Charles M. Grinstead and J. Laurie Snell:
INTRODUCTION to PROBABILITY
Published by AMS
Solutions to the exercises
SECTION 1.1
1.
As
n
increases, the
proportion
of heads gets closer to 1/2, but the
di
ff
erence
between the number
of heads and half the number of flips tends to increase (although it will occasionally be 0).
3.
(b) If one simulates a su
ffi
ciently large number of rolls, one should be able to conclude that the
gamblers were correct.
5.
The smallest
n
should be about 150.
7.
The graph of winnings for betting on a color is much smoother (i.e. has smaller fluctuations) than
the graph for betting on a number.
9.
Each time you win, you either win an amount that you have already lost or one of the original
numbers 1,2,3,4, and hence your net winning is just the sum of these four numbers. This is not a
foolproof system, since you may reach a point where you have to bet more money than you have.
If you and the bank had unlimited resources it would be foolproof.
11.
For two tosses, the probabilities that Peter wins 0 and 2 are 1/2 and 1/4, respectively. For four
tosses, the probabilities that Peter wins 0, 2, and 4 are 3/8, 1/4, and 1/16, respectively.
13.
Your simulation should result in about 25 days in a year having more than 60 percent boys in the
large hospital and about 55 days in a year having more than 60 percent boys in the small hospital.
15.
In about 25 percent of the games the player will have a streak of five.
SECTION 1.2
1.
P
(
{
a,b,c
}
) = 1
P
(
{
a
}
) = 1
/
2
P
(
{
a,b
}
) = 5
/
6
P
(
{
b
}
) = 1
/
3
P
(
{
b,c
}
) = 1
/
2
P
(
{
c
}
) = 1
/
6
P
(
{
a,c
}
) = 2
/
3
P
(
φ
) = 0
3.
(b), (d)
5.
(a) 1/2
(b) 1/4
(c) 3/8
(d) 7/8
7.
11/12
9.
3
/
4
,
1
11.
1 : 12
,
1 : 3
,
1 : 35
13.
11:4
15.
Let the sample space be:
ω
1
=
{
A,A
}
ω
4
=
{
B,A
}
ω
7
=
{
C,A
}
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
ω
2
=
{
A,B
}
ω
5
=
{
B,B
}
ω
8
=
{
C,B
}
ω
3
=
{
A,C
}
ω
6
=
{
B,C
}
ω
9
=
{
C,C
}
where the first grade is John’s and the second is Mary’s. You are given that
P
(
ω
4
) +
P
(
ω
5
) +
P
(
ω
6
) =
.
3
,
P
(
ω
2
) +
P
(
ω
5
) +
P
(
ω
8
) =
.
4
,
P
(
ω
5
) +
P
(
ω
6
) +
P
(
ω
8
) =
.
1
.
Adding the first two equations and subtracting the third, we obtain the desired probability as
P
(
ω
2
) +
P
(
ω
4
) +
P
(
ω
5
) =
.
6
.
17.
The sample space for a sequence of
m
experiments is the set of
m
tuples of
S
’s and
F
’s, where
S
represents a success and
F
a failure. The probability assigned to a sample point with
k
successes
and
m

k
failures is
1
n
k
n

1
n
m

k
.
(a) Let
k
= 0 in the above expression.
(b) If
m
=
n
log2, then
lim
n
→∞
1

1
n
m
= lim
n
→∞
1

1
n
n
log2
=
lim
n
→∞
( 1

1
n
n
log2
=
e

1
log2
=
1
2
.
(c) Probably, since 6log2
≈
4
.
159 and 36log2
≈
24
.
953.
19.
The leftside is the sum of the probabilities of all elements in one of the three sets. For the right
side, if an outcome is in all three sets its probability is added three times, then subtracted three
times, then added once, so in the final sum it is counted just once. An element that is in exactly
two sets is added twice, then subtracted once, and so it is counted correctly. Finally, an element in
exactly one set is counted only once by the right side.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 BORZOVA
 Probability, Probability theory, probability density function, Conditional expectation

Click to edit the document details