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
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
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
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
7.
B
Let us first determine k:
1=
Then
1
0
1
0
kxdxdy =
1k
1 2 1
k
kx |0 dy = dy =
0 2
0 2
2
1
k =2
1 1
1
E [ X ] = 2 x 2 dydx = 2 x 2 dx =
0
0 0
1 1
1
E [Y ] = y 2 x dxdy = ydy =
0
0 0
E [ XY ] =
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 ) + .
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 l
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
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) + (
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 t
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 o
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%
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
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)
Ques
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 (
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
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 mo
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 o
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
2
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
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)
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% watc
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 =
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
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 + +