Chapter Six
Probability
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.1
Random Experiment…
…a random experiment is an action or process that leads to
one of several possible outcomes. For example:
Experiment
Outcomes
Flip a coin
Heads, Tails
Exam Marks
Numbers: 0, 1, 2, ..., 100
Assembly Time
t > 0 seconds
Course Grades
F, D, C, B, A, A+
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.2
Probabilities…
List the outcomes of a random experiment…
List: “Called the Sample Space”
Outcomes: “Called the Simple Events”
This list must be exhaustive, i.e. ALL possible outcomes
included.
Die roll {1,2,3,4,5}
Die roll {1,2,3,4,5,6}
The list must be mutually exclusive, i.e. no two
outcomes can occur at the same time:
Die roll {odd number or even number}
Die roll{ number less than 4 or even number}
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.3
Sample Space…
A list of exhaustive [don’t leave anything out] and mutually
exclusive outcomes [impossible for 2 different events to
occur in the same experiment] is called a sample space and
is denoted by S.
The outcomes are denoted by O1, O2, …, Ok
Using notation from set theory, we can represent the sample
space and its outcomes as:
S = {O1, O2, …, Ok}
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.4
Requirements of Probabilities…
Given a sample space S = {O1, O2, …, Ok}, the probabilities
assigned to the outcome must satisfy these requirements:
(1) The probability of any outcome is between 0 and 1
i.e. 0 ≤ P(Oi) ≤ 1 for each i, and
(2) The sum of the probabilities of all the outcomes equals 1
i.e. P(O1) + P(O2) + … + P(Ok) = 1
P(Oi) represents the probability of outcome i
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.5
Approaches to Assigning Probabilities…
There are three ways to assign a probability, P(Oi), to an
outcome, Oi, namely:
Classical approach: make certain assumptions (such as
equally likely, independence) about situation.
Relative frequency: assigning probabilities based on
experimentation or historical data.
Subjective approach: Assigning probabilities based on the
assignor’s judgment. [Bayesian]
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.6
Classical Approach…
If an experiment has n possible outcomes [all equally likely
to occur], this method would assign a probability of 1/n to
each outcome.
Experiment:
Sample Space:
Probabilities:
Rolling a die
S = {1, 2, 3, 4, 5, 6}
Each sample point has a 1/6 chance of
occurring.
What about randomly selecting a student and observing their
gender? S = {Male, Female}
Are these probabilities ½?
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.7
Classical Approach…
Experiment: Rolling 2 die [dice] and summing 2 numbers
on top.
Sample Space: S = {2, 3, …, 12} What are the underlying,
unstated assumptions??
Probability Examples:
P(2) = 1/36
1
2
3
4
5
6
P(7) = 6/36
P(10) = 3/36
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
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6
7
8
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10
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12
6.8
Relative Frequency Approach…
Bits & Bytes Computer Shop tracks the number of desktop
computer systems it sells over a month (30 days):
For example,
10 days out of 30
2 desktops were sold.
Desktops Sold
# of Days
0
1
1
2
2
10
3
From this we can construct
the “estimated” probabilities of an event 4
(i.e. the # of desktop sold on a given day)…
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
12
5
6.9
Relative Frequency Approach…
Desktops Sold [X]
# of Days
Desktops Sold
0
1
1/30 = .03 =P(X=0)
1
2
2/30 = .07 = P(X=1)
2
10
10/30 = .33 = P(X=2)
3
12
12/30 = .40 = P(X=3)
4
5
5/30 = .17 = P(X=4)
∑ = 1.00
“There is a 40% chance Bits & Bytes will sell 3 desktops on any given
day” [Based on estimates obtained from sample of 30 days]
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.10
Subjective Approach…
“In the subjective approach we define probability as the
degree of belief that we hold in the occurrence of an event”
P(you drop this course)
P(NASA successfully land a man on the moon)
P(girlfriend says yes when you ask her to marry you)
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.11
Events & Probabilities…
An individual outcome of a sample space is called a simple
event [cannot break it down into several other events],
An event is a collection or set of one or more simple events
in a sample space.
Roll of a die: S = {1, 2, 3, 4, 5, 6}
Simple event: the number “3” will be rolled
Event: an even number (one of 2, 4, or 6) will be rolled
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.12
Events & Probabilities…
The probability of an event is the sum of the probabilities of
the simple events that constitute the event.
E.g. (assuming a fair die) S = {1, 2, 3, 4, 5, 6} and
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6
Then:
P(EVEN) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.13
Interpreting Probability…
One way to interpret probability is this:
If a random experiment is repeated an infinite number of
times, the relative frequency for any given outcome is the
probability of this outcome.
For example, the probability of heads in flip of a balanced
coin is .5, determined using the classical approach. The
probability is interpreted as being the long-term relative
frequency of heads if the coin is flipped an infinite number
of times.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.14
Joint, Marginal, Conditional Probability…
We study methods to determine probabilities of events that
result from combining other events in various ways.
There are several types of combinations and relationships
between events:
•Complement of an event [everything other than that event]
•Intersection of two events [event A and event B] or [A*B]
•Union of two events [event A or event B] or [A+B]
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.15
Example 6.1…
Why are some mutual fund managers more successful than
others? One possible factor is where the manager earned his
or her MBA. The following table compares mutual fund
performance against the ranking of the school where the
fund manager earned their MBA: Where do we get these
probabilities from? [population or sample?]
Mutual fund outperforms
the market
Mutual fund doesn’t
outperform the market
Top 20 MBA program
.11
.29
Not top 20 MBA program
.06
.54
E.g. This is the probability that a mutual fund
outperforms AND the manager was in a top20 MBA program; it’s a joint probability
[intersection].
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Venn Diagrams
6.16
Example 6.1…
Alternatively, we could introduce shorthand notation to
represent the events:
A1 = Fund manager graduated from a top-20 MBA program
A2 = Fund manager did not graduate from a top-20 MBA program
B1 = Fund outperforms the market
B2 = Fund does not outperform the market
A1
A2
B1
B2
.11
.29
.06
.54
E.g. P(A2 and B1) = .06
= the probability a fund outperforms the market
and the manager isn’t from a top-20 school.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.17
Marginal Probabilities…
Marginal probabilities are computed by adding across rows
and down columns; that is they are calculated in the margins
of the table:
P(A2) = .06 + .54
“what’s the probability a fund
manager isn’t from a top school?”
B1
B2
P(Ai)
A1
A2
.11
.29
.40
.06
.54
.60
P(Bj)
.17
.83
1.00
P(B1) = .11 + .06
“what’s the probability a fund
outperforms the market?”
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
BOTH margins must add to 1
(useful error check)
6.18
Conditional Probability…
Conditional probability is used to determine how two events
are related; that is, we can determine the probability of one
event given the occurrence of another related event.
Experiment: random select one student in class.
P(randomly selected student is male) =
P(randomly selected student is male/student is on 3rd row) =
Conditional probabilities are written as P(A | B) and read as
“the probability of A given B” and is calculated as:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.19
Conditional Probability…
Again, the probability of an event given that another event
has occurred is called a conditional probability…
P( A and B) = P(A)*P(B/A) = P(B)*P(A/B) both are true
Keep this in mind!
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.20
Conditional Probability…
Example 6.2 • What’s the probability that a fund will
outperform the market given that the manager graduated
from a top-20 MBA program?
Recall:
A1 = Fund manager graduated from a top-20 MBA program
A2 = Fund manager did not graduate from a top-20 MBA program
B1 = Fund outperforms the market
B2 = Fund does not outperform the market
Thus, we want to know “what is P(B1 | A1) ?”
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.21
Conditional Probability…
We want to calculate P(B1 | A1)
B1
B2
P(Ai)
A1
A2
.11
.29
.40
.06
.54
.60
P(Bj)
.17
.83
1.00
Thus, there is a 27.5% chance that that a fund will outperform the market
given that the manager graduated from a top-20 MBA program.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.22
Independence…
One of the objectives of calculating conditional probability is
to determine whether two events are related.
In particular, we would like to know whether they are
independent, that is, if the probability of one event is not
affected by the occurrence of the other event.
Two events A and B are said to be independent if
P(A|B) = P(A)
and
P(B|A) = P(B)
P(you have a flat tire going home/radio quits working)
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.23
Independence…
For example, we saw that
P(B1 | A1) = .275
The marginal probability for B1 is: P(B1) = 0.17
Since P(B1|A1) ≠ P(B1), B1 and A1 are not independent
events.
Stated another way, they are dependent. That is, the
probability of one event (B1) is affected by the occurrence of
the other event (A1).
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.24
Union…
Determine the probability that a fund outperforms (B1)
or the manager graduated from a top-20 MBA program (A1).
A1 or B1 occurs whenever:
A1 and B1 occurs, A1 and B2 occurs, or A2 and B1 occurs…
B1
B2
P(Ai)
A1
A2
.11
.29
.40
.06
.54
.60
P(Bj)
.17
.83
1.00
P(A1 or B1) = .11 + .06 + .29 = .46
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.25
Probability Rules and Trees…
We introduce three rules that enable us to calculate the
probability of more complex events from the probability of
simpler events…
The Complement Rule – May be easier to calculate the
probability of the complement of an event and then substract
it from 1.0 to get the probability of the event.
P(at least one head when you flip coin 100 times)
= 1 – P(0 heads when you flip coin 100 times)
The Multiplication Rule: P(A*B) “way I write it”
The Addition Rule: P(A+B) “way I write it”
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.26
Example 6.5…
A graduate statistics course has seven male and three female
students. The professor wants to select two students at
random to help her conduct a research project. What is the
probability that the two students chosen are female?
P(F1 * F2) = ???
Let F1 represent the event that the first student is female
P(F1) = 3/10 = .30
What about the second student?
P(F2 /F1) = 2/9 = .22
P(F1 * F2) = P(F1) * P(F2 /F1) = (.30)*(.22) = 0.066
NOTE: 2 events are NOT independent.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.27
Example 6.6…
The professor in Example 6.5 is unavailable. Her
replacement will teach two classes. His style is to select one
student at random and pick on him or her in the class. What
is the probability that the two students chosen are female?
Both classes have 3 female and 7 male students.
P(F1 * F2) = P(F1) * P(F2 /F1) = P(F1) * P(F2)
= (3/10) * (3/10) = 9/100 = 0.09
NOTE: 2 events ARE independent.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.28
Addition Rule…
Addition rule provides a way to compute the probability of
event A or B or both A and B occurring; i.e. the union of A
and B.
P(A or B) = P(A + B) = P(A) + P(B) – P(A and B)
Why do we subtract the joint probability P(A and B) from
the sum of the probabilities of A and B?
P(A or B) = P(A) + P(B) – P(A and B)
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6.29
Addition Rule…
P(A1) = .11 + .29 = .40
P(B1) = .11 + .06 = .17
By adding P(A) plus P(B) we add P(A and B) twice. To correct we subtract P(A and
B) from P(A) + P(B)
B1
A1
B1
B2
P(Ai)
A1
A2
.11
.29
.40
.06
.54
.60
P(Bj)
.17
.83
1.00
P(A1 or B1) = P(A) + P(B) –P(A and B) = .40 + .17 - .11 = .46
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.30
Addition Rule for Mutually Excusive Events
If and A and B are mutually exclusive the occurrence of one
event makes the other one impossible. This means that
P(A and B) = P(A * B) = 0
The addition rule for mutually exclusive events is
P(A or B) = P(A) + P(B)
Only if A and B are Mutually Exclusive.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.31
Example 6.7…
In a large city, two newspapers are published, the Sun and
the Post. The circulation departments report that 22% of the
city’s households have a subscription to the Sun and 35%
subscribe to the Post. A survey reveals that 6% of all
households subscribe to both newspapers. What proportion
of the city’s households subscribe to either newspaper?
That is, what is the probability of selecting a household at
random that subscribes to the Sun or the Post or both?
P(Sun or Post) = P(Sun) + P(Post) – P(Sun and Post)
= .22 + .35 – .06 = .51
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.32
Probability Trees [Decision Trees]…
A probability tree is a simple and effective method of applying the
probability rules by representing events in an experiment by lines. The
resulting figure resembles a tree.
This is P(F), the probability of
selecting a female student first
First selection
Second selection
This is P(F|F), the probability of
selecting a female student
second, given that a female was
already chosen first
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6.33
Probability Trees…
At the ends of the “branches”, we calculate joint
probabilities as the product of the individual probabilities
on the preceding branches.
First selection
Second selection
Joint probabilities
P(FF)=(3/10)(2/9)
P(FM)=(3/10)(7/9)
P(MF)=(7/10)(3/9)
Sample Space:[F1*F2, F1*M2, M1*F2, M1*M2]
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
P(MM)=(7/10)(6/9)
6.34
Probability Trees…
Note: there is no requirement that the branches splits be
binary, nor that the tree only goes two levels deep, or that
there be the same number of splits at each sub node…
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.35
Example 6.8…
Law school grads must pass a bar exam. Suppose pass rate for firsttime test takers is 72%. They can re-write if they fail and 88% pass
their second attempt [P(pass take 2/fail take 1)]. What is the
probability that a randomly grad passes the bar? [sample space?]
P(Pass) = .72
First exam
Second exam
P(Fail and Pass)=
(.28)(.88)=.2464
P(Fail and Fail) =
(.28)(.12) = .0336
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6.36
Bayes’ Law…
Bayes’ Law is named for Thomas Bayes, an eighteenth
century mathematician.
In its most basic form, if we know P(B | A),
we can apply Bayes’ Law to determine P(A | B)
P(B|A)
P(A|B)
for example …
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6.37
Breaking News: New test for early detection of cancer has been developed.
Let
C = event that patient has cancer
Cc = event that patient does not have cancer
+ = event that the test indicates a patient has cancer
- = event that the test indicates that patient does not have cancer
Clinical trials indicate that the test is accurate 95% of the time in detecting cancer for
those patients who actually have cancer: P(+/C) = .95
but unfortunately will give a “+” 8% of the time for those patients who are known not
to have cancer: P(+/ Cc ) = .08
It has also been estimated that approximately 10% of the population have cancer and
don’t know it yet: P(C) = .10
You take the test and receive a “+” test results. Should you
be worried? P(C/+) = ?????
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6.38
What we know.
P(+/C) = .95
P(+/ Cc ) = .08
P(C) = .10
From these probabilities we can find
P(-/C) = .05
P(-/ Cc ) = .92
P(Cc) = .90
Test Results
True State of Nature
Have Cancer: C
Do Not Have Cancer: CC
+
-
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6.39
Bayesian Terminology…
The probabilities P(A) and P(AC) are called prior
probabilities because they are determined prior to the
decision about taking the preparatory course.
The conditional probability P(A | B) is called a posterior
probability (or revised probability), because the prior
probability is revised after the decision about taking the
preparatory course.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.40
Students Work – Bayes Problem
The Rapid Test is used to determine whether someone has
HIV [H]. The false positive and false negative rates are 0.05
P(+/ Hc) and 0.09 P(-/H) respectively.
The doctor just received a positive test results on one of
their patients [assumed to be in a low risk group for HIV].
The low risk group is known to have a 6% P(H) probability
of having HIV. What is the probability that this patient
actually has HIV [after they tested positive]. Feel free to use
a table to work this problem
P(H) = 0.06
**** P(Hc) = ?
P(+/ Hc) = 0.05
**** P(-/ Hc) = ?
P(-/H) = 0.09
**** P(+/H) = ?
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.41
Students Work – Bayes Problem
Transplant operations for hearts have the risk that the body may reject
the organ. A new test has been developed to detect early warning signs
that the body may be rejecting the heart. However, the test is not
perfect. When the test is conducted on someone whose heart will be
rejected, approximately two out of ten tests will be negative (the test is
wrong). When the test is conducted on a person whose heart will not
be rejected, 10% will show a positive test result (another incorrect
result). Doctors know that in about 50% of heart transplants the body
tries to reject the organ.
*Suppose the test was performed on my mother and the test is positive
(indicating early warning signs of rejection). What is the probability that the
body is attempting to reject the heart?
*Suppose the test was performed on my mother and the test is negative
(indicating no signs of rejection). What is the probability that the body is
attempting to reject the heart?
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
6.42