1 Probability Please read sections 3.1 – 3.3 in your textbook Def: An experiment is a process by which observations are generated. Def: A variable is a quantity that is observed in the experiment. Def: The sample space (S) for an experiment is the set of all possible outcomes. Def: An event E is a subset of a sample space. It provides the collection of outcomes that correspond to some classification. Example: Note: A sample space does not have to be finite. Example: Pick any positive integer. The sample space is countably infinite. A discrete sample space is one with a finite number of elements, { }1,2,3,4,5,6 or one that has a countably infinite number of elements { }1,3,5,7,... . A continuous sample space consists of elements forming a continuum. { }x / 2 x 5< < 2 A Venn diagram is used to show relationships between events. A intersection B = (A ∩ B) = A and B The outcomes in (A intersection B) belong to set A as well as to set B. A union B = (A U B) = A alone or B alone or both Union Formula For any events A, B, P (A or B) = P (A) + P (B) – P (A intersection B) i.e. P (A U B) = P (A) + P (B) – P (A ∩ B) 3 cA complement not A A ' A A = = = = A complement consists of all outcomes outside of A. Note: P (not A) = 1 – P (A) Def: Two events are mutually exclusive (disjoint, incompatible) if they do not intersect, i.e. if they do not occur at the same time. They have no outcomes in common. When A and B are mutually exclusive, (A ∩ B) = null set = Ø, and P (A and B) = 0. Thus, when A and B are mutually exclusive, P (A or B) = P (A) + P (B) (This is exactly the same statement as rule 3 below) Axioms of Probability Def: A probability function p is a rule for calculating the probability of an event. The function p satisfies 3 conditions: 1) 0 ≤ P (A) ≤1, for all events A in the sample space S 2) P (Sample Space S) = 1 3) If A, B, C are mutually exclusive events in the sample space S, then P(A B C) P(A) P(B) P(C)∪ ∪ = + + 4 The Classical Probability Concept: If there are n equally likely possibilities, of which one must occur and s are regarded as successes, then the probability of success is s n . Example: Frequency interpretation of Probability: The probability of an event E is the proportion of times the event occurs during a long run of repeated experiments. Example: Def: A set function assigns a non-negative value to a set. Ex: N (A) is a set function whose value is the number of elements in A. Def: An additive set function f is a function for which f (A U B) = f (A) + f (B) when A and B are mutually exclusive. N (A) is an additive set function. Ex: Toss 2 fair dice. Let A be the event that the sum on the two dice is 5. Let B be the event that the sum on ...

1. 1
Probability
Please read sections 3.1 – 3.3 in your textbook
Def: An experiment is a process by which observations are
generated.
Def: A variable is a quantity that is observed in the experiment.
Def: The sample space (S) for an experiment is the set of all
possible outcomes.
Def: An event E is a subset of a sample space. It provides the
collection of outcomes
that correspond to some classification.
Example:

2. Note: A sample space does not have to be finite.
Example: Pick any positive integer. The sample space is
countably infinite.
A discrete sample space is one with a finite number of elements,
{ }1,2,3,4,5,6 or one that
has a countably infinite number of elements { }1,3,5,7,... .
A continuous sample space consists of elements forming a
continuum. { }x / 2 x 5< <
2
A Venn diagram is used to show relationships between events.
A intersection B = (A ∩ B) = A and B
The outcomes in (A intersection B) belong to set A as well as to
set B.
A union B = (A U B) = A alone or B alone or both

3. Union Formula
For any events A, B, P (A or B) = P (A) + P (B) – P (A
intersection B) i.e.
P (A U B) = P (A) + P (B) – P (A ∩ B)
3
cA complement not A A ' A A = = = =
A complement consists of all outcomes outside of A.
Note: P (not A) = 1 – P (A)
Def: Two events are mutually exclusive (disjoint,
incompatible) if they do not intersect,
i.e. if they do not occur at the same time. They have no
outcomes in common.

4. When A and B are mutually exclusive, (A ∩ B) = null set = Ø,
and P (A and B) = 0.
Thus, when A and B are mutually exclusive, P (A or B) = P (A)
+ P (B)
(This is exactly the same statement as rule 3 below)
Axioms of Probability
Def: A probability function p is a rule for calculating the
probability of an event. The
function p satisfies 3 conditions:
1) 0 ≤ P (A) ≤1, for all events A in the sample space S
2) P (Sample Space S) = 1
3) If A, B, C are mutually exclusive events in the sample space
S, then
P(A B C) P(A) P(B) P(C)∪ ∪ = + +
4
The Classical Probability Concept: If there are n equally likely
possibilities, of which one
must occur and s are regarded as successes, then the probability
of success is s
n

5. .
Example:
Frequency interpretation of Probability: The probability of an
event E is the proportion of
times the event occurs during a long run of repeated
experiments.
Example:
Def: A set function assigns a non-negative value to a set.
Ex: N (A) is a set function whose value is the number of
elements in A.
Def: An additive set function f is a function for which f (A U
B) = f (A) + f (B) when A and
B are mutually exclusive.
N (A) is an additive set function.
Ex: Toss 2 fair dice. Let A be the event that the sum on the
two dice is 5. Let B be the
event that the sum on the 2 dice is 6.
N(A) = 4 since A consists of (1,4), (2,3), (3,2), (4,1).
N (B) = 5 since B consists of (1,5), (2,4), (3,3), (4,2), (5,1)
N (A or B) = 4 + 5 = 9

6. 5
DeMorgan’s Laws:
6
Example: Problem 3.33, Pg. 74 Miller and Freund
In a group of 160 graduate engineering students, 92 are enrolled
in statistics, 63 are
enrolled in Operations Research and 40 are enrolled in both.
a. How many are enrolled in at least one course?

7. b. How many are enrolled in neither?
Example:
Doctor Jackson makes the following observation of new clients
entering a weight loss
seminar:
80% are considered overweight [O], (need to lose at least 20
pounds), and will be placed
on a low carbohydrate diet
70% do not exercise regularly, [NE] and will be joining a
weight training class.
55% are both overweight and do not exercise regularly.
a) Draw a Venn diagram.
b) Are the events O and NE mutually exclusive?
c) Find the probability that a client selected at random either
needs to lose weight
or does not exercise regularly.
7

8. d) Find the probability that a client selected at random is not
overweight.
We use the fact P (not A) = 1 – P (A)
e) Find the probability that a client selected at random is
overweight but (and) exercises
regularly
f) Find the probability that a client is not overweight and
exercises regularly.
Def: Two events are independent if the occurrence or non-
occurrence of one does not
change the probability that the other will occur.
Example of Independent events:
You roll 2 dice. Find the probability of rolling a 5 on each die.

9. If events A and B are independent, then P (A ∩ B) = P (A). P
(B)
8
Example of dependent events:
Draw 2 cards from a standard deck of 52 cards without
replacement. Find the probability
of drawing 2 Aces.
If events A and B are dependent, then P (A ∩ B) ≠ P (A). P (B)

10. Laws of Probability:
Multiplication Law of Probability:
If A, B, C, and D are mutually independent events, then
P (A ∩ B ∩ C ∩ D) = P (A).P (B).P(C).P (D)
Addition Law of Probability:
For any events A and B: P (A U B) = P (A) + P (B) – P (A ∩
B)
For A, B, mutually exclusive: P (A U B) = P (A) + P (B)
For A, B, independent: P (A U B) = P (A) + P (B) – P (A). P
(B)
P (A U B) = P (A) + P (B) [1 – P (A)]
9
Conditional Probability:
We use the notation P (A/B) to refer to the conditional
probability that event A occurs,
given the fact that event B is known to be true.

11. Ex: The probability that someone who is known to be
overweight is hypertensive would
be written as P (Hypertensive/Overweight) = P (H/O).
P(A B)P(A / B)
P(B)
∩
=
Venn diagram:
Given the fact that B has already occurred, makes the new
sample space the outcomes
in B. Thus, P (A/B) = the ratio of the outcomes in the
intersection of A and B to the
outcomes in B.
Similarly, P (B/A)= P(B A)
P(A)
∩ = P(A B)
P(A)

12. ∩
Venn diagram:
Conditional probabilities can also be thought of as limiting the
type of people you want to
include in your probability assessment.
10
Two events are independent if knowing that one occurs does not
change
the probability that the other occurs.
3 ways to check if two events A and B are independent:
1. If P (A/B) = P (A), then they are independent.
2. If P (B/A) = P (B), then they are independent.
3. If P (A ∩ B) = P (A). P (B)

13. If any of these is true, the others are also true.
If A, B are independent, cP(A /B) P(A) P(A /B )= =
P (B/A) = P (B) = cP(B/ A )
If A, B are dependent, P (A/B) ≠ P (A) ≠ cP(A /B )
P (B/A) ≠ P (B) ≠ cP(B/ A )
P (A ∩ B) ≠ P(A) P(B)
If we know the probabilities of any two events as well as the
probability of their
intersection we can calculate both P (A/B) as well as P (B/A).
In addition, for any 2 events A, B, if we know the probabilities
of two events and the
corresponding conditional probabilities we can calculate the
probability of their
intersection P (A and B) as follows:
P (A ∩ B) = P (B/A) P(A) = P(A/B) P(B)
This is known as the multiplication rule.

14. 11
Example:
Current estimates are that about 25% of all deaths are due to
cancer, and of the deaths
due to cancer, 30% are attributed to tobacco, 40% to poor diet
and 30% to other causes.
Find the probability that a death is due to cancer and tobacco.
Please use the
multiplication rule.
When A and B are mutually exclusive, are they dependent or
independent?
Answer: They are _______________, since
12

15. Example:
P (Smoker) = .44 P (Chronic Cough) = .4 P (Smoker and
Chronic Cough) = .37
a) Find the probability that a person is a smoker given the fact
that he has chronic
cough.
b) Find the probability that a person has a chronic cough given
the fact that he is a
smoker.
P (CC/S) =
c) Are the events S and CC dependent or independent?

16. 13
Example:
In an experiment to study the dependence of hypertension on
smoking habits, the
following data were (datum: singular, data: plural) collected on
180 individuals:
Smoking
Status
Nonsmoker Moderate
Smoker
Heavy
Smoker
Total
Hypertension
Status
Hypertension 21 36 30
No
Hypertension
48 26 19
Total

17. a. What is the probability that a randomly selected individual is
experiencing
hypertension?
b. Given that a heavy smoker is selected at random from this
group, what is the
probability that the person is experiencing hypertension?
c. Are the events hypertension and heavy smoker independent?
Give 3 different
supporting calculations.
d. Find the probability that a randomly selected individual is
either a moderate smoker
or is hypertensive.
14
Smoking
Status

18. Nonsmoker Moderate
Smoker
Heavy
Smoker
Total
Hypertension
Status
Hypertension 21 36 30
No
Hypertension
48 26 19
Total
e. Find the probability that a randomly selected individual is
either a nonsmoker or a
heavy smoker (OR, thus we use the union formula)
f. Find the probability that a randomly selected individual is
not hypertensive but (but
means AND) is a heavy smoker.

19. 15
Example:
Consider two events, A and B, of a sample space such that P (A)
= P (B) = .6
a. Is it possible that the events A and B are mutually exclusive?
Explain.
b. If the events A and B are independent, find the probability
that the two events occur
together, i.e., P (A and B)
c. If A and B are independent, find the probability that at least
one of the two events will
occur, i.e., P (A or B)
d. Suppose P (B/A) = .5, in this case are A and B independent or
dependent?

20. e. Suppose P (B/A) = .5. Find the P (A or B)
16
Example:
For two events A and B, P (A) = .6, P (B) = .4, P (B/A) = .6
a. Are A and B independent or dependent?
b. Find P (A/B)
c. At least one, i.e. P (A or B)
d. P(not A and not B)
e. P (A but not B)

21. 17
Def: In probability theory, a set of events is collectively
exhaustive if it incldes all
possible outcomes.
Ex: A fair die is rolled. The outcomes 1, 2, 3, 4, 5, and 6 are
collectively exhaustive,
because they encompass the entire range of possible outcomes.
Bayes Theorem
Def: Events 1 2 kB , B , .B … are said to partition a sample
space S if two conditions
exist:
a. i jB B ∩ = ∅ = the null set for any pair i and j
b. 1 2 kB U B U .U B … = Sample Space S

22. Bayes Theorem: If 1 2 kB , B , .B … partition S, and if A is
any event in the sample space
S, then
k
i 1
P(A / Bj).P(Bj)P(Bj / A)
P(A / Bi).P(Bi)
=
=
∑
B1 B2 B3
B4

23. 18
Further explanation:
19
Example of Bayes Theorem: In New York City, 51% of all

24. adults are males. One
masked adult was seen holding up an Apple Bank branch.
a) What is the probability that the person is a male?
b) It is known that immediately after leaving the bank branch,
the person was
smoking a cigar. Also, it is known that 9.5% of males smoke
cigars whereas 1.7% of
females smoke cigars based on data from the Substance Abuse
and Mental Health
Services Administration. Use this additional information to
find the probability that
the person holding up the bank was a female.
c) If the person was smoking a cigar, what is the probability
that the person who
held up the bank is male?
d) Use a table to obtain the same answers.

25. 20
A screening or diagnostic test for some condition is Positive
(+T) if it states that the
condition is present and negative (-T) if it indicates that the
condition is absent.
The validity of a test is defined as the ability of a test to
distinguish between those who
have the disease and those who do not. Validity has 2
components:
1. The sensitivity of the test is defined as the ability to correctly
identify those who
have the disease. i.e. P(+T/D)
2. The specificity of the test is defined as the ability to
correctly identify those who
do not have the disease. P(-T/D’)
Note: In order to calculate the sensitivity and specificity of a
test, we must know who
really has the disease and who does not from some other source
than the diagnostic
test. This external source of truth is the gold standard,
identifying the disease status of
each member of the population.
The prevalence of a disease is the probability of currently
having the disease. It is the
(# of people who currently have the disease) ÷ (# of people in

26. the population)
Term Brief Definition Notation
Sensitivity The probability of a positive test result given
that the individual tested actually has the
disease
P( +T | +D )
Specificity The probability of a negative test result given
that the individual tested does not have the
disease
P( −T | −D )
P(False Positive) Probability of a positive test result given the
individual does not have the disease
P( +T | −D )
P(False Negative) Probability of a negative test result given the
individual does have the disease
)|( +− DTP
Prevalence Proportion of individuals that have a disease at
a given point in time
)( +DP
Positive predictive
value
Probability of disease given a positive test result )|( ++ TDP

27. Negative predictive
value
Probability of not having disease given a
negative test result
)|( −− TDP
21
Example: Air traffic controllers are required to undergo random
drug testing. A urine
test detects the presence of amphetamines and barbiturates. The
sensitivity of the test
is known to be .96 and the specificity is .93. Based on past
drug testing of air traffic
controllers, the FAA reports that the probability of drug use at a
given time is .007.
Tree Diagram for Air traffic controllers
Contingency Table for Air traffic controllers

28. a. What is the probability that a randomly selected air traffic
controller will test positive?
(If the test comes back positive, the individual will have to
undergo a second test that
is more accurate, but more expensive than the urine test.)
b. Find the probability that a person truly uses drugs, given that
his test is positive, i.e.,
find the positive predictive value.
22
c. Find the probability that the individual is not a drug user
given that the test result is
negative i.e. negative predictive value
d. How does the positive predictive value change if the
prevalence rate increases to 15%?

29. e. How does the negative predictive value change if the
prevalence is 15%?
23
Example: Typical values reported for the mammogram which is
used to detect breast
cancer are sensitivity = .86, specificity = .88. Of the women
who undergo mammograms
at any given time, about 1% is estimated to actually have breast
cancer.
Tree Diagram for Mammogram
Contingency Table for Mammogram

30. A. Prevalence = .01
a. Find the probability of a positive test
b. Of the women who receive a positive mammogram, what
proportion actually
have breast cancer?
c. If a woman tests negative, what is the probability that she
does not have breast
cancer?
24
B. The prevalence rate for all women undergoing a mammogram
was changed to .001
Tree diagram for young women

31. Contingency table for young women
a. Find the probability that a randomly selected young woman
who tests positive
actually has breast cancer.
b. Find the probability that a randomly selected young woman
who tests negative
does not have the disease.