A Solution Manual for:
A First Course In Probability
by Sheldon M. Ross.
John L. Weatherwax
September 19, 2012
Introduction
Here youll nd some notes that I wrote up as I worked through this excellent book. Ive
worked hard to make these notes as good as I
4.
E
Let X1, X2, X3, and X4 denote the four independent bids with common distribution
function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is
given by
G ( y ) = Pr [Y y ]
= Pr ( X 1 y ) ( X 2 y ) ( X 3 y ) ( X 4 y )
= P
9.
C
The Venn diagram below summarizes the unconditional probabilities described in the
problem.
In addition, we are told that
P[A B C]
1
x
= P [ A B C | A B] =
=
3
P [ A B]
x + 0.12
It follows that
1
1
x = ( x + 0.12 ) = x + 0.04
3
3
2
x = 0.04
3
x = 0.0
Course 1
May 2001
Answer Key
1
2
3
4
5
E
C
C
E
C
21
22
23
24
25
E
C
A
E
B
6
7
8
9
10
D
B
A
C
E
26
27
28
29
30
B
A
C
D
B
11
12
13
14
15
D
D
E
A
D
31
32
33
34
35
C
E
D
A
C
16
17
18
19
20
A
B
D
B
D
36
37
38
39
40
B
E
D
A
B
A
B
C
D
E
Course 1 Solutions
7
8
8
1.
E
We are given that
5e 5b = p5 = p6 = 6e 6b
It follows that
5 e 6 b
= 5b = e 6 b e5 b = e 6 b+5 b = e b
6 e
5
ln = b
6
5
6
b = ln = ln
6
5
2.
C
First, solve for m such that
500 = 8 + 8 (1.07 ) + . + 8 (1.07 )
m 1
1 (1.07 )m
(1.07 )m 1
= 8
= 8
1 1
2
',ii;1"_ _. i /'
Suppose that the number ofminutes required to serve a
customer at the checkout counter ofa supermarket has an
exponential distribution for which the mean is 3. Using the
central limit theorem, approximate the probability that
the tola
5.
C
The domain of X and Y is pictured below. The shaded region is the portion of the domain
over which X<0.2 .
Now observe
1 x
0.2
1 2
6
1
x
+
y
dydx
=
6
(
)
0 y xy 2 y 0 dx
0 0
0.2
0.2
1
1
2
2
2
= 6 1 x x (1 x ) (1 x ) dx = 6 (1 x ) (1 x ) dx
0
0
2
2
31.
C
A Venn diagram for this situation looks like:
We want to find w = 1 ( x + y + z )
1
1
5
We have x + y = , x + z = , y + z =
4
3
12
Adding these three equations gives
1 1 5
(x + y) + (x + z) + ( y + z) = + +
4 3 12
2(x + y + z) = 1
x+ y+ z =
1
2
1 1
COURSE 1
MAY 2001
2.
A stock pays annual dividends. The first dividend is 8 and each dividend thereafter is 7%
larger than the prior dividend.
Let m be the number of dividends paid by the stock when the cumulative amount paid
first exceeds 500 .
Calculate
1.
The price of an investment at the end of month n is modeled by pn = nebn where b is a
constant. The model predicts that the price at the end of the sixth month is the same as
the price at the end of the fifth month.
Determine b .
May 2001
(A)
5
ln
6
(
Chapter 8. Confidence interval for the mean of N (, 2 )
Example Cell phone usage
A sample of 8 cell phone users revealed the following numbers (hours of
operation): 5, 3, 10, 13, 6, 1, 9, 7
Find a 95% confidence interval for the average hours of operation
The multiplication Principle
If one experiment has m outcomes and another experiment has n outcome,
then there are mn possible outcomes for the two experiments.
()1 . . . bn
(11
aibj
am
In general if there are p experiments and the rst one has 711 possibl
MVUE and sufcient Statistics
f (3L6) (Model based)
Let 6 be an unbiased estimator of 6,
19(6) : 6, Var(6) : E(6 6)2
It is desirable to nd an unbiased estimator of 6 with smallest variance.
(MVUE)
Question: How do we nd an unbiased estimator of a parameter
4109 Fall 2012
In this class we reView
1 Unbiasedness and introduce
2 Convergence in probability
3 Convergence in distribution
4 the Central Limit Theorem Example (an unbiased estimator of 6) X; N U (U, 6), 9 unknown (want
to estimate), then the p.d.f. of
Matrix Differential Calculus Cheat Sheet
Blue Note 142 (started on 27-08-2013)
Stefan Harmeling
compiled on 30-8-2013 15:45
Rules for the differentials1
Let , a, A be constants and , , u, v, x, f , U , V , F be functions.
d = 0
da = 0n
d() = d
d(u) = du
d
Matrix Calculus:
Derivation and Simple Application
HU, Pili
March 30, 2012
Abstract
Matrix Calculus[3] is a very useful tool in many engineering problems. Basic rules of matrix calculus are nothing more than ordinary
calculus rules covered in undergraduat
1.
The price of an investment at the end of month n is modeled by pn = nebn where b is a
constant. The model predicts that the price at the end of the sixth month is the same as
the price at the end of the fifth month.
Determine b .
May 2001
(A)
5
ln
6
(
15.
D
At the point (0, 5),
0 = 2t 2 + t 1 = ( 2t 1)(t + 1)
5 = t 2 3t + 1
1
The first equation says t = or t = 1 and the second says t = 1 .
2
The slope of the tangent line to C at (0, 5) is then
dy
2t 3
dy
dx
=
=
dx ( x , y )=(0,5) dt t =1 dt t =1 4t + 1
24.
E
The domain of s and t is pictured below.
Note that the shaded region is the portion of the domain of s and t over which the device
fails sometime during the first half hour. Therefore,
1/ 2 1
1 1/ 2
1
1
Pr S T = f ( s, t ) dsdt + f ( s, t ) dsdt
0
2.
Each of the graphs below contains two curves.
Identify the graph containing a curve representing a function y = f ( x ) and a curve
representing its second derivative y = f ( x ) .
(A)
(B)
(C)
(D)
(E)
May 2003
7
Course 1
1.
A survey of a groups viewing habits over the last year revealed the following
information:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
28% watched gymnastics
29% watched baseball
19% watched soccer
14% watched gymnastics and baseball
12% watched baseball and so
10.
E
Let
W = event that wife survives at least 10 years
H = event that husband survives at least 10 years
B = benefit paid
P = profit from selling policies
Then
and
Pr [ H ] = P [ H W ] + Pr H W c = 0.96 + 0.01 = 0.97
Pr [W | H ] =
Pr [W H ]
Pr W | H =
c
21.
E
The differential equation that we are given is separable. As a result, the general solution
is given by
1
Q ( N Q ) dQ = dt = t + C
where C is a constant. Now in order to calculate the integral on the lefthand side of this
equation, we first need t
19.
B
Let X1, Xn denote the life spans of the n light bulbs purchased. Since these random
variables are independent and normally distributed with mean 3 and variance 1, the
random variable S = X1 + + Xn is also normally distributed with mean
= 3n
and sta
13.
E
Let
X = number of group 1 participants that complete the study.
Y = number of group 2 participants that complete the study.
Now we are given that X and Y are independent.
Therefore,
P ( X 9 ) (Y < 9 ) ( X < 9 ) (Y 9 )
cfw_
= P ( X 9 ) (Y < 9 ) + P
11.
D
Observe that x and y follow the constraint equation
x + y = 160, 000
x = 160, 000 y where 0 y 160, 000
Now this constraint equation can be used to express policy sales g(x, y) as a function f(y)
of marketing y alone:
f ( y ) = g (160, 000 y, y ) = 0
38.
D
From f , observe that
4 x + c1 for 0 < x < 10
f ( x ) = kx + c2 for 10 < x < 30
3 x + c for x > 30
3
As a result, 200 = f(50) = 3(50) + c3 = 150 + c3 implies c3 = 50
And 0 = f ( 0 ) = 4 ( 0 ) + c1 = c1 ,
Then due to the continuity requirement,
10k +
26.
B
Let
u be annual claims,
v be annual premiums,
g(u, v) be the joint density function of U and V,
f(x) be the density function of X, and
F(x) be the distribution function of X.
Then since U and V are independent,
1
1
g (u , v ) = ( e u ) e v / 2 = e
30.
B
Let
x = number of ice cream cones sold
p(x) = price of x ice cream cones
C(x) = cost of selling x ice cream cones
R(x) = revenue from selling x ice cream cones
P(x) = profit from selling x ice cream cones
We are told that p(x) satisfies the followin