Chapter 5: Probability

1
Chapter 5 Probability
5.1 Probability Rules 5.2 The Addition Rule and
Complements 5.3 Independence and the
Multiplication Rule 5.4 Conditional Probability
and the General Multiplication Rule 5.5 Counting
Techniques
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November 4, 2008
2
Example
Question When you submit your 2007 IRS tax
return in 2008, what are your chances of having
it audited by the IRS if your income is less than
25,000? That is, what is the probability of
being audited? That is, what is the chance that
you will be audited by the government. Information
In 1997, 1.5 were audited.
2007 Tax Auditing
Income Filed Examined
lt 25K 59,211,700 1,076,945
25K-50K 27,263,000 259,794
50K-100K 17,019,200 196,582
gt 100K 4,540,800 129,320
3
How can Probability Quantify Randomness?
Question What does the word probability
mean? Possible Answer Probability is a branch of
mathematics that deals with calculating the
likelihood of a given event's occurrence, which
is expressed as a number between 1 and 0. An
event with a probability of 1 can be considered a
certainty for example, the probability of a coin
toss resulting in either "heads" or "tails" is 1,
because there are no other options, assuming the
coin lands flat. An event with a probability of
.5 can be considered to have equal odds of
occurring or not occurring for example, the
probability of a coin toss resulting in "heads"
is .5, because the toss is equally as likely to
result in "tails." An event with a probability of
0 can be considered an impossibility for
example, the probability that the coin will land
(flat) without either side facing up is 0,
because either "heads" or "tails" must be facing
up. A little paradoxical, probability theory
applies precise calculations to quantify
uncertain measures of random events. http//whatis
.techtarget.com/definition/0,,sid9_gci549076,00.ht
ml
4
Another Definition
Probability is the branch of mathematics that
studies the possible outcomes of given events
together with the outcomes' relative likelihoods
and distributions. In common usage, the word
"probability" is used to mean the chance that a
particular event (or set of events) will occur
expressed on a linear scale from 0
(impossibility) to 1 (certainty), also expressed
as a percentage between 0 and 100. The analysis
of events governed by probability is called
statistics. http//mathworld.wolfram.com/Probabil
ity.html
5
Randomness

• Randomness is often observed in the outcomes of a
  response variable in either an observational or
  experimental study.
• All the possible outcomes are known, but it is
  uncertain which outcome will occur for any given
  observation.
• Randomness is the opposite of deterministic where
  a given input doesnt always produces the same
  result.

6
Creating Random Events

• A machine or procedure that produces random
  events is called a randomizer.
• Examples of Randomizers
• Rolling dice
• Wheel of Fortune
• Flipping a coin
• Drawing a card from a shuffled deck

7
Applets
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Terminology
The process of rolling the die several times with
varying results is called a probability
experiment. Each roll of the die or dice is
called a trial or outcome or event. The number
of times that a certain event (outcome) occurs
divided by the total number so trials is called
the cumulative proportion or relative frequency
of the probability experiment. Suppose that in
rolling one die 200 times, the number 2 occurs 45
times. The cumulative proportion of this event
is 45/200 0.225.
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Events and Sample Space
Definition A simple event is an outcome from a
probability experiment that is observed on a
single repetition of the experiment. The sample
space of a probability experiment is the set of
all possible simple events from the probability
experiment. An event is a collection of simple
events in other words, it is a subset of the
sample space. An event that consists of more
than one outcome is called a compound event.
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Examples
Probability Experiment Sample Space
Coin Toss H,T
Roll a single Die 1,2,3,4,5,6
True/False Quiz Question T,F
Roll two Dices (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),,(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
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Simple and Compound Events
Probability Experiment Sample Space Simple Event Compound Event
Roll two Dice (1,1),(1,2),,(6,6) Die1 1 Die2 4 (1,4) Sum of Dice is 5 (1,4),(4,1),(2,3),(3,2)
Choose a single Card (A,H),(2,H),(3,H),, (A,S)(2,S),,(K,S) Ace of Hearts (A,H) Queen (Q,H),(Q,D),(Q,C),(Q,S)
H heart, D diamond, C club, S spade
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Examples
Probability Experiment Event Sample Space
Birth of a single child Male(m) or female(f) male m is a simple event. S m,f
Three births 2 males and 1 female is not a simple event. mmf,mfm,fmm are each simple events. S mmm,fff,ffm,mmf,mff,mfm,fmf,fmm
Note The event, 2 males and 1 female, is not
simple, because it can happen as mmf, mfm or fmm.
We would call this a compound event.
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Probability
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Remark
If the relative frequencies of events are known
in a population, then the probability of these
events are exactly the relative frequencies. For
example, if we know that a bag contains 2 red
balls, 3 green balls, 4 white balls and 1 blue
ball, then the probability of randomly selecting
a red ball from the bag is the relative frequency
of red balls i.e., 2/10 1/5 or 0.2. If the
relative frequency of an event in a population is
unknown, then it can be approximated by
simulating the event.
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Rules of Probabilities
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Probability Model
Definition A probability model is a table that
lists all possible outcomes of a probability
experiment and their probabilities. The form of
this table is shown below.
Outcome Probability



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Example (probability model)
The following table contains the probabilities of
choosing different color balls out of a bag.
Could this be a probability model for the
experiment?
Ball Color Probability
red 0.25
yellow 0.35
white 0.20
blue 0.25
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Example (probability model)
Suppose that we have a bag of jelly beans in six
colors (green, orange, blue, red, yellow and
brown). We have determined by performing a
number of trials that the probability of picking
a particular color is
Color Probability
green 0.16
orange 0.20
blue 0.24
red 0.13
yellow 0.14
brown 0.13
Note This last equation is the probability of
picking a jelly bean that is of the colors green
or orange or blue or .
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Example
Probability Experiment Gender of children when a
couple has three children. Event Exactly 2 boys
among the three children. E 2 boys in 3
births. Sample Space S fff, ffm, fmf, fmm,
mff, mfm, mmf, mmm. n 8. Event for 2 boys
fmm, mfm, mmf. m 3. Probability P(E) m/n
3/8 0.375.
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Law of Large Numbers
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Two Types of Probability

1. Classical Method Using mathematics to compute
   the exact probability of an event.
2. Empirical Method Approximate the probability of
   an event by using a probability experiment and
   using the relative frequency of the event to
   estimate the exact probability

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Example (Empirical)
Find the long term number for heads in the
probability experiment of flipping a coin. S
h,t. N 10. Trial t, h, t, h, h, t, t, t,
h, t. fh 4/10 0.4 N 50. Trial h, h, h,
h, t, t, h, t, h, t, h, t, t, h, t, t, t, t, h,
h, t, t, h, t, t, h, h, h, t, t, h, h, h, h, t,
t, t, t, h, t, t, h, h, h, t, h, h, h, t, t. fh
25/50 0.5 N 100. Trial h, h, h, t, h, t,
t, t, h, t, t, h, h, h, h, h, t, t, t, t, h, t,
t, t, t, t, h, h, t, t, h, h, t, h, h, t, h, t,
t, h, t, h, h, h, t, h, t, h, t, h, t, h, t, h,
h, t, t, h, t, t, h, h, t, h, h, t, t, t, t, h,
t, h, t, t, t, t, t, h, h, t, t, t, h, h, t, h, h,
t, t, t, h, h, t, h, h, h, t, t, h, h. fh
48/100 0.48 As N gtgt 1, fh approaches 1/2.
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Caution
Over the short run, we do not expect the
proportion of a particular event to the same as
the proportion in the long run. For example,
if I flip a coin 6 times, then I do not necessary
expect that heads will occur exactly 3 times.
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Examples (classical)
Find the probability of select a heart out from a
deck of cards each of which have equal chance of
being selected. E heart P(E) 13/52
1/4 Find the probability of selecting a red card
from a deck of cards each of which have equal
chance of being selected. E red card P(E)
26/52 1/2 Find the probability of selecting a
face card from a deck of cards each of which have
equal chance of being selected. E face
card P(E) 12/52 3/13
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Example (classical)
The West Meade Golf Shop sells used golf balls.
This past Saturday I went to the shop and was
told that the bag of used golf balls coned 35
Titlists, 25 Maxflis, and 20 Top-Flites. I was
told that I could reach into the bag and select a
ball. What is the probability that I would
select a Titleist? Solution Let E
Titleist. P(E) 35/(352520) 35/80
0.4375.
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Example (classical)

• Suppose a probability model has the following
  sample space
• S 1,2,3,4,5,6,7,8,9,10
• i.e., there are ten possible outcomes in the
  probability experiment.
• (a) Compute the probability of the event of
  selecting three numbers with the outcome of
  3,4,7
• (b) Compute the probability of the event of
  selecting one number with the outcome of it being
  an even integer.
• Answers (a) P(E) 3/10 (b) P(E) 1/2.

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Example
Example Three question quiz. Problem Find the
possible outcomes for the student taking this
three question quiz. Sample Space
CCC,CCI,CIC,CII,ICC,ICI,IIC,III
Question Chances of getting all three questions
correct? Note that is a simple event. Answer
1/8 0.125
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Question What are the chances of getting one
out of the three questions correct? Answer If A
CII,ICI,IIC, then P(A) 3/8 0.375. Note
that A is a compound event.
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Probability Estimates from Survey Data
200 Vanderbilt students were surveyed about their
main recreational habits listening to music,
watch television, playing cards, exercising,
other. The following table summarizes the
frequencies and relative frequencies in the
survey.
Recreation Frequency Relative Frequency
music 75 3/8
TV 50 1/4
card 15 3/40
exercise 35 7/40
other 25 1/8
The probability that a Vanderbilt student watches
TV for his or her main recreation is
approximately 0.25.
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Tree Diagram

• An nice way of visualizing a sample space with a
  small number of outcomes.
• As the number of possible outcomes for each trial
  increases, the tree diagram becomes impractical.

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Tree Example
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Example
Background An experimental study by the
University of Wisconsin to determine if Echinacea
is an effective treatment for the common cold.

• Medical Experiment
• Multi-center randomized experiment
• Half of the volunteers are randomly chosen to
  receive the herbal remedy and the other half will
  receive the placebo
• Clinic in Madison, Wisconsin has four volunteers
• Two men Jamal and Ken
• Two women Linda and Mary
• Probability Experiment
• Randomly pairing the four volunteers
• Sample Space to receive the herbal remedy
• (Jamal, Ken), (Jamal, Linda), (Jamal, Mary),
  (Ken, Linda), (Ken, Mary), (Linda, Mary)

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Assumption The six possible outcomes in this
sample space for receiving the Echinacea are
equally likely. Hence, the probability that any
simple event in the sample space will occur is
1/6. Hence, if S (Jamal, Ken), (Jamal,
Linda), (Jamal, Mary), (Ken, Linda), (Ken, Mary),
(Linda, Mary), then the probability of picking
Ken and Linda is 1/6. The probability of picking
one man and one women in a simple event is the
probability of picking (Jamal, Linda) or (Jamal,
Mary) or (Ken, Linda) or (Ken, Mary) i.e., 4(1/6)
2/3. The probability of picking a simple event
containing only women is (Linda, Mary) which has
the probability 1/6.
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Example
Problem Suppose we roll two dice (1 2)
once. What is the probability that the sum of
the numbers on the dice is 7? Sample Space Let
(x,y) denote the ordered pair where x is the
number form die 1 and y is the number from die
2. Then S (1,1),(1,2),(1,3),(1,4),(1,5),(1,6
),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4
,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(
5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) a
nd N 36. Event E (1,6),(2,5),(3,4),(4,3),(5,
2),(6,1) Probability P(E) 6/36 1/6.
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Problem What is the probability of rolling box
cars? Event E (6,6) (simple
event) Probability P(E) 1/36
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Example
Problem What are the chances of a taxpayer
being audited by the IRS in 2003? Solution The
problem can be solved with a contingency table
for the audits according to income level. We can
compute the relative frequencies of being audited
for each income level.
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Sample Space We define the sample space to be
ordered pairs (x,y) where x is the income range
and y is yes (audited) or no (not audited). For
x we introduce some notation lt 25K x
1 25K-49.999K x 2 50K-99.999K x 3 100K
lt x 4. S (1,yes),(1,no),(2,yes),(2,no),(3,
yes),(3,no),(4,yes),(4,no)
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Income/Audited Yes No Total
lt 25K (1) 90 14,011 14,100
25K-49.999K (2) 71 30,629 30,700
50K-99.999K (3) 69 24,631 24,700
100K (4) 80 10,620 10,700
Total 310 79,890 80,200

• Probability of being audited (any income)
  310/80200 0.004 or 0.4
• Probability of being audited and making an
  income over 100K 80/10700 0.007 or 0.7
• Probability of being audited and making an
  income less than 25K 90/14100 0.006 or 0.6.
• Probability of not being audited and making an
  income less than 25K 14010/14100 0.994 or
  99.6.

Remark In this example we did not use the
sample space, but rather a contingency table of
incomes and audits.
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Subjective Probability
The probability of an event that is obtained on
the basis of personal judgment is called
subjective probability. This type of calculation
is the opposite of objective probability (for
example, empirical probability calculations). Exa
mple What was the probability of landing a man
on the moon in the 1960s? Any estimate of this
probability would be subjective since we have no
prior history of the event.
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The Addition Rule and Complements
Definition Two events in a probability
experiment are said to be disjoint if they have
no common outcomes. Another term for the same
concept is mutually exclusive i.e., both events
cannot happen simultaneously. Suppose that A and
B are events. If they are disjoint, then the
probability and A and B happening is zero i.e.,
P(A and B) 0.
Section 5.2
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Intersection and Union of Sets
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Venn Diagrams
There is a graphical way of looking at the
intersection and union of sets. They are called
Venn Diagrams.
http//kt2.exp.sis.pitt.edu8080/venn/andor.jsp
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Picture of Intersection/Union
44
Complement of a Set
45
Sets and Events
Suppose that A is a set of events (possibly a
single event in the sample space) and B is
another set of events (again, possibly a single
event) in a common sample space. We consider A
and B to be subsets of the sample space. We can
perform the set operations intersection and
union. For example, the union of A and B is set
of events that arise in A or B. The intersection
of A and B is the set of events that is common to
both events. Similarly, we can talk about the
complement of Ac i.e., the set of events that are
not in A, but are in the sample space.
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Union/Intersection of Events

• The union of two events A and B is the new event
  consisting of events that are either in A or B.
• The intersection of two events A and B is the
  new event consisting of events that are in A and
  B.

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Sets, Events and Probabilities
Suppose that we want to calculate the probability
that a particular event in the same space will
occur. For example in our three question pop
quiz illustration, what is the probability that a
student will have two and only two correct
answers to the three questions. As we already
know, the sample space can be viewed as a set and
events are subsets of the sample space. That is,
we have the event A CCI,CIC,ICC. Notice that
the event B CII,ICI,IIC is a disjoint event
from the event A. In fact, A is disjoint from
its complement Ac CCC,CII,CIC,IIC,III.
The event, C, of getting one or two correct
answers on the quiz is given by the set C which
is the union of A and B i.e., C
CII,ICI,IIC,CCI,CIC,ICC.
We would like to determine the probability of the
event C by using information about the
probabilities of events A and B.
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Notation
49
Disjoint Events
50
Probability of A or B for Disjoint Events
51
Example
Consider a deck of 52 cards (spades, hearts,
clubs and diamonds 2,3,,10,J,Q,K,A). Consider
the problem of drawing one card from this deck.
The sample space has 52 simple events (drawing
one of the cards). We characterize the sample
space as ordered pairs (count,suit) e.g., 10 of
diamonds is (10,D).
Question What is the probability of drawing the
ace of diamonds? Answer P(A,D) 1/52 0.019 .
Question What is the probability of a
king? Answer P(king) P(K,H) P(K,D) P(K,C)
P(K,S) 4/52 1/13 0.077 Note that (K,H),
(K,D), (K,C) and (K,S) are mutually exclusive
events.
52
Question What is the probability of drawing a
heart? Answer P(heart) P(A,H) P(2,H)
P(K,H) 13/52 1/4 0.25 Note that
(A,H),(2,H),,(K,H) are mutually exclusive events.
Question What is the probability of drawing a
king or queen? Answer P(king or queen)
P(king) P(queen) P(king or queen) 4/52
4/52 8/52 2/13 0.154
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Probability of A or B for any two Events
54
Example
Consider a deck of 52 cards (spades, hearts,
clubs and diamonds 2,3,,10,J,Q,K,A). Consider
the problem of drawing one card from this deck.
The sample space has 52 simple events (drawing
one of the cards).
Question What is the probability of drawing the
ace or a diamond? Answer Let A ace and B
diamond. These are not disjoint events. P(A)
4/52 1/13, P(B) 13/52 P(A or B) P(A) P(B)
- P(A and B) P(A and B) P(ace and diamond)
1/52 Hence, P(ace or diamond) 1/13 13/52 -
1/52 (413-1)/52 16/52 4/13
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Example
Two hundred and fifty Vanderbilt students were
analyzed for the IQ and their ability to do a
certain mathematical puzzle. The results are
summarized in the following contingency table.
Puzzle/IQ Average (90-120) High (gt120 )
Couldnt do Puzzle 75 30
Could do Puzzle 20 125
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Let A high IQ. Then P(A) (30125)/250
155/250 0.62 Let B could. Then P(B)
(20125)/250 145/250 0.58 Question What is
the probability that a student has a high IQ or
could do the puzzle? Answer P(A or B) P(A)
P(B) - P(A and B) 0.62 0.58 - ? From the
table, P(A and B) 125/250 0.50 Hence, P(A
or B) 0.62 0.58 - 0.50 0.70
Puzzle/IQ Average (90-120) High (gt120 ) Total
Couldnt do Puzzle 75 30 105
Could do Puzzle 20 125 145
Total 95 155 250
57
Another Approach
Let us set the situation as ordered pairs S
(couldnt,average),(couldnt,high),(could,average
),(could high). We would like to find the
probable of choosing the ordered pair (could,
high). There are 250 order pairs. We want to
calculate the probability of choosing ordered
pairs (could,---),(---,high).
Then P(could,---) (20125)/250 145/250
0.58 P(---,high) (30125)/250 155/250
0.62 P(could,high) 125/145 0.50
Then P(could,---) or (---,high) 0.58 0.62
- 0.50 0.70
58
Example (gender/marriage)
Consider the following contingency table for
present marital status and gender of people in
the U.S. over the age of 18 in 2003.
Males Females Total
Never Married 28.6 23.3 51.9
Married 62.1 62.8 124.9
Widowed 2.7 11.3 14.0
Divorced 9.0 12.7 21.7
Total 102.4 110.1 212.5
We want to calculate the probability that a
person in the U.S. over the age of 18 has some
particular characteristic. We use the relative
frequencies from this table to compute the
probabilities.
59
Question What is the probability that a person
in this census is a female? Answer P(female)
110.1/212.5 0.518 Question What is the
probability that a person in this census is
widowed? Answer P(widowed) 14.0/212.5 0.066
Question What is the probability that a
person in this census is a widowed or
divorced? Answer P(widowed or divorced)
P(widowed) P(divorced) - P(widowed and
divorced) P(widowed or divorced) 14.0/212.5
21.7/212.5 - 0 0.168
Males Females Total
Never Married 28.6 23.3 51.9
Married 62.1 62.8 124.9
Widowed 2.7 11.3 14.0
Divorced 9.0 12.7 21.7
Total 102.4 110.1 212.5
60
Question What is the probability that a person
in this census is a married or female? Answer
P(married or female) P(married) P(female) -
P(married and female) P(married or female)
124.9/212.5 110.1/212.5 - 62.8/215.5
172.2/212.5 0.799 Question What is the
probability that a person in this census is a
male or divorced? Answer P(male or divorced)
P(male) P(divorced) - P(male and
divorced) P(male or divorced) 102.4/212.5
21.7/212.5 - 9.0/212.5 115.1/212.5 0.542
Males Females Total
Never Married 28.6 23.3 51.9
Married 62.1 62.8 124.9
Widowed 2.7 11.3 14.0
Divorced 9.0 12.7 21.7
Total 102.4 110.1 212.5
61
Complement of an Event
Example Three question quiz. S
CCC,CCI,CIC,CII,ICC,ICI,IIC,III Let E be the
event of having one correct answer E
CII,ICI,IIC. Then Ec CCC,CCI,CIC,ICC,III.
62
Probability of the Complement of an Event
Example Three question quiz. Find the
probability of not having one, and only one,
correct answer on the quiz. Let E be the event of
having one correct answer, E CII,ICI,IIC,
then Ec CCC,CCI,CIC,ICC,III. If each event
in the sample space is equally likely, then P(E)
3/8 and P(Ec) 5/8 1 - 3/8.
63
Example
The following table shows the relative
frequencies of the size of farms in the U.S.
Size in Acres (x) Relative Frequency
x lt 10 0.084
10 x lt 50 0.265
50 x lt 100 0.161
100 x lt 180 0.149
180 x lt 260 0.077
260 x lt 500 0.106
500 x lt 1000 0.076
1000 x lt 2000 0.047
2000 x 0.035

1. What is the probability that a farm in the U.S.
   will be between 100 and 500 acres?
2. What is the probability that a farm will be
   greater than or equal to 10 acres?

(1) P(100 x lt500) P(100 x lt180) P(100 x
lt180) P(180 x lt260) P(260 x lt500)
0.149 0.077 0.106 0.332 (2) P(x 10)
1 - P(x lt 10) 1 - 0.084 0.916
64
Example
Roulette consists of a wheel with 38 slots,
numbered 0,1,2,,36,00 i.e., 38 slots. The
odd-numbered slots are red and the even-numbered
slots are black. The slots, 0 00, are colored
green. The wheel is spun and a metal ball
circles the wheel until it lands in a numbered
slot. (a) What is the probability that the ball
lands on a green or red slot? (b) What is the
probability that it does not land in a green slot?

• Let A event of landing on red slot, B event
  of landing on green slot. Then P(A) 18/38
  9/19 and P(B) 2/38 1/19.
• Then P(A or B) P(A) P(B) - P(A and B) 9/19
  1/19 - 0 10/19
• (b) P(Bc) 1 - P(B) 1 - 1/19 18/19.

65
Independence and the Multiplication Rule
Definition Two events A and B are called
independent events if the fact that A occurs does
not affect the probability of B occurring and
vice-versa. When the probability of A effects the
probability of B, then we say that the events are
dependent.

• Examples Independent
• Rolling a die and getting a 4 and then rolling it
  a second time and getting a 2.
• Drawing a card from a deck and get a 10 of
  diamonds and then replacing the card, shuffling
  and then drawing a 10 of diamonds again.
• Suppose a patient at the Vanderbilt Hospital is
  selected at random from a group of patients. Let
  A be the event that the patient has
  atherosclerosis. Let B be the event that the
  patient is a smoker. A and B are not independent
  i.e., they are dependent since smoker has a high
  incidence of heart disease.

Section 5.3
66
Dependent Events

• Examples Dependent Events
• Being a lifeguard and getting a suntan.
• Parking in a no-parking zone and getting a
  parking ticket.
• Being a Vanderbilt student and getting a good
  education.

67
Remark
Disjoint events and independent events are
different concepts. Two events are disjoint if
they one event occurs, then the other one will
not happen. Two events are independent if the
probability of one event does not event the
probability of the other event. Hence, if two
events are disjoint, then they cannot be
independent
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Probability of A and B for Independent Events
Let A and B be two independent events in a sample
space. Then P(A and B) P(A)P(B). Note If A
and B are disjoint events, then P(A and B) 0.
69
Example
Suppose we flip a coin twice in a row. What is
the probability that we will see two
heads? Sample Space S h,t. Note that the
events are independent. Let A h and B h.
Then P(A and B) P(A)P(B) (0.5)(0.5) 0.25
. Question Is this the same as asking the
probability of the event given the first flip is
head, then the second flip is also a head?
70
Example
Suppose we have a two question pop quiz. We
perform a probability experiment and find the
following data about the outcome of taking the
quiz.
Outcomes II IC CI CC
Probabilities 0.26 0.11 0.05 0.58
Suppose that we consider the events (answering
the questions) are not independent. For example,
if you answer the first question correct, the
probability of answer the second question correct
is not necessarily the same as if you had
answered the first question incorrectly. Let A
event that the first question is answered
correctly, irregardless of the answer to the
second question. Let B event that the second
question is answered correctly, irregardless of
the answer to the first question.
71
Outcomes II IC CI CC
Probabilities 0.26 0.11 0.05 0.58
P(A) P(CI or CC) P(CI) P(CC) 0.05 0.58
0.63 P(B) P(IC or CC) P(IC) P(CC) 0.11
0.58 0.69 P(A and B) P(CC) 0.58 If A
and B were independent, then P(A and B) P(A)
P(B) (0.63)(0.69) 0.43 Notice that we
computed different values for P(A and B),
depending on the whether we assume independence
or not of the two events.
72
Example

• The E.P.T. Pregnancy Test states that the test is
  99 accurate in detecting typical pregnancy
  hormone levels. Suppose that we randomly select
  12 pregnant women.
• What is the probability that all 12 of them will
  test positively?
• (b) What is the probability that at least one
  will not test positively?

• Let A event that one will test positively.
  Hence, P(A) 0.99
• (a) P(all test positively) P(A and A and and
  A) P(A)12 (0.99)12 0.886385
• (b) P(at least one test negatively) 1 - P(all
  test positively) 0.113615

73
Caution
Dont assume that events are independent unless
you have given this assumption careful thought
and it seems plausible.
74
Summary
75
Conditional Probability and the General
Multiplication Rule
Suppose that we have two events, A and B. (a)
P(A or B) P(A) P(B) - P(A and B) (b) If they
are independent events, then P(A and B)
P(A)P(B). Recall that two events are independent
if the the probability of A is independent of B
and vice-versa. Question What happens if the
events are not independent i.e., they are
dependent. That is, the probability of one
dependents on the probability of the
other. Definition The symbol P(AB) means the
probability of an event A given that an event B
has occurred. This probability is called a
conditional probability.
Section 5.4
76
Conditional Events
P(EF) the probability of E given F In other
words, the probability of E, given the F has
already happened.
77
Condition Probability Formula
78
Example
Twenty-five percent of Vanderbilt professors in
the age range 50-60 years old have hypertension.
In this same group, five percent also have
diabetes. Given that an individual in this age
group has hypertension, what is the probability
that he or she will also have diabetes?
79
Example
Here we consider two dependent events with the
events being income level and the other being
audited.
Contingency Table for IRS filers.
Probabilities of audit for different income
classes P(yes and lt25K) 90/80200
0.0011221 P(no and lt25K) 14011/80200
0.1747007 P(yes and 25K-50K) 71/80200
0.0008862 P(yes) 310/80200 0.0038653 (overall)
Income/Audited Yes No Total
lt 25K 90 14,011 14,100
25K-49.999K 71 30,629 30,700
50K-100K 69 24,631 24,700
100K lt 80 10,620 10,700
Total 310 79,890 80,200
Probability Table for IRS filers.
S (lt25K,yes),(lt25K,no),(25-50K,yes),
(25-50K,no),(50-100K,yes),(50-100K,no),(100Klt,
yes),(100Klt,no). P(lt25K,yes) 0.0011
P(100Klt,no) 0.1324 The sum of events in the
sample space 0.00110.17470.00090.38190.00090
.30710.0010.1324 1.0
Income/Audited Yes No Total
lt 25K 0.0011221 0.1747007 0.1758104
25K-49.999K 0.0008852 0.3819077 0.3827930
50K-100K 0.0008603 0.3071119 0.3079800
100K lt 0.0009975 0.1324189 0.1334164
Total 0.0038653 0.9961346 1.000
80
Question Let A be the event audityes and let
B be the event incomegt100K. Find P(AB) i.e.,
the probability that an individual will be
audited, given that his or her income is greater
than 100K. Answer We note that P(A and B)
P(yes,100Klt) 0.0009975. Furthermore, the
probability that taxpayer has an income greater
than 100K is P(100Klt) 0.1344164. Therefore,
P(yes 100Klt) P(yes,100Klt) /P(100Klt)
0.0009975/0.1334164 0.0074766
Probability Table for IRS filers (from previous
page)
Income/Audited Yes No Total
lt 25K 0.0011221 0.1747007 0.1758104
25K-49.999K 0.0008852 0.3819077 0.3827930
50K-100K 0.0008603 0.3071119 0.3079800
100K lt 0.0009975 0.1324189 0.1334164
Total 0.0038653 0.9961346 1.0
81
Probability Table for IRS filers (from previous
page)
Income/Audited Yes No Total
lt 25K 0.0011221 0.1747007 0.1758104
25K-49.999K 0.0008852 0.3819077 0.3827930
50K-100K 0.0008603 0.3071119 0.3079800
100K lt 0.0009975 0.1324189 0.1334164
Total 0.0039 0.9961 1.0
With some rounding, we can form a conditional
probability table.
Income/Audited Yes No Total
lt 25K 0.006 0.994 1.000
25K-49.999K 0.002 0.998 1.000
50K-100K 0.003 0.997 1.000
100K lt 0.007 0.993 1.000
Given income is less than 25K,
0.0011221/0.1758104 0.0062 0.1747007/0.1758104
0.9937 Given income is greater than
100K, 0.0009975/0.1334164 0.0074 0.1324189/0.13
34164 0.9925
82
Remark
We cam use the contingency table directly i.e.,
without finding the conditional proportions.
Income/Audited Yes No Total
lt 25K 90 14,011 14,100
25K-49.999K 71 30,629 30,700
50K-100K 69 24,631 24,700
100K lt 80 10,620 10,700
Total 310 79,890 80,200
83
Example
A survey asked 100 people their opinions about
gender and combat in the military. Here is the
results of this survey.
Gender/Combat Yes (y) No (n) Total
Male (M) 32 18 50
Female (F) 8 42 50
Total 40 60 100
P(y and F) 8/100 P(n and M) 18/100
Question What is the probability the answer is
yes, given this it is a female? Answer P(yF)
P(y and F)/P(F) (8/100)/(50/100) 8/50 4/25
0.16 Question What is the probability the
subject is male, given the answer it is
no? Answer P(Mn) P(M and n)/P(n)
(18/100)/(60/100) 18/60 3/10 0.30
84
Example

• Problem Analyzing the Triple Blood Test for Down
  Syndrome. Take blood from a pregnant woman and
  perform the biochemical analysis (see if there is
  an extra copy of chromosome 21). The test can be
  either positive or negative. Unfortunately, it
  is not always accurate.
• Results of Test
• True Positive Test is positive and baby has
  extra chromosome.
• False Positive Test is positive and baby does
  not have extra chromosome.
• True Negative Test is negative and baby does not
  have extra chromosome.
• False Negative Test is negative and baby has
  extra chromosome.

85
Study 5,282 women of age 35 or older took the
Triple Blood Test and after they had their child,
the accuracy of the test was analyzed. Here is a
contingency table for the study.
Syndrome/Test Positive (p) Negative (n) Total
Yes (D) 48 6 54
No (Dc) 1307 3921 5228
Total 1355 3927 5282
P(p) 1355/5282 0.2565 P(n) 3927/5282
0.7434 P(D and p) 48/5282 0.0091 P(D and n)
6/5282 0.0011 P(Dp) P(D and p)/P(p)
0.0091/0.2565 0.035 P(Dn) P(D and n)/P(n)
0.0011/0.7434 0.0015
86
Probability Table P(A), P(B), P(A and B)
Syndrome/Test Positive (p) Negative (n) Total
Yes (D) 0.009 0.001 0.010
No (Dc) 0.247 0.742 0.990
Total 0.257 0.743 1.0
Conclusion If the sample of 5,282 is
representative of all women who take the test,
then for women who test positively, only
approximately 4 P(Dp) of the fetuses have
Down Syndrome. However, there is 0.9 chance of
the fetus having the disease and 0.1 chance of a
negative test still having the syndrome P(Dn).
Hence, a negative test is a good indicator of
the disease not be present.
87
Example

• The following table show the results of a study
  of 137,243 men in the U.S. The study
  investigated the association between cigar
  smoking and death from cancer.
• What is the probability that a randomly select
  man form the study who died from cancer was a
  former cigar smoker?
• (b) What is the probability that a randomly
  select man who was a former cigar smoker died
  from cancer?

Died from Cancer Did not die from cancer Total
Never smoked cigars 782 120,747 121,529
Former cigar smoker 91 7,757 7,848
Current cigar smoker 141 7,725 7,866
Total 1,014 136,229 137,243
88
Counting Techniques
In order to calculate probabilities using the
classical method, one must be able to count how
many times a particular event will occur. For
sample spaces that are small, this is usually an
easy task. However, for sample spaces that are
large, we will need some counting formulas.
Section 5.5
89
Multiplication Rule of Counting
This rule is often useful in counting outcomes
(events) that have a tree structure. For
example, how many different ways can the births
of three children occur?
90
Generalization
Suppose that we have 3 branch points with the
first branch point 2 possibilities, then 3
possibilies, then 2 possibilities.
91
Example
Problem A girl has five blouses and three
skirts. How many different outfits can she put
together? Answer (5)(3) 15 Problem A girl
has five blouses and three skirts and 4 pairs of
shoes. How many different outfits (including the
shoes) can she put together? Answer (5)(3)(4)
60
92
Factorials and Sequences
1. How many sequences can be constructed in the
above way from the set S a,b,c,d. Answer 4!
(1)(2)(3)(4) 24 2. A salesman must travel to
five cities to promote his products. How many
different trips are possible if any route between
two cities is possible? Answer Sample trip
city1, city2, city3, city4, city5. Hence, 5!
120.
93
Permutations
94
Example
How many different ways (permutations) can four
letters, a,b,c,d be arranged, taking two a
time, e.g., a,b, a,d, c,d, etc?
a,b,b,a,a,c,c,a,a,d,d,a,b,c,c,b,c
,d,d,c,b,d,d,b --- 12 arrangements
95
Permutations with Duplicates
96
Combinations
97
Examples
98
Examples
99
Example