (sum = i | first = j) = P (second = i − j | first = j) =
P (second = i − j) = (1/6), by independence. Divide the two results to find that
P ( first = 6 | sum = i) =
1
.
i−1
What does this mean? Is this a surprise?
#7, p. 104. The problem, stated more carefully, asks the following: We have 2 children, all sexassignments are equally likely. Given that at least one of the children is a boy, find
the conditional probability that the other child is a girl. This conditional probability
is, by definition, the ratio of P {a boy and a girl} to P {at least one boy}. But
P {a boy and a girl} = P (b1 g2 ) + P (g1 b2 ) =
P {at least one boy} = 1 − P (g1 g2 ) =
1
,
2
whereas
3
.
4
Therefore, the probability that the other child is the King’s sister is (1/2) ÷ (3/4) =
(2/3).
#11, p. 104. Let E denote {ecotopic pregnancy}, and S denote {smoker}. We are told that
P (E | S) = 2P (E | S c ),
P (S) = 0.32.
We want P (S | E). First of all, notice that
P (E) = P (E | S)P (S) + P (E | S c )P (S c ) = 1.68P (E | S).
Now,
P (S | E) = P (E | S)
P (S)
0.32
0.32
= P (E | S)
=
,
P (E)
P (E)
1.68
which is approximately 0.19.
1
#26, p. 106.
(a) Relabeling the cards does not affect the probabilities. Therefore, this is the same
as asking for the conditional probability of getting an ace of spades given that we
have dealt an ace on the first card. But
P ( A♠ second | A first) =
(3/52) · (1/51)
3
P (A first and A♠ second)
=
=
.
P (A first)
4/52
204
(b) This one is easier: P (2♣ second | A first) =
1
51 .
Theoretical Exercises
#20, p. 118. We are told that P0 = 1. Now,
Pn = P {dry n days later}
= P ( dry n days later | dry n − 1 days later) Pn−1
+ P ( dry n days later | wet n − 1 days later) (1 − Pn−1 )
= pPn−1 + (1 − p)(1 − Pn−1 )
= (2p − 1)Pn−1 + (1 − p).
Subtract
1
2
from both sides, and then multiply both sides by 2, to find that
1
1
1
Pn − = (2p − 1)Pn−1 + 1 − p −
= (2p − 1) Pn−1 −
.
2
2
2
Iterate to find that
1
1
1
n
Pn − = (2p − 1) P0 −
= (2p − 1)n .
2
2
2
That is, Pn =
1
2
+ 12 (2p − 1)n , as asserted.
#28, p. 119. Let En denote the event that, in the first n flips, there are r heads and (n−r) tails. Let
Ci denote the event that the ith coin was selected. We know that P (Ci ) = 1/(k + 1),
and
n−r
r i
n
i
1−
.
P (En | Ci ) =
k
k
r
[Binomial probabilities.] Let Hi denote the event that, in the ith flip, we got heads.
Then
k
X
P (Hn+1 | En ) =
P (Hn+1 ∩ Ci | En )
i=0
=
k
X
P (Hn+1 | Ci ∩ En )P (Ci | En )
i=0
=
k X
i
k
i=0
2
P (Ci | En ),
since the (n + 1)st toss is independent of the first n tosses. Now,
P (Ci )
P (En )
r n−r
n
i
i
1/(k + 1)
=
1−
.
r
k
k
P (En )
P (Ci | En ) = P (En | Ci )
By the Bayes’s formula,
P (En ) =
k
X
P (En | Cj )P (Cj )
j=0
n−r
k r j
j
1 X n
=
1−
.
k + 1 j=0 r
k
k
Therefore, combine terms to find that
i n−r
i r+1
1
−
i=0 k
k
Pk
j r
j n−r
1
−
j=0 k
k
R 1 r+1
n−r
x (1 − x)
dx
0
,
R1
xr (1 − x)n−r dx
0
Pk
P (Hn+1 | En ) =
→
as k → ∞. Suppose we could prove that
Z
C(n, m) :=
1
xn (1 − x)m dx =
0
n!m!
.
(n + m + 1)!
(1)
Then we have shown that
(r + 1)!(n − r)!/(n + 2)!
C(r + 1, n − r)
=
C(r, n − r)
r!(n − r)!/(n + 1)!
r+1
=
,
n+2
lim P (Hn+1 | En ) =
k→∞
which is (essentially) what we are asked to verify. It remains to prove (1).
R1
Obviously, C(0, m) = 0 (1 − x)m dx = 1/(1 + m). Now integrate by parts (u =
xn , dv = (1 − x)m dx) to find that du = nxn−1 dx, v = −(1 − x)m+1 /(m + 1), and
x=1 Z 1
C(n, m) =
u dv = uv −
v du
x=0
0
0
Z 1
n
n
=
xn−1 (1 − x)m+1 dx =
C(n − 1, m + 1).
m+1 0
m+1
Z
1
3
Therefore,
n
n
n−1
C(n − 1, m + 1) =
·
C(n − 2, m + 2)
m+1
m+1 m+2
n
n−1
n−j
= ··· =
·
···
C(n − j − 1, m + j + 1)
m+1 m+2
m+1+j
n
n−1
1
= ··· =
·
···
C(0, m + n)
m+1 m+2
m+n
n
n−1
1
=
·
···
m+1 m+2
m+n+1
n!m!
=
,
(n + m + 1)!
C(n, m) =
as desired.
4