Lec 1 probability

9/22/2018
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SBE 304: Bio-Statistics
Probability
Dr. Ayman Eldeib
Systems & Biomedical
Engineering Department
Fall 2018
SBE 304: Probability
Outline
• Introduction
• Review of Venn Diagrams & Set Notation
• A Probabilistic Model for an Experiment
• The Probability of an Event
• Theory of Probability
• Calculating the Probability of an Event
• Tools for counting sample points/events
• Characterizing a set of Measurements

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Statistics
Descriptive Inferential
Correlational GeneralisingOrganising,
summarising &
describing data
Significance
Introduction
Relationships
General Definition
 The mathematical theory of probability deals
with patterns that occur in random events.
 Probability is a way of expressing knowledge
or belief that an event will occur. The
probability of a random event denotes the
relative frequency of occurrence of an
experiment's outcome, when repeating the
experiment.
 Probability is a way to represent an
individual’s degree of belief in a statement, or
an objective degree of rational belief, given
the evidence.

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Venn Diagrams & Set Notation
Universal Set
Let's say that our universe contains the numbers 1, 2, 3,
and 4; i.e. S = {1, 2, 3, 4}.
Subset
Let A be the set containing the numbers 1 and 2; that is, A =
{1, 2}.
A ⊆ S
Let B be the set containing the numbers 2 and 3; that is, B =
{2, 3}.
B ⊆ S
The Venn diagram is made up of two or more non- or
overlapping circles/shapes. It is often used in mathematics to
show relationships between sets.
Venn Diagrams & Set Notation Cont.
Set notation pronunciation meaning Venn diagram answer
A U B "A union B"
Everything that is in
either of the sets {1, 2, 3}
A ^ B
or
( A ∩ B)
"A intersect B"
only the things that are in
both of the sets {2}
Ac
or ~A or A
"A complement",
or "not A"
Everything in the universe
outside of A {3, 4}
A – B
"A minus B", or
"A complement B"
everything in A except for
anything in its overlap
with B
{1}
~(A U B) "not (A union B)"
Everything outside
A and B {4}
~(A ^ B)
or
~( A ∩ B)
"not (A intersect B)"
everything outside of the
overlap of A and B {1, 3, 4}
-

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Null Set
The Null, or empty, set denoted by Ø, is the set consisting
of no points/members. Thus, Ø is a subset of every set.
Mutually Exclusive or Disjoint
Two sets, A and B are said to be disjoint if A ∩ B = Ø.
Cont.
Given the following Venn diagram, shade in
A ∩ C
Examples
Given the following Venn diagram, shade in
A U (B – C)
(B – C) U A
Cont.
A ∩ C

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 (A‘) ‘= A
 The distributive law for set operations implies that
 (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)
 (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)
 DeMorgan’s laws imply that
 (A ∪ B) ‘ = A‘ ∩ B‘
 (A ∩ B) ‘ = A‘ ∪ B‘
 Also, remember that
 A ∪ B = B ∪ A
 A ∩ B = B ∩ A
Cont.
A Probabilistic Model for an Experiment
A spinner has 4 equal sectors colored yellow, blue, green and
red. What are the chances of landing on blue after spinning
the spinner? What are the chances of landing on red?
Problem
Solution
The chances of landing on blue are 1 in 4, or one fourth.
The chances of landing on red are 1 in 4, or one fourth.

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A Probabilistic Model for an Experiment
Definition Example
An experiment is a situation involving
chance or probability that leads to
results called outcomes. It is the process
by which an observation is made
In the problem above, the experiment
is spinning the spinner.
An outcome is the result of a single trial
of an experiment.
The possible outcomes are landing on
yellow, blue, green or red.
An event is one or more outcomes of an
experiment.
One event of this experiment is landing
on blue.
Probability is the measure of how likely
an event is.
The probability of landing on blue is
one fourth.
Cont.
The Probability of an Event
Probability Of An Event
P(A) =
The Number Of Ways Event A Can Occur
The total number Of Possible Outcomes
The probability of event A is the number of ways
event A can occur divided by the total number of
possible outcomes.

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A spinner has 4 equal sectors colored yellow, blue, green
and red. After spinning the spinner, what is the probability
of landing on each color?
Experiment
Outcomes
The possible outcomes of this experiment are yellow, blue,
green, and red
Cont.
Probabilities
P(yellow) =
# of ways to land on yellow
total # of colors
=
1
4
P(blue) =
# of ways to land on blue
total # of colors
=
1
4
P(green) =
# of ways to land on green
total # of colors
=
1
4
P(red) =
# of ways to land on red
total # of colors
=
1
4
Cont.

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A single 6-sided die is rolled. What is the probability of
each outcome? What is the probability of rolling an even
number? of rolling an odd number?
Experiment
Outcomes
The possible outcomes of this experiment are 1, 2, 3, 4, 5
and 6.
Cont.
P(3) =
# of ways to roll a 3
total # of sides
=
1
6
P(even) = # ways to roll an even number
total # of sides
=
3
6
=
1
2
P(odd) = # ways to roll an odd number
total # of sides
=
3
6
=
1
2
Probabilities
Cont.

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Theory of Probability
A set of all possible outcomes. It is usually denoted by S
The Sample Space
A Discrete Sample Space
It contains either a finite or a countable number of distinct
sample points/events.
Event Probability
A
not A
A and B
A or B
A given B
The Multiplicative Law
The Additive Law
Bayes’ Rule
Cont.

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 B∪B‘ = Ω, the universal event.
 B∩B‘ = Φ, an empty event.
 A = A ∩ (B ∪ B‘)
 A = A∩B ∪ A∩B‘
since A∩B and A∩B ‘ are mutually exclusive
 P(A) = P(A∩B ∪ A∩B ‘)
 P(A) = P(A∩B) + P(A∩B ‘)
 P(A) = P(A|B)P(B) + P(A|B ‘)P(B ‘)
Cont.
 A doctor is called to see a sick child. The doctor has prior
information that 90% of sick children in that neighborhood
have the flu, while the other 10% are sick with measles. Let
F stand for an event of a child being sick with flu and M
stand for an event of a child being sick with measles.
Assume for simplicity that F ∪ M = Ω, i.e., that there no other
maladies in that neighborhood.
 A well-known symptom of measles is a rash (the event of
having which we denote R). P(R|M) = .95. However,
occasionally children with flu also develop rash, so that
P(R|F) = 0.08.
 Upon examining the child, the doctor finds a rash. What is
the probability that the child has measles?.
Example I
Cont.

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 P(M|R) = .95 × .10 / (.95 × .10 + .08 × .90)
 P(M|R) ≈ 0.57
Solution I
Cont.
P(R|M) P(M) / (P(R|M) P(M) + P(R|F) P(F)) P(M|R) =
 In a study, physicians were asked what the odds of breast
cancer would be in a woman who was initially thought to
have a 1% risk of cancer but who ended up with a positive
mammogram result (a mammogram accurately classifies
about 80% of cancerous tumors and 90% of benign tumors.)
 95 out of a hundred physicians estimated the probability of
cancer to be about 75%. Do you agree?
Example II
Introduce the events:
P - mammogram result is positive,
B - tumor is benign,
M - tumor is malignant.
Cont.

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Introduce the events:
P - mammogram result is positive,
B - tumor is benign,
M - tumor is malignant.
Bayes' formula in this case is
P(M|P) = P(P|M) P(M) / (P(P|M) P(M) + P(P|B) P(B))
= .80 × .01 / (.80 × .01 + .10 × .99)
≈ 0.075
≈ 7.5%.
A far cry from a common estimate of 75%.
Solution II
Cont.
Calculating the Probability of an Event
I. Define the experiment.
II. List all the simple outcomes associated with the
experiment. These outcomes constitute the sample
space.
III. Assign reasonable probabilities to these outcomes.
Make sure each probability is greater than or equal to
zero and the sum of the probabilities is 1. If all
outcomes are equally likely the probability of each will
be 1/n.
IV. Define the event of interest A and see the outcomes in
the sample space which makes A occur.
V. Find P(A) by adding the probabilities of these outcomes
which make A occur.

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 If the sample space contains a very large number of
sample points, complete itemization will be tedious, time
consuming and might be practically impossible.
 When this occurs we need not list the points but just use
combinatorial analysis to know the number of points in the
sample space (N) and the event space (n).
 If outcomes are equally likely the probability of the event is
n/N.
Notes:
Cont.
P(A ∪ B ∪ C) = P((A ∪ B) ∪ C)
= P(A ∪ B) + P(C) – P((A ∪ B) ∩ C)
= P(A) + P(B) – P(A ∩ B) + P(C)
– P((A ∩ C) ∪ (B ∩ C))
= P(A) + P(B) + P(C) – P(A ∩ B)
-[P(A ∩ C) + P(B ∩ C) – P(A ∩B ∩ C)]
= P(A) + P(B) + P(C) – P(A ∩ B)
- P(A ∩ C) - P(B ∩ C) + P(A ∩B ∩ C)
Three or More Events
Cont.
 P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

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 A collection of events, E1, E2, …Ek, is said to be mutually
exclusive for all pairs:
Ei ∩ Ej = Φ, an empty event
Three or More Events
For a collection of mutually exclusive events,
P(E1 ∪ E2 ∪ ….. ∪ Ek) = P(E1) + P(E2) + …… + P(Ek)
Cont.
 P(A) = P(A∩B) + P(A∩B ‘)
 P(A) = P(A|B)P(B) + P(A|B ‘)P(B ‘)
 In general, a collection of mutually exclusive sets, E1, E2,
…Ek, such that E1 ∪ E2 ∪ ….. ∪ Ek = S, is said to be
exhaustive.
 P(A) = P(A∩E1) + P(A∩E2) + ….. P(A ∩ Ek)
 P(A) = P(A| E1)P(E1) + P(A| E2)P(E2) + ….. + P(A| Ek)P(Ek)
 P(Ej| A) = P(A| Ej)P(Ej) / ∑ P(A| Ei)P(Ei) {i=1,2, ….,k)}
Three or More Events Total Probability Rule
Cont.

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Tools for counting sample points/events
Multiplication of Choices
With m elements a1, a2, ….., am and n elements b1, b2,
….., bn, it is possible to form mn = m * n pairs containing
one element from each group
Multiplication of Choices
Example:
There are seven different trails to the top of the mountain. In how
many different ways can a person hike up and down the
mountain if
1.he must take the same trail both ways;
2.he can, but need not, take the same trail both ways;
3.he does not want to take the same trail both ways.
7 * 7
7 * 6
7
Cont.

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Permutation
An ordered arrangement of r distinct objects is called a
permutation (n!). The number of ways of ordering n distinct
objects taken r at a time will be designed by the symbol
Pr
n
Pr
n = n(n - 1)(n - 2)…..(n – r + 1) = n! / (n - r)!
Tools for counting sample points/events Cont.
Permutation
Consider a set of elements, such as S {a, b, c}. A
permutation of the elements is an ordered sequence of
the elements. For example, abc, acb, bac, bca, cab, and cba
are all of the permutations of the elements of S.
Example:
n! = n(n - 1)(n - 2)…..(1)
3! = 3 * 2 * 1 = 6
Cont.

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Permutation
A printed circuit board has eight different locations in which a
component can be placed. If four different components are to be
placed on the board, how many different designs are possible?
Example:
Each design consists of selecting a location from the eight locations
for the first component, a location from the remaining seven for the
second component, a location from the remaining six for the third
component, and a location from the remaining five for the fourth
component. Therefore,
P4
8 = 8*7*6*5= 8! / (8 - 4)! = 1680
1680 different designs are possible.
Cont.
Partitioning n distinct objects into k distinct groups
The number of ways N of partitioning n distinct objects into
k distinct groups containing n1, n2, ….., nk, objects,
respectively, where each object appears in exactly one
group and ∑ ni = n, where i=1, 2, ….., k, is
N = ( )n
n1 n2 ….. nk =
________________
n!
n1! n2! ….. nk!
Tools for counting sample points/events Cont.

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Partitioning n distinct objects into k distinct groups
A part is labeled by printing with four thick lines, three
medium lines, and two thin lines. If each ordering of the
nine lines represents a different label, how many different
labels can be generated by using this scheme?
Example:
( )N =
n
n1 n2 ….. nk =
________________
n!
n1! n2! ….. nk!
= 9!/4!3!2!
Cont.
Combinations
The number of unordered subsets of size r chosen
(without replacement) from n available objects is
( )n
r =
C
n
r = ________________
n!
r! (n – r)!
Cont.

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Combinations
Consider the problem of choosing two patient monitoring systems for
ICU out of a group of five and imagine that the monitoring systems
vary in competence, 1 being the best, 2 second best, and so on, for 3,
4, and 5. These rating are of course unknown to the nurse.
A- Find the number of ways of selecting two systems out of five.
B- Let A denote the event that exactly one of the two best systems
appears in a selection of two out of five. Find the number of sample
points in A and P(A).
Example:
(5
2) = 5!/2!3! = 10
n = (2
1) (3
1) = 2*3=6 P(A) = 0.6
Cont.
When male and female workers become sick, what do they do? A
survey was carried out and the responses of 500 workers were
tabulated
Example:
Action Male (M) Female
Consult a doctor (D) 115 70
Consult a pharmacist for a pain killer medication 60 140
Go to a hospital 75 40
1. What is the probability that a man would consult a pharmacist?
2. What proportion of workers consults a doctor?
3. If a worker is known to be a female, what is the probability that she will go to a
hospital?
4. Among workers consulting a pharmacist, what proportion are males?
5. Are gender and consulting a doctor independent? Verify your answer.
Dependent: P(D) ≠ P(D|M)
60/200 = 0.3
60/500 = 0.12
185/500 = 0.37
40/250 = 0.16

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Characterizing a set of Measurements
The Mean
The mean of a sample of n measured responses
y1, y2, ….., yn is given by
Y =
__
___1
n
∑
n
i = 1
yiµ =
The Variance
The variance of a sample of measurements y1, y2, ….., yn
is the sum of the differences between the measurements
and their mean, divided by n – 1.
S2 =
_______1
n - 1
∑
n
i = 1
(yi – µ)2
σ 2 =

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The standard deviation
The standard deviation of a sample of measurements is
the positive square root of the variance; that is,
S = S2 = σ 2 = σ
Graphical Methods – Frequency Histogram
A frequency distribution is a
tabulation of the values that one
or more variables take in a
sample. Each entry in the table
contains the frequency or count
of the occurrences of values
within a particular group or
interval, and in this way the
table summarizes the
distribution of values in the
sample.
Rank Degree of agreement Number
1 Strongly agree 20
2 Agree somewhat 30
3 Not sure 20
4 Disagree somewhat 15
5 Strongly disagree 15

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Graphical Methods – Frequency Histogram
0
5
10
15
20
25
30
35
Strongly agree Agree
somewhat
Not sure Disagree
somewhat
Strongly
disagree
1 2 3 4 5
Number
Number
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