Download Chapter5 Probability. Introduction. We will consider random

STP226 Brief Class Notes Instructor: Ela Jackiewicz
Chapter5 Probability.
Introduction.
We will consider random experiments with chance outcomes. Events are outcomes that may or
may not occur.
Notation:
Capital letters like E will denote events
Probability of event E=P(E)
Frequency interpretation of probability:
P(E) can be approximated by the relative frequency of occurrences of E in a very large number
of repetitions of our chance experiment.
Ex. In a 1024 single baby births in a Phoenix hospital 507 babies were girls. If A=event that a
girl is born in that hospital, then
P( A)≈
507
1024
PROBABILITY (classical approach)
The Sample Space (S) associated with any experiment is the set of all possible outcomes that can
occur as a result of the experiment. So naturally, we will call each element of the sample space an
outcome.
An event (E) is any subset of the sample space.
The probability of an event E (written as P(E)) in a sample space (S) with equally likely outcomes
is given by
number of outcomes in E
P(E) =
number of outcomes in S
Consider the experiment of rolling a fair die two times (or a pair of dice once) The figure
below gives a representation of all the 36 equally likely outcomes of the sample space associated with
this experiment.
EXAMPLE1
(1,1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
(3,1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
(4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
(5,1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
(6,6)
If for example E=sum of both rolls is 7, then E={(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
P(E)= 6/36=1/6
STP226 Brief Class Notes Instructor: Ela Jackiewicz
Outcomes in a sample space do not have to be equally likely, next example will consider that
case:
EXAMPLE2
Suppose we roll a die that is not balanced and probabilities of each “face showing are as follows:
P(1)=0.1, P(2)=0.1, P(3)=0.2, P(4)=0.5, P(5)=0.05, P(6)=0.05
Let A=event that we will get a number larger than 4, A={5, 6}
P(A)=P(5)+P(6)=0.05+0.05=0.1
In general if sample space S={ e1, e 2,. .... e n } and events e i do not have to be equally likely,
but sum of the probabilities of all outcomes in S still equals 1. For event E that consists of some
outcomes in S, P(E)=sum of the probabilities of these outcomes:
P( E)=∑ P( ei )
Some Probability Rules.
1)
0≤P (E)≤1 P(impossible event)=0 and P(certain event)=1
2)
P (Ec )=1−P ( E)
, where
Ec =not E is an opposite event to E
3) Sum of probabilities of all outcomes in a sample space is 1
P ( Aor B)=P ( A)+ P( B)−P ( Aand B) or another notation:
4)
Notation:
A∪B= Aor B
A and B= A∩B= A & B
These rules can be illustrated using Venn Diagrams.
STP226 Brief Class Notes Instructor: Ela Jackiewicz
Mutually Exclusive Events
Two or more events are said to be mutually exclusive if at most one of them can occur when the
experiment is performed, that is, if no two of them have outcomes in common.
On the following diagram a) represents mutually exclusive events and b) not mutually exclusive events
Example1 continue: Consider an experiment with rolling a balanced die twice (or pair of fair dice
once)
a) What is the probablity that sum of both dice is not 7. That is P(notE)=1-1/6=5/6
b) P(sum=7 and number on the first roll is 2)=P((2,5))=1/36
since F=number on the first roll is 2={(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}
c) P( sum=7 or number on the first roll is 2)=6/36+6/36-1/36=11/36
Example3 . Blood type distribution (in %) in the USA is given in the following table:
Blood Type
A
B AB
O
%
42 10 4 44
Suppose we select randomly one American, Compute the probabilities of the following events:
a) P(AB)=.04 b) P(A or B) = .52
c) P(
Ac )= 1-.42=.58 d) P(A and O)=0
(Events A , O are mutually exclusive)
Example4 Consider a random family with 3 children, denote the sex of each child: B=boy, G=girl
Let A, B, C be the following events:
A=family has exactly 1 boy C=family has no more than 1 girl D= family has at least 2 girls
a) Give sample space :
S={GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB}
STP226 Brief Class Notes Instructor: Ela Jackiewicz
b) Compute P(A and C)=0
c) Compute P(A or C)=7/8
d) Compute P(A and D)=3/8
e) Compute P(A or D)=4/8=0.5
f) Compute P(C and D)=0
g) Compute P(C or D)=1
Example5 Following table represents distribution of X=number of children in a family for some
community in Tempe:
X
0
1
2
3
4
Relative
frequency
0.03
0.17
0.50
0.25
0.05
Suppose one family is randomly selected from that community, compute following probabilities:
P(family has no children)=0.03
P(family has some children)=1-0.03=0.97
P(family has at most 2 or more than 3 children)=1-0.25=0.75
P(family has at most 2 and more than 3 children)=0
P(family has at least 2 and no more than 3 children)=0.5+0.25=0.75
P(family has at least 2 or no more than 3 children)=1
Example6 In a study of the relationship between health risks and income levels a large group of
people reported that they are “stresses “ (which means extremely or quite stressful most days) or
“not stressed” ( which means most days are a bit stressful, not very stressful or not stressful at all).
Their income level was also recorded. Summary is presented in the table below:
Low
Income Level
Medium
High
total
stressed
526
274
216
1016
Not stressed
1954
1680
1899
5533
total
2480
1954
2115
6549
If the random person from this study is selected, compute the following:
a) probability that person is stressed= 1016/6549=.155
b) probability that person is low income=2480/6549=.3787
Probabilities in a and b are called marginal probabilities, events are related to one variable only.
c) probability that person is stressed and low income= 526/6549=.080
d) probability that person is stressed or is low income= .155+.3787-.080=.4537
Probability in c is called joint probability, it involves both variables.
STP226 Brief Class Notes Instructor: Ela Jackiewicz
Following is OPTIONAL :
P ( Aand B)
, provided
P ( B)
5)
P ( A/ B)=
6)
P ( Aand B)=P ( A) P ( B) and
P(B)≠0 (Conditional probability)
P ( A/B)=P ( A) for independent events only
We can use these rules to answer following questions:
e) conditional probability that low income person is stressed= 526/2480=.2121 (we consider only low
income people and out of these we count how many are stressed)
(or using the formula:
526 / 6549
gives the same answer)
2480/ 6549
f) are events “ person has low income” and “person is stressed” independent?
P(stressed)=.155 ,
independent
P(low income)=.3787,
.155(.3787)=.059≠.080 , so events are not
In general:
To check independence of two events A, B you either check if P(A|B)=P(A) or if
P( A∩B)=P (A ) P( B)
In case of equality you have independence, otherwise events are not independent.