Download Statistics 510: Notes 1

Statistics 510: Notes 3
Reading: Sections 2.5, 1.1-1.2.
Notes: Homework will be posted on the web site www-statwharton.upenn.edu/~dsmall/stat510-f05 by tonight.
Reminder: I have office hours Tuesday and Thursday 4:455:45, Tuesday 1-2 and by appointment. Our TA Abishek
has office hours Monday 1-3.
I. Proposition 2.4.3 and Example:
Proposition 2.4.3: P( E  F )  P( E )  P( F )  P( E  F ) .
This is called the inclusion-exclusion identity for two
events. Proposition 2.4.4 is an inclusion-exclusion identity
for more than two events.
Example 6 from last class: Winthrop, a premed student, has
been summarily rejected by all 126 U.S. medical schools.
Desperate, he sends his transcripts and MCATs to the two
least selective campuses he can think of, the two branch
campuses ( X and Y ) of Swampwater Tech. Based on the
success his friends have had there, he estimates that his
probability of being accepted at X is 0.7, and at Y , 0.4.
He also suspects that there is a 75% chance that at least one
of his applications will be rejected. What is the probability
that at least one of the schools will accept him?
An article in the Los Angeles Times (August 24, 1987)
discusses the statistical risks of AIDS infection:
“Several studies of sexual partners of people infected
with the virus show that a single act of unprotected vaginal
intercourse has a surprisingly low risk of infecting the
uninfected partner – perhaps one in 100 to one in 1000.
For an average, consider the risk to be one in 500. If there
are 100 acts of intercourse with an infected partner, the
odds of infection increase to one in five.
Statistically, 500 acts of intercourse with one infected
partner or 100 acts with five partners lead to a 100%
probability of infection (statistically, not necessarily in
reality).”
What is wrong with the article’s use of probability?
II. Sample Spaces Having Equally Likely Outcomes
For many experiments, it is natural to assume that all
outcomes in the sample space are equally likely to occur.
That is, consider an experiment whose sample space S is a
finite set, say S  {1, 2, , N } . Then it is often natural to
assume that
P({1})  P({2})   P({N }) .
This implies that
1
P({i})  , i  1, , N
N
because P( S )  1 by Axiom 2 and
P( S )  P({1})   P({N }) by Axiom 3.
For a sample space with equally likely outcomes, for any
event E ,
number of outcomes in E
P( E ) 
(1.1)
number of outcomes in S
(this follows from Axiom 3).
In other words, if we assume that all outcomes of an
experiment are equally likely to occur, then the probability
of an event E equals the proportion of outcomes in the
sample space that are contained in E .
Example 1: A green and a red die are rolled. The dice are
“fair,” meaning that all faces have the same probability of
turning up. What is the probability that the sum of the
faces on the dice will equal 7?
Solution: There are 36 outcomes in the sample space.
There are 6 outcomes in event
E  {sum of two dice equals 7} , namely (1,6), (2,5), (3,4),
(4,3), (5,2), (6,1) where the first number corresponds to the
green die and the second number corresponds to the red
die. Thus, P( E )  6 / 36  1/ 6 by (1.1).
Example 2: Your friend at Penn has a phone number that
starts with 498. Suppose that the remaining four digits in
your friend’s phone number are equally likely to be any
sequence. What is the probability that your friend’s phone
number contains seven distinct digits.
To solve this problem using (1.1), we need to how many
possible sequences of four digits there are and how many
such sequences contain four distinct digits, that are all
distinct from 498.
We now return to Chapter 1 to develop methods for
counting.
II. Combinatorics (Section 1.2)
Combinatorics is the branch of mathematics concerned
with counting, arranging and ordering.
A fundamental principle in combinatorics is:
The basic principle of counting: Suppose that two
experiments are to be performed. Then if experiment 1 can
result in any one of m possible outcomes and if for each
outcome of experiment 1 there are n possible outcomes of
experiment 2, then together there are mn possible outcomes
of the two experiments.
Proof: The basic principle may be proved by enumerating
all the possible outcomes of the two experiments as
follows:
(1,1), (1,2), ..., (1,n)
(2,1), (2,2), ..., (2, n)
.
.
.
(m,1), (m,2), ..., (m,n)
where we say that the outcome is (i,j) if experiment 1
results in its ith possible outcome and experiment 2 results
in its jth possible outcome. Hence the set of possible
outcomes consists of m rows, each row containing n
elements, which proves the result.
Example 3: A small community consists of 10 women,
each of whom has 3 children. If one woman and one of her
children are to be chosen as mother and child of the year,
how many different choices are possible?
When there are more than two experiments to be
performed, the basic principle of counting can be
generalized as follows:
Generalized basic principle of counting: If r experiments
that are to be performed are such that the first one may
result in any of n1 possible outcomes and if for each of
these n1 possible outcomes, there are n2 possible outcomes
of the second experiment, and if for each of the possible
outcomes of the first two experiments, there are n3 possible
outcomes of the third experiment, and if ... then there is a
total of n1 * n2 * * nr possible outcomes of the r
experiments.
Example 4: How many computer login passwords are
possible if a password must consist of a letter followed by a
letter followed by a number followed by a number?
Example 5: A fast-food restaurant offers customers a
choice of eight toppings that can be added to a hamburger.
How many different hamburgers can be ordered?
Return to Example 2: Your friend at Penn has a phone
number that starts with 498. Suppose that the remaining
four digits in your friend’s phone number are equally likely
to be any sequence. What is the probability that your
friend’s phone number contains seven distinct digits.
Example 6 -- The Birthday Problem (Example 2.5i in
book):
If n people are present in a room, what is the probability
that two of them celebrate their birthday on the same day of
the year? How large need n be so that this probability is
greater than ½?
Assume that no person is born on February 29th and that a
randomly chosen person has the same chance of being born
on any of the other 365 days of the year.
Comments on the birthday problem:
1. To facilitate the computation, it was assumed tha a
person had the same chance of being born on any given day
of the year. Birth certificates, however, show that the
assumption is not entirely valid, births being somewhat
more common during the summer than during the winter.
It has been proved, though, that any such nonuniformity
only serves to increase the chance that two people share the
same birthday in a room of n people.
2. There have been 43 US presidents. Two did have the
same birthday – Harding and Polk, were both born on
November 2nd. More interestingly, Adams, Jefferson and
Monroe all died on July 4th and Fillimore and Taft both
died on March 8th.