Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Lesson 3
Chapter 2: Introduction to Probability
Michael Akritas
Department of Statistics
The Pennsylvania State University
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
1
Sample Spaces, Events and Set Operations
2
Equally Likely Outcomes
The Probability Mass Function and Probability Sampling
Counting Techniques
3
Axioms and Properties of Probability
4
Conditional Probability
The Law of Total Probability and Bayes Theorem
5
Independent Events
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Overview
This chapter presents the basic ideas and tools of classical
probability, which is the branch of probability that arose
from games of chance. This includes
an introduction to combinatorial theory, and
an introduction to the concepts of conditional probability
and independence.
Probability evolved to deal with modeling the randomness
of phenomena such as the number of earthquakes, the
amount of rainfall, the life time of a given electrical
component, or the relation between education level and
income, etc. Such probability models will be discussed in
Chapters 3 and 4.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Sample Spaces
Definition
The set of all possible outcomes of a random experiment is
called the sample space of the experiment, and will be
denoted by S.
Example
a) Give the sample space of the experiment which selects two
fuses and classifies each as non-defective or defective.
b) Give the sample space of the experiment which selects
two fuses and records how many are defective.
c) Give the sample space of the experiment which records
the number of fuses inspected until the second defective is
found.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
Three undergraduate students from a particular university are
selected and their opinions about a proposal to expand the use
of solar energy are recorded on a scale from 1 to 10.
a) Give the sample space of this experiment. What is the size
of this sample space?
b) Describe the sample space if only the average of the three
responses is recorded. What is the size of this sample
space?
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Solution:
a) When the opinions of three students are recorded the set of
all possible outcomes consist of the triplets (x1 , x2 , x3 ), where
x1 = 1, 2, . . . , 10 denotes the response of the first student,
x2 = 1, 2, . . . , 10 denotes the response of the second student,
and x3 = 1, 2, . . . , 10 denotes the response of the third student.
Thus, the sample space is described as
S1 = {(x1 , x2 , x3 ) : x1 = 1, 2, . . . , 10, x2 = 1, 2, . . . , 10,
x3 = 1, 2, . . . , 10}.
There are 10 · 10 · 10 = 1000 possible outcomes.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
b) The easiest way to describe the sample space, S2 , when the
three responses are averaged is to say that it is the collection of
distinct averages (x1 + x2 + x3 )/3 formed from the 1000 triplets
of S1 . The word ”distinct” is emphasized because the sample
space lists each individual outcome only once, whereas several
triplets might result in the same average. For example, the
triplets (5, 6, 7) and (4, 6, 8) both yield an average of 6.
Determining the size of S2 is can be done, most easily, with the
following R commands:
S1=expand.grid(x1=1:10,x2=1:10,x3=1:10)
length(table(rowSums(S1)))
Michael Akritas
# lists all triplets in S1
# gives the number
of different sums
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Events
In experiments with many possible outcomes, investigators
often classify individual outcomes into distinct categories.
For example, the opinion ratings may be classified into low
(L = {0, 1, 2, 3}), medium (M = {4, 5, 6}) and high
(H = {7, 8, 9, 10}).
Definition
Such collections of individual outcomes, i.e. subsets of the
sample space, are called events. An event consisting of only
one outcome is called a simple event.
Events are denoted by letters such as A, B, C, etc.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
1
In selecting one card at random from a deck of cards, the
event
A = {the card is a spade} consists of 13 outcomes.
2
The event E = {at most 3 heads in five tosses of a coin}
consists of the outcomes 0, 1, 2, 3.
We say that a particular event A has occurred if the
outcome of the experiment is a member of (i.e. contained
in) A.
The sample space of an experiment is an event which
always occurs when the experiment is performed.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Set Operations
The union, A ∪ B, of events A and B, is the event
consisting of all outcomes that are in A or in B or in both.
The intersection, A ∩ B, of A and B, is the event
consisting of all outcomes that are in both A and B.
The complement, A0 or Ac , of A is the event consisting of
all outcomes that are not in A.
The events A and B are said to be mutually exclusive or
disjoint if they have no outcomes in common. That is, if
A ∩ B = ∅, where ∅ denotes the empty set.
The difference A − B is defined as A ∩ B c .
A is a subset of B, A ⊂ B, if e ∈ A implies e ∈ B.
Two sets are equal, A = B, if A ⊂ B and B ⊂ A.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Union of A and B
A∪B
A
B
Intersection of A and B
A∩B
A
B
Figure: Venn diagrams for union and intersection
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The complement of A
Ac
The difference operation
A−B
A
A
B
Figure: Venn diagrams for complement and difference
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
A
A
B
B
Figure: Venn diagram illustrations of A, B disjoint, and A ⊂ B
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Commutative Laws: a) A ∪ B = B ∪ A,
b) A ∩ B = B ∩ A
Associative Laws: a) (A ∩ B) ∩ C = A ∩ (B ∩ C)
b) (A ∪ B) ∪ C = A ∪ (B ∪ C)
Distributive Laws: a) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C),
b) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)
De Morgan’s Laws: a) (A ∪ B)c = Ac ∩ B c
b) (A ∩ B)c = Ac ∪ B c
• Two types of proof: (a) By Venn diagram (informal), and
(b) Show formally the equality of the two sides, i.e. show that
A ⊂ B and B ⊂ A, where A is the set on the left and B is the set
on the right of each equality.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
The following table classifies a population of 100 plastic disks in
terms of their scratch and shock resistance.
shock resistance
high
low
scratch
high 70
9
resistance low
16
5
Suppose that one of the 100 disks is randomly selected. What
is the sample space? Give the outcomes in the events a) ”the
selected disk has low shock resistance”, and b) ”the selected
disk has low shock resistance or low scratch resistance”.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Definition of Probability
The probability of an event E, denoted by P(E), is used to
quantify the likelihood of occurrence of E by assigning a
number from the interval [0, 1].
Higher numbers indicate that the event is more likely to
occur.
A probability of 1 indicates that the event will occur with
certainty, while a probability of 0 indicates that the event
will not occur.
Read Section 2.3.1.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Assignment of Probabilities
It is simplest to introduce probability in experiments with a
finite number of equally likely outcomes, such as those
used in games of chance, or simple random sampling.
Probability for equally likely outcomes
If the sample space consists of N outcomes which
are equally likely to occur, then the probability of
each outcome is 1/N.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Population Proportions as Probabilities
A unit is selected by s.r. sampling from a finite statistical
population of a categorical variable. If category i has Ni units,
then the probability the selected unit came from category i is
pi = Ni /N.
where N is the total number of units (so N =
P
Ni ). Thus,
(a) In rolling a die, the probability of a three is p = 1/6.
(b) If 160 out of 500 tin plates have one scratch, and one tin
plate is selected at random, the probability that the selected
plate has one scratch is p = 160/500.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Efron’s Dice
Die A: four 4s and two 0s
Die B: six 3s
Die C: four 2s and two 6s
Die D: three 5’s and three 1’s
Specify the events A > B, B > C, C > D, D > A.
Find the probabilities that A > B, B > C, C > D, D > A.
Hint: When two dice are rolled, the 36 possible outcomes are
equally likely.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Outline
1
Sample Spaces, Events and Set Operations
2
Equally Likely Outcomes
The Probability Mass Function and Probability Sampling
Counting Techniques
3
Axioms and Properties of Probability
4
Conditional Probability
The Law of Total Probability and Bayes Theorem
5
Independent Events
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Even when population units are selected with equal
probability, the outcomes of the random variable recorded
may not be equally likely.
For example, when die is rolled twice, each of the 36
possible outcomes are equally likely. But if we record the
sum of the two rolls, these outcomes are not equally likely.
Definition
The probability mass function, or pmf, of a discrete random
variable X , is a list of the probabilities p(x) for each value x of
the sample space SX of X .
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Example
Roll a die twice. Find the pmf of X = the sum of the two die
rolls.
Solution: SX = {2, 3, . . . , 12}. This list of possible values,
together with the corresponding probabilities, can be found with
the R commands:
S=expand.grid(X1=1:6,X2=1:6) ; table(S$X1+S$X2)
Try also S[which(S$X1+S$X2==7),].
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
It is useful to think of a random experiment as a sampling
from the sample space.
Example
1
Sampling in the tin plate example can be thought of as
sampling from S = {0, 1, 2}. But it should not be simple
random sampling (why?).
2
One US citizen aged 18 and over is selected by
s.r.sampling and his/her opinion regarding solar energy is
rated on the scale 0, 1, . . . , 10. This can be thought of as
random (but not simple random!) sampling from
S = {0, 1, . . . , 10}.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Definition
When a sample space is thought of as the set from which we
sample, we refer to it as sample space population. The
random sampling from the sample space population (which is
need not be simple random sampling) is called probability
sampling, or sampling from a pmf.
This idea makes it possible to think of different experiments
as sampling from the same population. For example
Inspecting 50 products and recording the number of
defectives, and
Interviewing 50 people and recording if they read New York
Times
can both be thought as probability sampling from their
common sample space S = {0, 1, 2, . . . , 50}.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Example (Simulating an Experiment with R)
Use the pmf of X = sum of two die rolls to simulate 1000
repetitions of the experiment which records the sum of two die
rolls. Take their mean and use it to guess the population mean.
The R commands are
S=expand.grid(X1=1:6,X2=1:6) ; pmf=table(S$X1+S$X2)/36
mean(sample(2:12, size=1000, replace=T, prob=pmf))
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Outline
1
Sample Spaces, Events and Set Operations
2
Equally Likely Outcomes
The Probability Mass Function and Probability Sampling
Counting Techniques
3
Axioms and Properties of Probability
4
Conditional Probability
The Law of Total Probability and Bayes Theorem
5
Independent Events
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Why Count?
In classical probability counting is used for calculating
probabilities. For the probability of an event A we need to
know
the number of outcomes in A, N(A), and
if the sample space consists of a finite number of equally
likely outcomes, also the total number of outcomes, N(S),
because
P(A) =
N(A)
N(S)
Some counting questions are difficult (e.g. how many
different five-card hands are possible from a deck of 52
cards?) and thus we need specialized counting techniques.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Some Counting Questions
1
How many samples of size n can be formed from N units?
The answer is the number of combinations
of n objects
selected from N, denoted by Nn , and equals
N
N!
, where k! = 1 × 2 × · · · × k .
=
n!(N − n)!
n
52
52!
For example,
= 2, 598, 960 is the number of
=
5
5!47!
hands of n = 5 cards that can be formed from a deck of
N = 52 cards.
Knowing that N(S) = 2, 598, 960 we can calculate the
probability of individual hands such as the hand with 4 aces
and the king of hearts. What is it?
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
1
What is the probability of A = {the hand has 4 aces}? We
need to learn how to determine N(A).
2
When inspecting n items as they come off the assembly
line, the probability of the event
E = {k of the n inspected items are defective}
is calculated using the concept of independence and the
answer to the question
How many different n-long sequences consisting of k 1s (for
defective) and n − k 0s (for non defective) can be formed?
The answer is (again) the number of combinations kn .
In what follows we will justify the formula for kn . In the
process we will learn how to answer question 2.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
The Product Rules
The Simple Product Rule: Suppose a task can be
completed in two stages. If stage 1 has n1 outcomes, and
if stage 2 has n2 outcomes regardless of the outcome in
stage 1, then the task has n1 × n2 outcomes.
Example
If A = {the hand has 4 aces}, find N(A).
Solution. The task is to form a hand with 4 aces. It can be
completed in two stages: First select the 4 aces, and then
select one additional card. Here n1 = 1 and n2 = 48 (why?).
Thus, N(A) = 48.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Example (1)
1
In how many ways can we select the 1st and 2nd place
winners from the four finalists Niki, George, Sophia and
Martha?
Answer: 4 × 3 = 12.
2
In how many ways can we select two from Niki, George,
Sophia and Martha?
12
4
Answer:
(Why?) Note: 6 = # of combinations =
.
2
2
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
The General Product Rule: If a task can be completed in
k stages and stage i has ni outcomes, regardless of the
outcomes the previous stages, then the task has
n1 × n2 × · · · × nk outcomes
Example (2)
1
In how many ways can we select a 1st, 2nd and 3rd place
winners from Niki, George, Sophia and Martha?
Answer: 4 × 3 × 2 = 24.
2
In how many ways can we select three from Niki, George,
Sophia and Martha?
24
Answer:
(Why?) Note:
6
4
4 = # of combinations =
.
3
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Permutations
The answer to Example (1), part 1, i.e. 12, is the number of
permutations of 2 items selected from 4.
The answer to Example (2), part 1), i.e. 24, is the number
of permutations of 3 items selected from 4.
Definition
The number of ordered selections (i.e. when we keep track of
the order of selection) of k items from n is called the number of
permutations of k items selected from n, it is denoted by Pk,n ,
and equals
Pk,n = n × (n − 1) × . . . × (n − k + 1) =
Michael Akritas
n!
(n − k)!
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Combinations
In the answer to Example (1), part 2, i.e. 12
2 , the 2 in the
denominator is the number of permutations of 2 items
selected from 2 (P2,2 = 2 × 1).
In the answer to Example (2), part 2), i.e. 24
6 , the 6 in the
denominator is the number of permutations of 3 items
selected from 3 (P3,3 = 3 × 2 × 1).
Extending the rational used to obtain these answers, we have
The number of combinations of k items selected from a
group of n is
Pk,n
n
n!
=
=
k
k!
k!(n − k)!
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
The numbers kn are called binomial coefficients because
of the Binomial Theorem:
n X
n k n−k
n
(a + b) =
a b
.
k
k=0
Example
a) How many n-long sequences consisting of k 1s and n − k
zeros can be formed?
b) How many paths going from the lower left corner of a 4 × 3
grid to its upper right corner? Assume one is allowed to
move either to the right or upwards.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Multinomial Coefficients
Suppose we want to assign 8 engineers to work on
projects A, B, and C, so that 3 work on project A, 2 work on
B, and 3 work on C. In how many ways can this be done?
The number of ways n units can be divide in r groups of
specified sizes is given by
n
n1 , n2 , . . . , nr
=
n!
n1 !n2 ! · · · nr !
These numbers are called multinomial coefficients
because of the Multinomial Theorem.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Example
An order comes in for 5 palettes of low grade shingles. In the
warehouse there are 10 palettes of high grade, 15 of medium
grade, and 20 of low grade shingles. An inexperienced shipping
clerk is unaware of the distinction in grades of asphalt shingles
and he ships 5 randomly selected palettes.
1
How
many different groups of 5 palettes are there?
45
5 = 1, 221, 759.
2
What is the probability
45 that all of the shipped palettes are
20
low grade? 5 / 5 = 15, 504/1, 221, 759 = 0.0127.
3
What is the probability that 2 of the shipped palettes are of
grade and 3 are from low grade?
hmedium
i 45
15 20
/ 5 = (105 × 2280)/1, 221, 759 = 0.0127 =
2
3
0.1959.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Example
A communication system consists of 15 indistinguishable
antennas arranged in a line. The system functions as long as
no two non-functioning antennas are next to each other.
Suppose six antennas stop functioning.
a) How many different arrangements of the six
non-functioning antennas result in the system being
functional? (Hint: The 9 functioning antennas, lined up
among themselves, define 10 possible locations for the 6
non-functioning antennas so the system functions.)
b) If the arrangement of the 15 antennas is random, what is
the probability the system is functioning?
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Example
What is the probability that 5 randomly dealt cards form a full
house?
Solution: First, the number of all 5-card hands is
52!
52
5 = 5!47! = 2, 598, 960. Next, think of the task of forming a
full house as consisting of two stages. In Stage 1 choose two
cards of the same kind, and in stage 2 choose three cards of
the same kind. Since
there are 13 kind of cards, stage 1 can be
13 4
completed in 1 2 = (13)(6) = 78 ways (why?). For each
outcome of stage 1, the task of stage 2 becomes that of
selecting three of a kind fromone
of the remaining 12 kinds.
12 4
This can be completed in 1 3 = 48 ways. Thus there are
(78)(48) = 3, 744 possible full houses, and the desired
probability is 0.0014.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Probability Mass Function and Probability Sampling
Counting Techniques
Reading assignment
Read Examples 2.3.8 - 2.3.14
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Axioms of Probability
The axioms governing any assignment of probabilities are:
Axiom 1: P(A) ≥ 0, for all events A
Axiom 2: P(S) = 1
Axiom 3: If A1 , A2 , . . . are disjoint
P(A1 ∪ A2 ∪ . . .) =
∞
X
P(Ai )
i=1
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Properties of Probability
Proposition
1
If A and B are disjoint, P(A ∩ B) = 0.
2
If E1 , . . . , Em are disjoint, then
P(E1 ∪ · · · ∪ Em ) = P(E1 ) + · · · + P(Em )
3
4
5
If A ⊂ B then P(A) ≤ P(B).
P(A) = 1 − P(Ac ), for any event A.
P
P(A) = {all simple events Ei in A} P(Ei )
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Proposition
1
If S = {s1 , . . . , sn } and the n outcomes are equally likely,
then P(si ) = 1/n, for all i.
2
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
3
P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
− P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
The probability that a firm will open a branch office in Toronto is
0.7, that it will open one in Mexico City is 0.4, and that it will
open an office in at least one of the cities is 0.8. Find the
probabilities that the firm will open an office in:
1
neither of the cities
2
both cities
3
exactly one of the cities
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
Use the R commands attach(expand.grid(X1=0:1,X2=0:1,
X3=0:1,X4=0:1)); table(X1+X2+X3+X4)/length(X1) to find the
pmf of the random variable X = number of heads in four flips of
a coin. (The answer is
x
p(x)
0
0.0625
1
0.25
2
0.375
3
0.25
4
.)
0.0625
(a) What can we say about the sum of all probabilities?
(b) What is P(X ≥ 2)?
Read also Examples 2.4.2, 2.4.3.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Updating Probabilities
In experiments with multivariate outcome variable, knowledge
of the value of one variable may help predict another.
For now, the word prediction will mean update the probabilities
of events regarding the other variable. The updated
probabilities are called conditional probabilities.
Knowing a man’s height helps update the probability that
he weighs over 170lb.
Knowing a person’s education level helps update the
probability of that person being in a certain income
category.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Given partial information regarding the outcome of simple
random selection restricts the population. The outcome can be
regarded as s.r.s. from the restricted population.
Example
If the outcome of rolling a die is known to be even, what is
the probability it is a 2?
If the selected card from a deck is known to be a figure
card, what is the probability it is a king?
Given event A = {household has a cat}, what is the
probability of B = {household has a dog}?
http:
//stat.psu.edu/˜mga/401/fig/Venn_Square.pdf
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
The Multiplication Rule
The conditional probability of the event A given the
information that event B has occurred is denoted by P(A|B)
and equals
P(A|B) =
P(A ∩ B)
, provided P(B) > 0
P(B)
THE MULTIPLICATION RULE: The definition of P(A|B) yields
an alternative formula for P(A ∩ B):
P(A ∩ B) = P(A|B)P(B) or P(A ∩ B) = P(B|A)P(A)
The rule extends to more than two events. For example,
P(A ∩ B ∩ C) = P(A)P(B|A)P(C|A ∩ B)
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
1
40% of bean seeds come from supplier A and 60% come
from supplier B. Seeds from supplier A have 50%
germination rate while those from supplier B have a 75%
rate. What is the probability that a randomly selected seed
came from supplier A and will germinate?
ANSWER: P(A ∩ G) = P(G|A)P(A) = 0.5 × 0.4 = 0.2
2
Three players are dealt a card in succession. What is the
probability that the 1st gets an ace, the 2nd gets a king,
and the 3rd gets a queen?
ANSWER: P(A ∩ B ∩ C) = P(A)P(B|A)P(C|A ∩ B) =
4 4 4
52 51 50 = 0.000454
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
Fifteen percent of all births involve Cesarean (C) section.
Ninety-eight percent of all babies survive delivery (S), whereas,
when a C section is performed the baby survives with
probability 0.96. What is the probability that a baby will survive
delivery if a C section is not performed?
Solution.
P(S ∩ C) = 0.96 × 0.15 = 0.144 (why?)
P(S ∩ C c ) = 0.98 − 0.144 = 0.836 (why?)
P(S|C c ) = 0.836/0.85 = 0.9835.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
Of the customers entering a department store 30% are men
and 70% are women. The probability a male shopper will
spend more than $50 is 0.4, and the corresponding probability
for a female shopper is 0.6. The probability that at least one of
the items purchased is returned is 0.1 for male shoppers and
0.15 for female shoppers. Find the probability that the next
customer to enter the department store is a woman who will
spend more than $50 on items that will not be returned.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Solution.
Let W = {customer is a woman},
B = {the customer spends >$50} and
R = {at least one of the purchased items is returned}. We
want the probability of the intersection of W , B and R c . By the
formula for the intersection of three events, this probability is
given by
P(W ∩ B ∩ R c ) = P(W )P(B|W )P(R c |W ∩ B)
= 0.7 × 0.6 × 0.85 = 0.357.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
The multiplication rule typically applies in situations where the
events whose intersection we wish to compute are associated
with different stages of an experiment. For example, in the
previous example there are three stages:
a) record customer’s gender,
b) record amount spent by customer, and
c) record whether any of the items purchased is subsequently
returned.
Therefore, by the generalized fundamental principle of
counting, this experiment has 2 × 2 × 2 = 8 different outcomes.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
> 50
0.4
M
0.6
< 50
0.1
R
0.9
Rc
0.1
R
0.9
Rc
0.15
R
0.85
Rc
0.15
R
0.85
Rc
0.3
0.7
W
0.4
< 50
0.6
> 50
Figure: Tree diagram for last example
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Outline
1
Sample Spaces, Events and Set Operations
2
Equally Likely Outcomes
The Probability Mass Function and Probability Sampling
Counting Techniques
3
Axioms and Properties of Probability
4
Conditional Probability
The Law of Total Probability and Bayes Theorem
5
Independent Events
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
The Law of Total Probability
Let the events A1 , A2 . . . , Ak be disjoint and make up the entire
sample space, and let B denote an event whose probability we
want to calculate, as in the figure
A1
B
A2
A3
A4
If we know P(B|Aj ) and P(Aj ) for all j = 1, 2, . . . , k , the Law of
Total Probability gives
P(B) = P(A1 )P(B|A1 ) + · · · + P(Ak )P(B|Ak )
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
1
40% of bean seeds come from supplier A and 60% come
from supplier B. Seeds from supplier A have 50%
germination rate while those from supplier B have a 75%
rate. What is the probability that a randomly selected seed
will germinate?
ANSWER: P(G) = P(A)P(G|A) + P(B)P(G|B)
= 0.4 × 0.5 + 0.6 × 0.75 = 0.65
2
Three players are dealt a card in succession. What is the
probability that the 2nd gets a king?
ANSWER:
4
Why?
52
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
Two dice are rolled and the sum of the two outcomes is
recorded. What is the probability that 5 happens before 7?
Solution 1: If En =
{no 5 or 7 appear on the first n − 1 rolls and a 5 appears on the nth},
then
P(∪∞
n=1 En )
=
∞
X
P(En )
n=1
=
∞
X
n=1
Michael Akritas
(1 −
2
10 n−1 4
)
=
36
36
5
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Solution 2: Let B be the desired event,
A1 = {first roll results in 5}, A2 = {first roll results in 7},
A3 = {first roll results in neither 5 not 7}. Then
P(B) = P(B|A1 )P(A1 ) + P(B|A2 )P(A2 ) + P(B|A3 )P(A3 )
= P(A1 ) + 0 + P(B)P(A3 ).
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
Two consecutive traffic lights have been synchronized to make
a run of green lights more likely. In particular, if a driver finds
the first light to be red, the second light will be green with
probability 0.9, and if the first light is green the second will be
green with probability 0.7. The probability of finding the first
light green is 0.6.
(a) Find the probability that a driver will find the second traffic
light green.
(b) Recalculate the probability of part (a) through a tree
diagram for the experiment which records whether or not a
car stops at each of the two traffic lights.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Solution.
(a) Let A and B denote the events that a driver will find the first,
respectively the second, traffic light green. Because the events
A and Ac constitute a partition of the sample space, according
to the Law of Total Probability
P(B) = P(A)P(B|A) + P(Ac )P(B|Ac )
= 0.6 × 0.7 + 0.4 × 0.9 = 0.42 + 0.36 = 0.78.
(b) The experiment has two outcomes resulting in the second
light being green which are represented by the paths with the
pairs of probabilities (0.6,0.7) and (0.4,0.9). The sum of the
probabilities of these two outcomes is as above.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
R
R
0.4
0.6
G
0.1
0.9
G
0.3
R
0.7
G
Figure: Tree diagram for previous example
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Bayes Theorem
Consider events B and A1 , . . . , Ak as in the Law of Total
Probability. Now, however, we ask a different question:
Given that B has occurred, what is the probability that a
particular Aj has occurred?
The answer is provided by the Bayes theorem:
P(Aj |B) =
P(Aj ∩ B)
P(Aj )P(B|Aj )
= Pk
P(B)
j=1 P(Ai )P(B|Ai )
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Bayes Theorem
Consider events B and A1 , . . . , Ak as in the Law of Total
Probability. Now, however, we ask a different question:
Given that B has occurred, what is the probability that a
particular Aj has occurred?
The answer is provided by the Bayes theorem:
P(Aj |B) =
P(Aj ∩ B)
P(Aj )P(B|Aj )
= Pk
P(B)
j=1 P(Ai )P(B|Ai )
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
1
40% of bean seeds come from supplier A and 60% come
from supplier B. Seeds from supplier A have 50%
germination rate while those from supplier B have a 75%
rate. Given that a randomly selected seed germinated,
what is the probability that it came from supplier A?
ANSWER:
P(A|G) =
2
0.2
P(A)P(G|A)
=
0.65
P(A)P(G|A) + P(B)P(G|B)
Given that the 2nd player got an ace, what is the
3
probability that the 1st got an ace? ANSWER:
(Why?)
51
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
The Law of Total Probability and Bayes Theorem
Example
Seventy percent of the light aircraft that disappear while in flight
in a certain country are subsequently discovered. Of the aircraft
that are discovered, 60% have an emergency locator, whereas
10% of the aircraft not discovered have such a locator.
Suppose a light aircraft has disappeared.
1
What is the probability that it has an emergency locator
and it will not be discovered?
2
What is the probability that it has an emergency locator?
3
If it has an emergency locator, what is the probability that it
will not be discovered?
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Probability of an Intersection
The formula for the probability of A ∪ B yields
P(A ∩ B) = P(A) + P(B) − P(A ∪ B).
A simpler formula is possible if A and B are independent:
For independent events,
P(A ∩ B) = P(A)P(B).
The above also serves as the definition of independent
events.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Independent events arise in connection with independent
experiments or independent repetitions of the same
experiment.
Two experiments are independent if there is no mechanism
through which the outcome of one experiment will
influence the outcome of the other.
A die is rolled twice. Are the two rolls independent?
Two cards are drawn without replacement from a deck of
cards. Are the two draws independent?
In two independent repetitions of an experiment, any event
associated with the first repetition will be independent of
any event associated with the second repetition.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
Toss a coin twice. Find the probability of two heads.
Solution. Since the two tosses are independent,
P([H in toss 1] ∩ [H in toss 2])
= P([H in toss 1])P([H in toss 2]) =
11
1
= .
22
4
Alternatively, since P([H in toss 1] ∩ [H in toss 2]) =
1
4
(why?),
and also P(H in toss 1)P(H in toss 1) = 14 ,
we can conclude that the two events are independent.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example (Fair game with unfair coin)
A biased coin results in heads with probability p (e.g. p = 0.3).
Flip this coin twice. If the outcome is (H,H) or (T,T) ignore the
outcome and flip the coin two more times. Repeat until the
outcome of the two flips is either (H,T) or (T,H). In the first case
you say you got tails, and in the second case you say you got
heads. Prove that now the probability of getting heads equals
0.5.
Solution: Ignoring the outcomes (H,H) and (T,T), is equivalent
to conditioning on the the event B = {(H, T ), (T , H)}. Thus,
P((T , H)|B) =
P((T , H))
P((T , H) ∩ B)
=
P(B)
P(B)
=
(1 − p)p
= 0.5.
p(1 − p) + (1 − p)p
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
Consider again Efron’s dice. That is Die A: four 4s and two 0s;
Die B: six 3s; Die C: four 2s and two 6s; Die D: three 5’s and
three 1’s. Find the probabilities that A > B, B > C, C > D, and
D > A using the properties of probability and the concept of
independence.
Solution (partial): Note that P(C > D) equals
P(C = 2 and D = 1) + P(C = 6 and D = 5 or 1) why?
= P(C = 2)P(D = 1) + P(C = 6)P(D = 5 or 1) why?
=
21 1
2
+ = .
32 3
3
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Independence of Multiple Events
When there are several independent experiments, events
associated with distinct experiments are independent.
But the definition is a bit more complicated:
Definition
The events A1 , . . . , An are mutually independent if
P(Ai1 ∩ Ai2 ∩ . . . ∩ Aik ) = P(Ai1 )P(Ai2 ) . . . P(Aik ) for any
sub-collection Ai1 , . . . , Aik of k events chosen from A1 , . . . , An
All conditions are needed because, for example,
P(A ∩ B ∩ C) = P(A)P(B)P(C)
does not imply that A, B, C are independent.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
Consider rolling a die and define the events A = {1, 2, 3},
B = {3, 4, 5}, C = {1, 2, 3, 4}. Verify that
P(A ∩ B ∩ C) = P(A)P(B)P(C),
but that A, B are not independent (and thus A, B, C are not
mutually independent).
Solution: First, since A ∩ B ∩ C = {3}, it follows that
P(A ∩ B ∩ C) =
114
1
= P(A)P(B)P(C) =
.
6
226
Next, A ∩ B = {3}, so P(A ∩ B) =
Michael Akritas
1
6
6= P(A)P(B) =
11
2 2.
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
The three components of the series system shown in the figure
fail with probabilities p1 = 0.1, p2 = 0.15 and p3 = 0.2,
respectively, independently of each other. What is the
probability the system will fail?
1
2
3
Figure: Components connected in series
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Example
The three components of the parallel system shown in figure
function with probabilities p1 = 0.9, p2 = 0.85 and p3 = 0.8,
respectively, independently of each other. What is the
probability the system functions?
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
1
2
3
Figure: Components connected in parallel
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Read Example 2.6.9.
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability
Outline
Sample Spaces, Events and Set Operations
Equally Likely Outcomes
Axioms and Properties of Probability
Conditional Probability
Independent Events
Go to previous lesson http://www.stat.psu.edu/
˜mga/401/course.info/lesson2.pdf
Go to next lesson http://www.stat.psu.edu/˜mga/
401/course.info/lesson4.pdf
Go to the Stat 401 home page http:
//www.stat.psu.edu/˜mga/401/course.info/
http://www.google.com
Michael Akritas
Lesson 3 Chapter 2: Introduction to Probability