Chapter 4. Probability
Chapter Problem: Are polygraph instruments effective as
“lie detector”?
Table 4-1 Results from Experiments with Polygraph Instruments
Did the Subject Actually Lie?
No (Did Not Lie) Yes (Lied)
Positive Test Result
(Polygraph test indicated
that the subject lied.)
15
(False positive)
42
(True positive)
Negative Test Result
(Polygraph test indicated
that the subject did not lie)
32
(True negative)
9
(False negative)
1
4.1 Review and Preview
• How statistician think:
Rare Event Rule for Inferential Statistics
If, under a given assumption, the probability of a particular
observed event is extremely small, we conclude that the
assumption is probably not correct.
Re-phrase:
1. Assumption
2. Find the probability of an event, under the assumption, is very
small
3. Conclude that the assumption is probably not correct
2
4.2 Basic Concepts of Probability – Part 1
Definition
• An event is any collection of outcomes of a procedure.
• A simple event is an outcome or an event that cannot be further
broken down into simpler components.
• The sample space for a procedure consists of all possible simple
events.
e.g.1
Procedure
Single birth
Example of event
female (simple event)
Complete sample space
{f, m}
3 births
2 female and a male
{fff, ffm, fmf, fmm, mff,
mfm, mmf, mmm}
3
4.2 Basic Concepts of Probability – Part 1
Notion for Probabilities
P denotes a probability
A, B, and C denote specific events
P(A) denotes the probability of event A occurring.
Three different approaches to find the probability of an event:
1. Relative frequency approximation of probability
2. Classical approach to probability (requires equally likely
outcome)
3. Subjective probabilities
4
4.2 Basic Concepts of Probability – Part 1
1: Relative Frequency Approximation of Probability
Conduct (or observe) a procedure, and count the number times
that event A actually occurs. Based on these actual result, P(A)
is estimated as follows
number of times A occurred
P( A) 
number of times the trial was repeated
When trying to determine the probability that an individual car
Crashes in a year, we must examine past results to determine the
number of cars in use in a year and the number of them that
crashed, then we find the ratio of the number of cars that crashed
to the total number of cars. For a recent year, the result is a
probability of 0.0480. (see example 2)
5
4.2 Basic Concepts of Probability – Part 1
2: Classical Approach to Probability
Assume that a given procedure has n different simple event and
that each of those simple events has an equal chance of
occurring. If event A can occur is s of these n ways, then
P( A) 
number of ways A can occurr
s

number of different simple events n
When trying to determine the probability of winning the grand
prize in a lottery by selecting 6 numbers between 1 and 60, each
combination has an equal chance of occurring. The probability
of winning is 0.0000000200, which can be found by using
methods presented later in this chapter.
6
4.2 Basic Concepts of Probability – Part 1
3: Subjective Probabilities
P(A), the probability of event A, is estimated by using
knowledge of the relevant circumstances.
When trying to estimate the probability of an astronaut
surviving a mission in space shuttle, experts consider past
event along with changes in technologies and conditions to
develop an estimate of the probability. As of this writing,
that probability has be estimated by NASA scientists as
0.99.
7
4.2 Basic Concepts of Probability – Part 1
Law of Large Numbers. As a procedure is repeated again and
again, the relative frequency probability (from Rule 1) of an
event tends to approach the actual probability.
Probability and Outcomes that are not always equally likely. One
common mistake is to incorrectly assume that outcomes are
equally likely just because we know nothing about the
likelihood of different outcomes.
8
4.2 Basic Concepts of Probability – Part 1
e.g.2 Probability of a car crash. Find the probability that a randomly
selected car in the U.S. will be in crash this year.
Sol. For a recent year, there were 6,511,100 cars that crashed among
the 135,670,000 cars registered in the U.S. (Statistical Abstract
of the U.S.). We use the relative frequency approach:
P(Crash) 
number of cars that crashed
6,511,100

 0.048
total number of cars
135,670,000
9
4.2 Basic Concepts of Probability – Part 1
e.g.3 Probability of a Positive Test Result. Refer to table 4-1
included in the chapter problem. Assuming that one of the 98
test results summarized in table 4-1 is randomly selected, find
the probability that it is a positive test result.
Sol. The sample space consists of the 98 test results in table 4-1.
Among the 98 results, 57 of them are positive results (42 + 15).
Since each test result is equally likely to be selected, we can
apply the classical approach:
number of positive test results
total number of results
57

 0.582
98
P(positive test result from table 4 - 1) 
10
4.2 Basic Concepts of Probability – Part 1
e.g.4 Genotype When studying the affect of heredity on height, we
can express each individual genotype, AA, Aa, aA, and aa, on
an index card and shuffle the four cards randomly and select
one of them. What is the probability that we select a genotype
in which the two components are different.
Ans. Classical approach, 0.5
e.g.5 Probability of a President from Alaska. Find the probability
that the next President of the U.S. is from Alaska.
Ans. Subjective estimate, 0.001
e.g.6 Stuck in an Elevator. What is the probability that you will get
stuck in the next elevator that you ride?
Ans. Subjective estimate, 0.0001
11
4.2 Basic Concepts of Probability – Part 1
Finding the total number of outcomes.
e.g.7 Gender of children. Find the probability that when a couple has
3 children, they will have exactly 2 boys. Assume that the boys
and girls are equally likely and that the gender of any child is
not influenced by the gender of any other child.
P(2 boys in 3 birth) 
3
 0.375
8
12
4.2 Basic Concepts of Probability – Part 1
Finding the total number of outcomes.
e.g.8 American Online Survey. AOL asked users this question
about KFC, “will KFC gain or lose business after eliminating
trans fats?” Among the responses received, 1941 said that KFC
would gain business, 1260 said that KFC business would
remain the same, and 204 said that KFC would lose business.
Find the probability that a randomly selected response states
that KFC would gain business.
P(response of a gain in business) 
1941
 0.570
3405
Interpretation. There is a 0.570 probability that if a response is randomly
selected, it was a response of a gain in business. Important: Note that the survey
involves a voluntary response sample because the AOL users themselves
decided whether to respond. Consequently, when interpreting the results of this
survey, keep in mind that they do not necessary reflect the opinions of the
general population. The response reflect only the opinions of those who chose
13
to respond.
4.2 Basic Concepts of Probability – Part 1
1
 Certain
 Likely
0.5
 50-50 Chance
 Unlikely
0
 Impossible
Figure 4 – 2 Possible Values for Probabilities
14
4.2 Basic Concepts of Probability – Part 1
• The probability of an impossible event is 0
• The probability of an event that is certain to occur is 1
• For any event A, the probability of A is between 0 and 1
inclusive. That is 0  P(A)  1
e.g.9 Find the probability that Thanksgiving Day will be on (a)
Wednesday, (b) Thursday.
P(Thanksgiving on Wednesday) = 0
P(Thanksgiving on Thursday) = 1
15
4.2 Basic Concepts of Probability – Part 1
Complementary Events – The complement of event A, denoted by
A, consists of all outcomes in which event A does not occur.
e.g.10 Guessing on an SAT Test. A typical question on an SAT test
requires the test taker to select one of five possible choices: A,
B, C, D, or E. Because only one answer is correct, if you make
a random guess, your probability of being correct is 1/5 or 0.2.
Find the probability of making a random guess and not being
correct.
Ans.
0.8
16
4.2 Basic Concepts of Probability – Part 1
Rounding off Probabilities. Either simple fractions or round the
decimal to three significant digits.
e.g.11
0.04788219  0.0480
(e.g.2)
1/3  0.333
½ = 0.5 (no need to write as 0.500)
1941/3405  0.570
(e.g.8)
17
4.2 Basic Concepts of Probability – Part 1
e.g.12 Unusual Events? In a clinical experiment of the Salk vaccine
for polio,
–
–
–
–
200,745 children were given a placebo
201,229 other children were treated with the Salk Vaccine
115 cases of polio among those in the placebo group
33 cases in the treatment group
If we assume that the vaccine has no effect, the probability of
getting such test results is found to be “less than 0.001.”
• Is an event with a probability less than 0.001 an unusual event?
Ans. Yes, the event is “unusual”
• What does that probability imply about the effectiveness of the
vaccine? Ans. we conclude that the vaccine appears to be effective
• Rare event rule for inferential statistics! (beginning of the
chapter)
18
4.2 Basic Concepts of Probability – Part 2
Part 2. Beyond the Basics of Probability: Odds – expressions of
likelihood are often given as odds, such as 50:1
Definition
The actual odds against event A occurring are the ratio P(A)/P(A).
It is usually expressed in the form a : b
The actual odds in favor of event A occurring are P(A)/P(A).
If odds against event A is a : b, then odds in favor of A is b : a.
The payoff odds against event A represents the ratio of net profit (if
you win) to the amount bet.
payoff odds against event A = (net profit):(amount bet)
19
4.2 Basic Concepts of Probability – Part 2
Odds
e.g.13 If you bet $5 on the number 13 in roulette, your probability of
winning is 1/38 and the payoff odds are given by the casino as 35:1.
a. Find the actual odds against the out come of 13
P( A) 
1
37
, P( A )  , so odds against wi nning is P( A ) : P( A)  37 : 1
38
38
b. How much net profit would you make if you win by betting 13?
By definition 35:1 = (net profit):5, therefore, net profit is 35(5) = $175
20
4.2 Basic Concepts of Probability – Part 2
Odds
e.g.9 If you bet $5 on the number 13 in roulette, your probability of
winning is 1/38 and the payoff odds are given by the casino as 35:1
c. If the casino were operating just for the fun of it, and the payoff
odds were changed to match the actual odds against 13, how much
you win if the outcome is 13?
Now 37:1 = (net profit):5
Net profit is 37(5) = $185
21
4.3 Addition Rule
Notation for Addition Rule
Definition. A compound event is any event combining two or more
simple events.
Notation for Addition Rule
P(A or B) = P(in a single trial, event A occurs or event B
occurs or they both occur)
Be careful about the notation P(A and B)
22
4.3 Addition Rule
Notation for addition rule
Table 4-1 Results from Experiments with Polygraph Instruments
Did the Subject Actually Lie?
No (Did Not Lie) Yes (Lied)
Positive Test Result
(Polygraph test indicated
that the subject lied.)
15
(False positive)
42
(True positive)
57
Negative Test Result
(Polygraph test indicated
that the subject did not lie)
32
(True negative)
9
(False negative)
41
47
51
23
4.3 Addition Rule
Notation for addition rule
e.g.1 Polygraph Test refer to table 4-1. if a subject is randomly
selected from 98 subjects, find the probability of selecting a
subject who had a positive test result or lied.
Count: how many tested positive, or lied, or both.
Three methods for counting
method 1.
15 + 42 + 9 = 66
method 2.
51 + 57 – 42 = 66
method 3.
57 + 9 = 66
Ans. P(positive test result or lied) = 66/98  0.673
24
4.3 Addition Rule
Notation for addition rule
Formal Addition Rule:
P(A or B) = P(A) + P(B) – P(A and B)
where P(A and B) denotes the probability that A and B both
occur at the same time as an outcome in a trial of a
procedure.
Caution: avoid counting outcomes more than once.
25
4.3 Addition Rule
Notation for addition rule
Intuitive Addition Rule
To find P(A or B)
– find the sum of the number of ways event A can occur
– find the sum of the number of ways event B can occur
– adding in such a way that every outcome is counted only
once
– P(A or B) is equal to that sum, divided by the total number
of outcomes in the sample space.
26
4.3 Addition Rule
Notation for addition rule
Definition
Event A and B are disjoint (or mutually exclusive) if they cannot
occur at the same time. (That is, disjoint events do not overlap.)
e.g.2 Polygraph Test (Table 4 –1). Selecting one subject from 98.
a. Determine whether the following events are disjoint:
Event A: getting a subject with negative test result
Event B: getting a subject who did not lie Not disjoint
b. Find the probability of selecting a subject who had negative test
result or did not lie.
56/98  0.571
27
4.3 Addition Rule
Notation for addition rule
AB
A
Figure 4 – 3 Venn Diagram
For Event That are Not Disjoint
B
Figure 4 – 3 Venn Diagram
For Event That are Disjoint
This figure explains why
P(A or B) = P(A) + P(B) – P(A and B)
28
4.3 Addition Rule
Complementary Events
Recall, The complement of event A, denoted by A, consists of all
outcomes in which event A does not occur.
– A and A must be disjoint; it is impossible for an event and its
complement to occur at the same time
– A either occur or does not occur, which means that either A
or A must occur. Therefore
P( A or A )  P( A)  P( A )  1
29
4.3 Addition Rule
Complementary Events
Rule of Complementary Events
P( A)  P( A )  1
P( A )  1  P( A)
P( A)  1  P( A )
e.g.3 FBI data show that 62.4% of murders are cleared by arrests.
We can express the probability of a murder being cleared by an
arrest as P( cleared ) = 0.624. Find P( cleared )
Ans. P( cleared ) = 1 – 0.624 = 0.376
A major advantage of the rule of complementary events is that its
use can greatly simplify certain problems.
30
4.4 Multiplication Rule
Notation
P(A and B) = P(event A occurs in a first trial and
event B occurs in a second trial)
Consider following two question (1st T or F, 2nd multiple choice)
1.
2.
True or False: A pound of feathers is heavier than a pound of gold.
Who said that “smoking is one of the leading causes of statistics”?
a.
b.
c.
d.
e.
Philip Morris
Smokey Robinson
Fletcher Knebel
R. J. Reynolds
Virginia Slims
Ten possible outcomes:
31
4.4 Multiplication Rule
T
F
Figure 4 – 6 Tree
Diagram of Test
Answers
2
a
Ta
b
Tb
c
Tc
d
Td
e
a
Te
Fa
b
Fb
c
Fc
d
Fd
e
Fe
5
=
In this case
P(T and c) = 1/10,
P(T) = ½,
P(c) = 1/5,
Note that
½ (1/5) = 1/10,
Is this suggesting
P(T and c) = P(T)P(c)?
10
32
4.4 Multiplication Rule
e.g.1 If two of the subjects included in the table are randomly
selected without replacement, find the probability that the 1st
selected person had a positive test result and the 2nd selected
person had a negative test result.
Table 4-1 Results from Experiments with Polygraph Instruments
Did the Subject Actually Lie?
No (Did Not Lie) Yes (Lied)
Positive Test Result
(Polygraph test indicated
that the subject lied.)
15
(False positive)
42
(True positive)
Negative Test Result
(Polygraph test indicated
that the subject did not lie)
32
(True negative)
9
(False negative)
33
4.4 Multiplication Rule
Solution: 1st selection
P(positive test result) = 57/98
2nd selection
P(negative test result) = 41/97
P(1st subject has positive test result
and 2nd subject has negative result) = (57/98)(41/97)
= 0.246
The key point is that we must adjust the probability of the second
event to reflect the outcome of the first event.
34
4.4 Multiplication Rule
Notion for Conditional Probability
P(B|A) represents the probability of event B occurring after it
is assumed that event A has already occurred. (We can read
B|A as “B given A” or as “event B occurring after event A
has already occurred.”)
Definition – Two events A and B are independent if the occurrence
of one does not affect the probability of the occurrence of the
other. If A and B are not independent, they are said to be
dependent.
35
4.4 Multiplication Rule
Formal Multiplication Rule
P(A and B) = P(A)  P(B|A)
start
P(A and B)
Multiplication Rule
P(A and B) = P(A)P(B)
yes
Are A and B
Independent?
no
P(A and B) = P(A)P(B|A)
Figure 4 – 7 Applying Multiplication Rule
Caution: when applying the multiplication rule, always consider
whether the events are independent or dependent.
36
4.4 Multiplication Rule
e.g.2 Quality Control in Manufacturing. Consider a small sample
of 5 pacemakers, including three that are good (denoted by G)
and two that are defective (denoted by D). A medical researcher
randomly select two of the pacemakers for further
experimentation. Find the probability that the first selected
pacemaker is good (G) and the second pacemaker is also good
(G). Use each of the following assumption:
(a) with replacement; Ans. Independent, (3/5)(3/5) = 9/25 = 0.36
(b) without replacement. Ans. Dependent, (3/5)(2/4) = 0.3
Interpretation. Note that in part (b) we adjusted the second probability, to take into
Account the selection in the first outcome.
Also note that a medical researcher would not sample with replacement, as in part
(a). However, in statistics we have a special interest in sampling with replacement.
(see section 6-4)
37
4.4 Multiplication Rule
Treating Dependent Events as Independent – it is common practice
to treat events as independent when small sample are draw from
large populations.
Guideline. If a sample size is no more than 5% of the population
treats the selection as being independent.
e.g.3 Quality Control in Manufacturing. Assume that we have a
batch of 100,000 pacemakers, including 99,950 good ones and
50 defective ones.
a. If two are randomly selected, find P(both are good).
Ans. P(both good) = 0.999
b. If 20 are randomly selected, find P(all 20 are good).
38
Ans. P(all 20 are good) = 0.990
4.4 Multiplication Rule
The following example illustrate the importance of carefully
identifying the events being considered.
e.g.4 Birthdays. Assume that two people are randomly selected and
also assume that birthdays occur on the days of the week with
equal frequencies.
a. Find the probability that two people are born on the same day of
the week.
1
P(Two are on the same day) 
7
b. Find the probability that two people are born on Monday.
P(Two are on Monday) 
1
49
39
4.4 Multiplication Rule
Important applications of the multiplication rule.
• E.g.5 Gives insight into hypothesis testing
• E.g.6 illustrates the principle of redundancy
40
4.4 Multiplication Rule
e.g.5 Effectiveness of Gender Selection. A geneticist develops a
procedure for increasing the likelihood of female babies. In an
additional test, 20 couples uses the method and the results
consist of 20 females among 20 babies. Assuming that gender
selection procedure has no effect, find the probability of getting
20 females among 20 babies by chance. Based on the result, is
there strong evidence to support the geneticist’s claim that the
procedure is effective in increasing the likelihood that babe’s
will female. (conclusion)
Sol. If no effect, the P(all 20 offspring are female) = (1/2)20 =
0.000000954
Therefore gender-selection process appears to be effective in
increasing the …
41
4.4 Multiplication Rule
e.g.6 Redundancy for Increased Reliability. Modern aircraft engines are now
highly reliable. One design feature contributing to that reliability is the use
of redundancy, whereby critical components are duplicated so that if one
fails, the other will work. For example, single-engine aircraft now have two
independent electrical systems so that if one electrical system fails, the other
can continue to work so that the engine does not fail. For the purpose of this
example, we assume that the probability of an electrical system failure is
0.001.
a. If the engine ins an aircraft has one electrical system, what is the probability
that it will work?
Ans. P(work) = 0.999
b.
If the engine in an aircraft has two independent electrical systems, what is the
probability that the engine can function with a working electrical system?
Ans. P(work) = 0.999999
42