Download Chapter 5: Regression - Memorial University of Newfoundland

Stat 1510:
General Rules of Probability
Agenda
2

Independence and the Multiplication Rule

The General Addition Rule

Conditional Probability

The General Multiplication Rule

Tree Diagrams

Bayes’ Rule
Probability Rules
3
Rule 1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1.
Rule 2. If S is the sample space in a probability model, then P(S) = 1.
Rule 3. If A and B are disjoint, P(A or B) = P(A) + P(B).
This is the addition rule for disjoint events.
Rule 4. For any event A, P(A does not occur) = 1 – P(A).
Venn Diagrams
4
Sometimes it is helpful to draw a picture to display relations among several
events. A picture that shows the sample space S as a rectangular area and
events as areas within S is called a Venn diagram.
Two disjoint events:
Two events that are not disjoint, and
the event {A and B} consisting of
the outcomes they have in common:
Venn diagram
5
A&B
A&B&C
A &C
B&C
Union and Intersection
6
The union of two events A and B is the event that occurs if either
A or B (or Both) occurs on a singer performance of the experiment.
We generally denoted this event as A U B
The intersection of two events A and B is the event that
occurs if both A and B on a single performance of the
experiment. We generally denoted this event as A  B
Example:
Consider the experiment of tossing a fair die in which following events are defined
A : Toss an even number
B: Toss a number less than or equal to 3.
Find the events AUB and AB and its probabilities?
Multiplication Rule for
Independent Events
7
If two events A and B do not influence each other, and if knowledge about
one does not change the probability of the other, the events are said to
be independent of each other.
Multiplication Rule for Independent Events
Two events A and B are independent if knowing that one occurs does
not change the probability that the other occurs. If A and B are
independent:
P(A and B) = P(A  B)= P(A)  P(B)
Multiplication Rule for Independent Events
Example
8



Suppose that about 20% of incoming male new undergrad
students smoke.
Suppose these undergrads are randomly assigned in pairs
to dorm rooms (assignments are independent).
The probability of a match
(both smokers or both non-smokers):
 both are smokers: 0.04 = (0.20)(0.20)
68%
 neither is a smoker: 0.64 = (0.80)(0.80)
 only one is a smoker: ?
32% (100%  68%)
}
What if pairs are self-selected?
The General Addition Rule
9
We know if A and B are disjoint events,
P(A or B) = P(A U B) = P(A) + P(B)
Addition Rule for Any Two Events
For any two events A and B:
P(A or B) = P(AUB) = P(A) + P(B) – P(A  B)
For any three events – A,B &C
P(A U B U C) = P(A) + P(B) +P© – P(A  B) P(A  C) – P(B  C) + P(A  B C)
Case Study
10
Student Demographics
At a certain university, 80% of the students were inprovince students (event A), 30% of the students
were part-time students (event B), and 20% of the
students were both in-province and part-time
students (event {A and B}). So we have that P(A) =
0.80, P(B) = 0.30, and P(A and B) = 0.20.
What is the probability that a student is either an inprovince student or a part-time student?
Case Study
11
All Students
Other
Students
Part-time
(B)
In-province
(A)
{A and B}
0.30
0.20
0.80
P(A or B) = P(A) + P(B)  P(A and B)
= 0.80 + 0.30  0.20 = 0.90
Conditional Probability
12
The probability we assign to an event can change if we know that some other
event has occurred. This idea is the key to many applications of probability.
When we are trying to find the probability that one event will happen under the
condition that some other event is already known to have occurred, we are trying
to determine a conditional probability.
The probability that one event happens given that another event is
already known to have happened is called a conditional probability.
When P(A) > 0, the probability that event B happens given that event A
has happened is found by:
P(B | A) =
P(A and B)
P(A)
The General Multiplication Rule
13
The definition of conditional probability reminds us that in principle all
probabilities, including conditional probabilities, can be found from the
assignment of probabilities to events that describe a random phenomenon.
The definition of conditional probability then turns into a rule for finding the
probability that both of two events occur.
The probability that events A and B both occur can be found using the
general multiplication rule
P(A and B) = P(A) • P(B | A)
where P(B | A) is the conditional probability that event B occurs given
that event A has already occurred.
Note: Two events A and B that both have positive probability are independent
if:
P(B|A) = P(B)
Case Study
14
Student Demographics
At a certain university, 20% of first year students
smoke, and 25% of all students are new students.
Let A be the event that a student is a first year
student, and let B be the event that a student
smokes.
So we have that P(A) = 0.25, and P(B|A) = 0.20.
What is the probability that a student smokes and
is a first year student?
Case Study
15
Student Demographics
P(A) = 0.25 , P(B|A) = 0.20
P(A and B)
= P(A)  P(B|A)
= 0.25  0.20
= 0.05
5% of all students are first year student smokers.
Tree Diagrams
16
We learned how to describe the sample space S of a chance process.
Another way to model chance behavior that involves a sequence of
outcomes is to construct a tree diagram.
Consider flipping a
coin twice.
What is the probability
of getting two heads?
Sample Space:
HH HT TH TT
So, P(two heads) = P(HH) = 1/4
Example
17
The Pew Internet and American Life Project finds that 93% of teenagers (ages
12 to 17) use the Internet, and that 55% of online teens have posted a profile
on a social-networking site.
What percent of teens are online and have posted a profile?
P(online) = 0.93
P(profile | online) = 0.55
P(online and have profile) = P(online)× P(profile | online)
= (0.93)(0.55)
= 0.5115
51.15% of teens are online and have
posted a profile.
Total Probability
18
The events A1 and A2 are mutually exclusive if no
two have any common outcomes
The events A1 and A2 are exhaustive if either A1 or
A2 must occur so that
A1 U A2 = S
If A1 and A2 are mutually exclusive and exhaustive,
and B is another event, then
P(B) = P(B|A1)*P(A1) + P(B|A2)*P(A2)
Bayes’ Theorem
19
Let A1 and A2 be mutually exclusive and
exhaustive events, and B is any other event with
P(B) > 0, then
P(A1 and B)
P(A1|B) 
P(B | A1)* P(A1) + P(B | A2)* P(A2)
Example
20
NCI estimates that 3.65% of women in their 60’s get breast
cancer. A mammogram can typically identify correctly 85% of
cancer and 95% of cases without cancer. If a women in her
60’s gets a positive mammogram, what is the probability that
she indeed has breast cancer?
P( cancer)=0.0365; P( no cancer)=0.9635
P( test +| cancer) = 0.85; P( test - | cancer) = 0.15
P( test + | no cancer) = 0.05; P( test - | no cancer) =0.95
What we want is P( cancer | test +)?
Example
21
P(Cancer | test +)

=
P(test + | Cancer) P(Cancer)
P(test
+ | Cancer) * P(cancer) + P(test + | no cancer) * P(no cancer)


0.85 * 0.0365
0.85 * 0.0365 + 0.05 * 0.9635
 0.392