36-225 INTRODUCTION TO PROBABILITY & STATISTICS I
Practice for the Final
You must show your work and/or explain your steps in order to get full credit or be considered for
partial credit.
This is a closed book/closed notes exam. You may use a two-sided sh
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 10
1) Exercise 6.26 (8 pts)
We are given that
m
1
my m1 ey /
f (y) =
for y > 0 and m > 0.
a) We know that U = Y m , or that Y = h1 (u) = U 1/m ; so dh1 /du = (1/m)u1/m1 . Thus
fU (
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 11
1) Exercise 6.76 (10 pts)
From the notes, we have the general formula
g(k) (y)
n!
k1
nk
f (y) [F (y)]
[1 F (y)]
(k 1)!(n k)!
(n + 1)
k1
nk
f (y) [F (y)]
[1 F (y)]
(k)(n k + 1)
=
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 9
1) Exercise 5.108 (10 pts)
We are given that f (y1 , y2 ) = ey1 , with 0 y2 y1 < . The region of integration is given by
The rst step is to nd E[Y1 Y2 ]:
E[Y1 Y2 ]
=
y1
dy1
0
dy2
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 8
1) Exercise 5.64 (10 pts)
We are given that Y1 Unif(0, 1) and Y2 Unif(0, 1).
P (Y1 < 2Y2 |Y1 < 3Y2 )
P (Y1 < 2Y2 Y1 < 3Y2 )
P (Y1 < 3Y2 )
P (Y1 < 2Y2 )
P (Y1 < 3Y2 )
=
=
Because
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 7
1) Exercise 4.90 (8 pts)
We are given that = E[Y ] = = 2.4.
First, we determine the probability P (Y > 5), given the exponential distribution:
5
P (Y > 5)
=
1 P (Y 5) = 1
0
1 y/
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 5
1) Exercise 3.38 (14 pts)
We are given that there are two glasses with formula A and one with formula B, so the probability
of picking B randomly is p = 1/3.
a) The probability d
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 4
1) Exercise 3.8 (9 pts)
Perhaps the easiest way to visualize the probabilities is to use a decision tree:
2
.9
=0
=0
.81 4 cells
)
S3
P(S2 P(S2 S3) + P(S2 S3)= 2(0.9)(0.1) = 0.1
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 6
1) Exercise 3.108 (10 pts)
As usual, we rewrite P (at least 1 defective) as 1 - P (no defectives). So we are looking for
the value of n that causes P (Y = 0) to become less than
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 3
1) Exercise 2.72 (7 pts)
a) The events A and M are independent i
P (A) = P (A|M ) =
P (A M )
P (M )
The table gives us that
P (A M ) = 0.24
P (A M ) = 0.16
P (A M ) = 0.36
M ) =
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 2
1) Exercise 2.28 (10 pts)
a) Lets label the three non-minority applicants as N1 , N2 , and N3 , and the minority applicant
as M . Noting that the order of selection to the positi
36-225 Introduction to Probability Theory Fall 2015
Solutions for Homework 1
1) (12 pts)
a) One can construct the histogram in a manner similar to that shown in the rst set of lecture
notes:
> d = scan("roeder_1990.dat")
Read 82 items
> hist(d,prob=T,xlab
36-225 — Introduction to Probability Theory — Fall 2015
Final
Name:
I have neither given nor received assistance on this exam.
Signature:
This exam is closed book and closed notes, except for cheat sheets that the instructor will provide. Calcu—
lators ar
36—225 — Introduction to Probability Theory — Fall 2015
Final
Name:
I have neither given nor received assistance on this exam.
Signature:
This exam is closed book and closed notes, except for cheat sheets that the instructor will provide. Calcu-
lators ar
Continuous Probability Distributions
Sample Statistics
Sample
Sample
n
Mean: y = n i=1 yi
1
n
1
Variance: s2 = n1 i=1 (yi
1
Variance (shortcut): s2 = n1
y
d
pdf: f (y) = dy F (y) cdf: F (y) = f (z)dz
f (y) 0; f (y)dy = 1; P (Y = y) = 0
y )2
n
2
b
( i=1
36-225 Introduction to Probability Theory Fall 2015
List of Problem Topics for Class Final
The nal exam will be held on Monday, December 14th, from 5:30 8:30 PM. It is worth 20% of your
nal grade. All ten problems on the exam will be drawn from the list o
Variance and Expected Value Review
Ex.
Consider that X assumes values in cfw_a, a. and pX(a) = pX(a) = 0.5. Find E[X] and V ar[X].
What is the behavior of V ar[X] as a funcion of a?
E[X] = a px(a) + a px(a)
= a 0.5 + a 0.5 = 0
V ar[X] = E[(X E[X])^2] = E[
Covariance and the Vector Space of Random Variables
-Let X and Y be two discrete random variables.
Cov[X, Y ] = E[(X E[X])(Y E[Y ])]
-Cov[X, Y ] = E[XY ] E[X]E[Y ].
Proof:
Cov[X, Y ] = E[(X E[X])(Y E[Y ])]
= E[XY XE[Y ] Y E[X] + E[X]E[Y ]
= E[XY ] E[Y ]E[
36-225 INTRODUCTION TO PROBABILITY & STATISTICS I Practice Exam
You must show your work in order to get full or be considered for partial credit.
You may use a one-sided sheet of notes (8.5 by 11 inches) and a calculator. You may not share a
calculator, p
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Test 3 (Version A)
1) (8 points) Because the function U = Y 3 is strictly increasing over the range 0 y 1, we use
the method of transformations:
fU (u) = fY (h1 (u)
u = h(y ) = y 3 , so h 1
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Test 1 (Version B)
1) (8 points) First note that order does not matter, so the answer will be in terms of combinations,
9
not permutations. There are C4 ways to choose three for one subcomm
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Test 1 (Version A)
1) (8 points) First note that order does not matter, so the answer will be in terms of combinations,
7
not permutations. There are C3 ways to choose three for one subcomm
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Test 3 (Version B)
1) (8 points) Because the function U = Y 3 is strictly increasing over the range 0 y 1, we use
the method of transformations:
fU (u) = fY (h1 (u)
u = h(y ) = y 3 , so h 1
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Test 2 (Version B)
1) (6 points) The key word here is approximate, as that means you are to use the Poisson
distribution to approximate this binomial probability.
We have that n = 1000 and
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Test 2 (Version A)
1) (6 points) The key word here is approximate, as that means you are to use the Poisson
distribution to approximate this binomial probability.
We have that n = 1000 and
Expected Value
-Let X be a discrete random variable and E and event. The expected value of X given A is denoted
by E[X|A] and
E[X|A] = sum w X(w)P(cfw_w|A)
-Let X be a discrete random variable. The expected value of X is denoted by E[X] and is equal to
E[
Random Variable Exercises
Ex.
Consider you pick a coin with equal probability among a fair coin and a coin with probability p of
landing heads. Next, you throw the coin you picked 3 times. Let X denote the number of heads you
observe. Find the pmf of X.
L
Variance
-Let X be a random variable. E[X] is a central value around which the possible values of X are
dispersed.
- The variance of X is a measure of the concentration of the possible values of X around E[X].
-The variance of a discrete random variable X