18.05 Problem Set 1, Spring 2014
Problem 1. (10 pts.) Poker hands.
After one-pair, the next most common hands are two-pair and three-of-a-kind:
Two-pair: Two cards have one rank, two cards have another rank, and
the remaining card has a third rank. e.g. c
18.05 Problem Set 4, Spring 2014 Solutions
Problem 1. (10 pts.) (a)
P (X
x) = 1
P (X < x) = 1
Z
x
e
x
dx = 1
(1
x
e
)=e
t, X2
t) = P (X1
Thus, FT (t) = P (T t) = 1
we nd
e
2 t
.
0
(b) For t
0, we know that T
t if and only if both X1
t and X2
P (T t) = P (
Exam 1 Practice Questions II, 18.05, Spring 2014
Note: This is a set of practice problems for exam 1. The actual exam will be much
shorter.
1. A full house in poker is a hand where three cards share one rank and two cards
share another rank. How many ways
18.05 Problem Set 2, Spring 2014 Solutions
Problem 1. (10 pts.)
(a) Listing the gender of older child rst, our sample space is cfw_BB, BG, GB, GG .
The event the older child is a girl is cfw_GB, GG and the event both children are
girls is cfw_GG . Thus th
18.05 Problem Set 3, Spring 2014 Solutions
Problem 1. (10 pts.)
(a) We have P (A) = P (B) = P (C) = 1/2. Writing the outcome of die 1 rst, we can
easily list all outcomes in the following intersections.
A \ B = cfw_(1, 1), (1, 3), (1, 5), (3, 1), (3, 3),
18.05 Problem Set 2, Spring 2014
Problem 1. (10 pts.) Boy or girl paradox.
The following pair of questions appeared in a column by Martin Gardner in Scientic
American in 1959.
(a) Mr. Jones has two children. The older child is a girl. What is the probabil
18.440 Midterm 1, Spring 2011: 50 minutes, 100 points.
SOLUTIONS
1. (20 points) Consider an in nite sequence of independent tosses of a coin
that comes up heads with probability p.
(a) Let X be such that the rst heads appears on the Xth toss. In other
wor
Exam 1 Practice Questions II solutions, 18.05,
Spring 2014
Note: This is a set of practice problems for exam 1. The actual exam will be much
shorter.
1.
We build a full-house in stages and count the number of ways to make each
stage:
13
Stage 1. Choose
Exam 1 Practice Questions I, 18.05, Spring 2014
Note: This is a set of practice problems for exam 1. The actual exam will be much
shorter.
1. There are 3 arrangements of the word DAD, namely DAD, ADD, and DDA. How
many arrangements are there of the word P
18.440 Midterm 1, Fall 2009: 50 minutes, 100 points
1. Carefully and clearly show your work on each problem without writing anything that is technically not true).
2. No calculators, books, or notes may be used.
1. (20 points) Evaluate the following expli
Exam 1 Practice Questions I solutions, 18.05,
Spring 2014
Note: This is a set of practice problems for exam 1. The actual exam will be much
shorter.
1.
Sort the letters: A BB II L O P R T Y. There are 11 letters in all. We build
arrangements by starting w
18.05 Problem Set 3, Spring 2014
Problem 1. (10 pts.) Independence. Three events A, B, and C are pairwise
independent if each pair is independent. They are mutually independent if they are
pairwise independent and in addition
P (A \ B \ C) = P (A)P (B)P (
18.05 Problem Set 5, Spring 2014
Problem 1. (10 pts.) Fitting a line to data using the MLE.
Suppose you have bivariate data (x1 , y1 ), . . . , (xn , yn ). A common model is that there
is a linear relationship between x and y, so in principle the data sho
Name
18.05 Exam 1
No books or calculators. You may have one 46 notecard with any information you like on it.
6 problems, 8 pages
Use the back side of each page if you need more space.
Simplifying expressions: You dont need to simplify complicated expressi
18.05 Problem Set 1, Spring 2014 Solutions
Problem 1. (10 pts.)
answer: (reasons below)
P (two-pair) = .047539, P (three-of-a-kind) = 0.021128,
two pairs is more likely
We create each hand by a sequence of actions and use the rule of product to count
how
18.05 Problem Set 4, Spring 2014
Problem 1. (10 pts.) Time to failure. Recall that an exponential random variable
X exp( ) has mean 1 and pdf given by f (x) = e x on x 0.
(a) Compute P (X
x).
(b) Suppose that X1 and X2 are independent exponential random v
18.05 Problem Set 5, Spring 2014 Solutions
Problem 1. (10 pts.) (a) We know that yi
yi = "i + axi + b N(axi + b, 2 ). That is,
1
p
f (yi | a, b, xi , ) =
2
b = "i N(0,
axi
e
(yi
axi b)2
2 2
2
). Therefore
.
(b) (i) The y values are 8, 2, 1. The likelihood