STAT346, section 002 Homework 1 solutions
1. Tenants in a large apartment complex are allowed to own dogs, cats, and rabbits, but
are not allowed to own more than two pets. Write down the sample space for the pets
owned by the tenants in a randomly select
STAT346, section 002 Homework 3 solutions
2-17-15 correction to 4b. The previous answer was correct, but the notation was not. Ive also
added a bit of explanation to the answer.
1. A box contains four bags with the following contents.
One bag has twelve
Midterm Exam 1
STAT 346, Spring 2011
Instructions: You cannot use any books or notes. You can use a calculator, but not a computer. You are
not allowed to communicate with anyone (verbally, in writing, or electronically), except for me, during the
exam pe
Homework 2 solutions
1. The area of the United States composed of the six states Maine, Massachusetts, New Hampshire, Vermont, Rhode Island, and Connecticut is called New England. If for a randomly
selected person from the US, the probability they were bo
1. Consider a series of contests between two teams (A and B) with the winner of the series the first team to win three games. The results can be described by the following tree diagram. The letters describe the winners of each game. The red circles indica
STAT346, fall 2015, Quiz 4 Solutions
1. Let X be a discrete random variable with distribution function
t<2
0
1
13 2 t < 4
4 4t<6
F (t) =
13
9
13 6 t < 8
1
t 8.
(a) What is the support?
cfw_2, 4, 6, 8
(b) Find the density function.
P(X = 2) = P(X 2) =
1
STAT346, fall 2015, Quiz 5 solutions
1. For each joint distribution function below, indicate if X and Y are independent or not independent, and show and/or explain how you know.
(a) f (x, y) = xex(y+1) ,
0 x < ,
0y<
X and Y are not independent because the
STAT346 2015 Midterm solutions
PLEASE LET ME KNOW IF YOU FIND ANY MISTAKES/TYPOS IN THIS.
Correction made to questions 2(c) 6:30pm 04/06/15
1. For each of the following inequalities, put a check in the box if it indicates that X and
Y are not independent
Fall 2016
Sample Midterm Exam II Solutions STAT346 Section 003
Name:
G #:
Answer all questions. Be precise and concise. Show all work. Answers without
work will NOT receive any credit.
Question 1: Two fair dice are rolled and the absolute value of the dif
STAT346, spring 2015, Quiz 3 solutions
1. P(X = x|Y = y, Z = z) =
P(Y = y|X = x, Z = z)C
P(Y = y|Z = z)
C = P(X = x|Z = z)
This is just Bayes rule. Without the conditioning on Z = z we have
P(X = x|Y = y) =
P(Y = y|X = x)P(X = x)
P(Y = y)
To use this form
STAT346, fall 2015, Quiz 1 solutions
(1) P(Ac ) = 1 P(A and B) P(A and B c )
P(Ac ) = 1 P(A B) P(A B c )
always true
P(Ac ) = 1 P(A)
= 1 P(A and B) + P(A and B c )
= 1 P(A and B) P(A and B c )
(2) P(A or B) = P(A and B) + P(A and B c ) + P (Ac and B)
P(A
STAT346, fall 2015, Quiz 2 solutions
1. Let A and B be events. You may use either logic or set theory notation to answer either (a)
or (b).
(a) Write P(A or B) as a function of P(B) and P(A and B c ).
(b) Write P(A B) as a function of P(B) and P(A B c ).
STAT346, section 002 Homework 10 solutions
When it exists, the moment generating function for a variable X is dened as
Mx (t) = E etX = E [exp(tX)] .
Moment generating function is commonly abbreviated as mgf. Many of the commonly used distributions have m
STAT346, section 002 Homework 11 solutions
Theorem 1
Let Y1 , Y2 , . . . , Yk be independent random variables, where for i = 1, . . . , k, Yi Binomial(ni , p).
Dene
k
W =
Yi .
i=1
Then W Binomial(n1 + n2 + + nk , p).
Theorem 2
Let Y be a normally distribu
STAT346, section 002 Homework 9 solutions
The density function for the Gamma distribution is
f (x) =
1
x1 ex/ for 0 x < ,
(a)
, > 0,
(1)
() is called the Gamma function. It is dened for all real numbers. For any a (, )
(a)
ua1 eu du,
0
where means is de
STAT346, section 002 Homework 7 solutions
1. Let X be a discrete random variable with support cfw_0, 4, 9, P(X = 0) = 1/6, and
P(X = 4) = 1/2.
(This is similar, but not identical to the question we did in class.)
(a) Draw a graph of the density function.
STAT346, section 002 Homework 6 (There are 6 questions, which start on page 3.)
Due at the beginning of class on March 26, 2015
There are probably typos in this. Please let me know if you nd any.
Any random variable with support cfw_0, 1 is called a Berno
STAT346, section 002 Homework 4 solutions
1. I have underlined the key words that indicate you are being given a conditional probability.
Suppose that 50% of the population are male, 5% of the men are color-blind, and 0.25%
of the women are color-blind. I
STAT346, section 002 Homework 5 solutions
This question is from A First Course in Probability, Ross, and is posted on blackboard.
However, I use dierent (and better) notation.
A plane is missing, and it is presumed that it is equally likely to have gone d
STAT346, section 002 Homework 8 Solutions
1. Let X be a discrete random variable with support cfw_1, . . . , n and density function
P(X = x) =
1
n
for x = 1, . . . , n,
(1)
where n is an integer. The distribution of a random variable with this density fun
Fall 2016
Midterm Exam II Solutions STAT346 Section 003
Name:
G #:
Answer all questions. Be precise and concise. Show all work. Answers without
work will NOT receive any credit.
Question 1: In the experiment of rolling a balanced die twice, let X be the m
Fall 2016
Sample Midterm Exam I Solutions STAT346 Section 003
Name:
G #:
Answer all questions. Be precise and concise. Show all work. Answers without
work will NOT receive any credit.
Question 1: Given two events A and B, verify that P (A) = P (AB)+P (AB
Fall 2016
Midterm Exam I Solutions STAT346 Section 003
Name:
G #:
Answer all questions. Be precise and concise. Show all work. Answers without
work will NOT receive any credit.
Question 1: Given two events A and B, verify that P (A B) = P (A) + P (B) P (A
Solutions for HW 1
STAT 346, Spring 2013
1)
(a) We have a sample space consisting of 52 equally-likely outcomes. Four of these outcomes belong to A,
and so P (A) = 4/52 = 1/13. It follows that for the desired probability we have
.
P (AC ) = 1 P (A) = 1 1/
Solutions for HW 2
STAT 346, Spring 2013
1) There are 9 = 84 possible subsets of three balls which are equally likely to be chosen. Of these,
3
2 3 4 = 24 of them contain one ball of each color, since the amber ball can be either of the two amber
balls, a
Solutions for HW 3
STAT 346, Spring 2013
1) Letting A be the event that at least one card is black, and E be the event that exactly one card is black,
for the desired probability we have
P (E |A) =
which is equal to
P (E A)
P (E )
=
=
P (A)
1 P (AC )
26
1
Solutions for HW 4
STAT 346, Spring 2013
1)
(a) The desired pmf is
0.3,
0.5,
pX (x) =
0.2,
0,
x = 5,
x = 1,
x = 1,
otherwise.
(b) We have that
P (X > 3) = 1 P (X 3) = 1 FX (3) = 1 0.7 = 0.3.
Alternatively, one can simply note that X has only one possibl
Solutions for HW 5
STAT 346, Spring 2013
1) We need
4 = Var(Y ) = Var(aX + b) = a2 Var(X ) = a2 /4,
which implies that a2 = 16, which means that we must have a = 4 (due to the requirement that a be
positive). We also need
10 = E (Y ) = E (aX + b) = aE (X
Solutions for HW 6
STAT 346, Spring 2012
1) Letting X be the number of times a black ball is obtain n 9 trials, X is a binomial (9, 1/3) random
variable. For parts (d) through (g), Y is a geometric random variable, and for part (h), V is a negative
binomi
Solutions for HW 7
STAT 346, Spring 2013
1)
(a) The support of X is (2, ) and so FX (x) = 0 for x 2. For x > 2,
x
FX (x) = P (X x) =
2
24t4 dt = 8t3 |x = 1 8x3 .
2
Altogether, we have
1
0,
FX (x) =
8
x3 ,
x > 2,
x 2.
(b) We have P (X > 4) = 1 P (X 4) = 1