Download Lecture 8 Probability

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Lecture 8
Probability
5-1 Introduction
5-2 Sample Spaces and Probability
5-3 The Addition Rules for
Probability
5-4 The Multiplication Rules and
Conditional Probability
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5-1 Introduction
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5-1 Introduction
Probability as a general concept
can be defined as the chance of an
event occurring.
Games of chance (card games, slot
machines, or lotteries).
Used in the fields of insurance,
investments, and weather
forecasting and many others.
Most important for us: probability
is the basis of inferential statistics.
Basic concepts of probability are
explained in this lecture.
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5-2 Tree Diagram for Tossing Two Coins
5-2 Sample Spaces and Probability
A probability experiment is a process
that leads to well-defined results called
outcomes.
An outcome is the result of a single trial
of a probability experiment.
NOTE: A tree diagram can be used as a
systematic way to find all possible
outcomes of a probability experiment.
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H
H
T
Second Toss
H
T
First Toss
T
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5-2 Formula for Classical
Probability
5-2 Sample Spaces - Examples
EX PER IM E N T SA M PLE SPA CE
Toss one coin
H, T
R oll a die
1, 2, 3, 4, 5, 6
A nsw er a truefalse question
Toss tw o coins
True, False
H H, HT, TH, TT
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© The McGraw-Hill Companies, Inc., 2000
5-2 Formula for Classical
Probability
5-2 Classical Probability - Examples
The probability of any event E is
number of outcomes in E
.
total number of outcomes in the sample space
This probability is denoted by
n( E )
P(E ) =
.
n(S )
This probability is called classical probability ,
and it uses the sample space S .
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(c) There are four 3s and 13
diamonds, but the 3 of diamonds is
counted twice in the listing. Hence
there are only 16 possibilities of
drawing a 3 or a diamond, thus
P(3 or diamond) = 16/52 = 4/13.
4/13
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For a card drawn from an ordinary deck,
find the probability of getting (a) a queen
(b) a 6 of clubs (c) a 3 or a diamond.
Solution: (a) Since there are 4 queens
and 52 cards, P(queen) = 4/52 = 1/13.
1/13
(b) Since there is only one 6 of clubs,
then P(6 of clubs) = 1/52.
1/52
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5-2 Classical Probability - Examples
Classical probability assumes that
all outcomes in a finite sample
space are equally likely to occur.
That is, equally likely events are
events that have the same
probability of occurring.
5-2 Classical Probability - Examples
When a single die is rolled, find the
probability of getting a 9.
Solution: Since the sample space is
1, 2, 3, 4, 5, and 6, it is impossible to
get a 9. Hence, P(9) = 0/6 = 0.
0
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5-2 Classical Probability
There are four basic probability rules.
These rules are helpful
in solving probability problems,
in understanding the nature of
probability, and
in deciding if your answers to the
problems are correct.
Probability Rule 1
The probability of an event E is a
number between 0 and 1:
0 £ P(E) £ 1
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Probability Rule 2
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Probability Rule 3
If an event E cannot occur (i.e.,
the event contains no members
in the sample space), the
probability of E is zero.
If an event E is certain, then
the probability of E equals 1.
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Probability Rule 4
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5-2 Complement of an Event
The sum of the probabilities of
the outcomes in the sample
space is 1.
The complement of an event E is the set of outcomes in the
sample space that are not included in the outcomes
of event E . The complement of E is denoted by E ( E bar ).
E
E
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5-2 Complement of an Event Example
Find the complement of each event.
Rolling a die and getting a 4.
Solution: Getting a 1, 2, 3, 5, or 6.
Selecting a letter of the alphabet and
getting a vowel.
Solution: Getting a consonant
(assume y is a consonant).
5-2 Complement of an Event Example
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5-2 Rule for Complementary Event
P(E ) = 1− P(E )
5-2 Empirical Probability
or
P(E ) = 1−P(E )
or
P ( E ) + P ( E ) = 1.
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the probability of an event being
in a given class is
frequency for the class
total frequencies in the distribution
f
= .
n
This probability is called the em pirical
P( E ) =
The difference between classical and
empirical probability is that classical
probability assumes that certain
outcomes are equally likely while
empirical probability relies on actual
experience to determine the
probability of an outcome.
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5-2 Formula for Empirical
Probability
Given a frequency distribution ,
Selecting a day of the week and
getting a weekday.
Solution: Getting Saturday or
Sunday.
Selecting a one-child family and
getting a boy.
Solution: Getting a girl.
5-2 Empirical Probability Example
In a sample of 50 people, 21 had
type O blood, 22 had type A blood,
5 had type B blood, and 2 had AB
blood. Set up a frequency
distribution.
probability and is based on observation .
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5-2 Empirical Probability Example
5-2 Empirical Probability Example
Type
A
B
AB
O
Frequency
22
5
2
21
50 = n
Find the following probabilities for
the previous example.
A person has type O blood.
Solution: P(O) = f /n = 21/50.
A person has type A or type B blood.
Solution: P(A or B) = 22/50+ 5/50
= 27/50.
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5-3 The Addition Rules for
Probability
5-3 The Addition Rules for
Probability
Two events are mutually exclusive
if they cannot occur at the same
time (i.e. they have no outcomes in
common).
A and B are mutually exclusive
A
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5-3 Addition Rule 1
5-3 Addition Rule 11- Example
When two events A and B are
mutually exclusive, the probability
that A or B will occur is
P ( A or B ) = P ( A ) + P ( B )
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B
At a political rally, there are 20
Republicans (R), 13 Democrats (D),
and 6 Independents (I). If a person is
selected, find the probability that he or
she is either a Democrat or an
Independent.
Solution: P(D or I) = P(D) + P(I)
= 13/39 + 6/39 = 19/39.
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5-3 Addition Rule 2
5-3 Addition Rule 11- Example
A day of the week is selected at
random. Find the probability that
it is a weekend.
Solution: P(Saturday or Sunday)
= P(Saturday) + P(Sunday)
= 1/7 + 1/7 = 2/7.
When two events A and B
are not mutually exclusive, the
probabilityy that A or B will
occur is
P( A or B ) = P( A) + P( B) − P ( A and B)
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5-3 Addition Rule 22- Example
5-3 Addition Rule 2
A and B
(common portion)
A
B
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5-3 Addition Rule 2 - Example
STAFF
STAFF
FEMALES
FEMALES
MALES
MALES
TOTAL
TOTAL
NURSES
NURSES
77
11
88
PHYSICIANS
PHYSICIANS
33
22
55
TOTAL
TOTAL
10
10
33
13
13
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In a hospital unit there are eight
nurses and five physicians. Seven
nurses and three physicians are
females. If a staff person is selected,
find the probability that the subject is
a nurse or a male.
The next slide has the data.
5-3 Addition Rule 2 - Example
Solution: P(nurse or male)
= P(nurse) + P(male) – P(male
nurse) = 8/13 + 3/13 – 1/13 = 10/13.
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5-3 Addition Rule 2 - Example
On New Year’s Eve, the probability that a
person driving while intoxicated is 0.32,
the probability of a person having a
driving accident is 0.09, and the
probability of a person having a driving
accident while intoxicated is 0.06. What is
the probability of a person driving while
intoxicated or having a driving accident?
5-3 Addition Rule 2 - Example
Solution:
P(intoxicated or accident)
= P(intoxicated) + P(accident)
– P(intoxicated and accident)
= 0.32 + 0.09 – 0.06 = 0.35.
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5-4 The Multiplication Rules and
Conditional Probability
5-4 Multiplication Rule 1
When two events A and B
Two events A and B are independent
if the fact that A occurs does not
affect the probability of B occurring.
Example: Rolling a die and getting a
6, and then rolling another die and
getting a 3 are independent events.
are independent , the
probability of both
occurring is
P ( A and B ) = P ( A ) ⋅ P ( B ).
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5-4 Multiplication Rule 1 Example
A card is drawn from a deck and
replaced; then a second card is
drawn. Find the probability of getting
a queen and then an ace.
Solution: Because these two events
are independent (why?), P(queen and
ace) = (4/52)⋅⋅(4/52) = 16/2704 = 1/169.
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5-4 Multiplication Rule 1 Example
A Harris pole found that 46% of
Americans say they suffer great stress
at least once a week. If three people
are selected at random, find the
probability that all three will say that
they suffer stress at least once a week.
Solution: Let S denote stress. Then
P(S and S and S) = (0.46)3 = 0.097.
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5-4 Multiplication Rule 1 Example
The probability that a specific medical
test will show positive is 0.32. If four
people are tested, find the probability
that all four will show positive.
Solution: Let T denote a positive test
result. Then P(T and T and T and T) =
(0.32)4 = 0.010.
5-4 The Multiplication Rules and
Conditional Probability
When the outcome or occurrence of the
first event affects the outcome or
occurrence of the second event in such
a way that the probability is changed,
the events are said to be dependent.
Example: Having high grades and
getting a scholarship are dependent
events.
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5-4 The Multiplication Rules and
Conditional Probability
5-4 Multiplication Rule 2
When two events A and B
are dependent , the
The conditional probability of an
event B in relationship to an event A
is the probability that an event B
occurs after event A has already
occurred.
The notation for the conditional
probability of B given A is P(B|A).
probability of both
occurring is
P ( A and B ) = P ( A) ⋅ P ( B| A).
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5-4 The Multiplication Rules and
Conditional Probability - Example
In a shipment of 25 microwave ovens,
two are defective. If two ovens are
randomly selected and tested, find the
probability that both are defective if the
first one is not replaced after it has
been tested.
Solution: See next slide.
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5-4 The Multiplication Rules and
Conditional Probability - Example
Solution: Since the events are
dependent, P(D1 and D2)
= P(D1)⋅⋅P(D2| D1) = (2/25)(1/24)
= 2/600 = 1/300.
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5-4 The Multiplication Rules and
Conditional Probability - Example
The WW Insurance Company found that
53% of the residents of a city had
homeowner’s insurance with its company.
Of these clients, 27% also had automobile
insurance with the company. If a resident
is selected at random, find the probability
that the resident has both homeowner’s
and automobile insurance.
5-4 The Multiplication Rules and
Conditional Probability - Example
Solution: Since the events are
dependent, P(H and A)
= P(H)⋅⋅P(A|H) = (0.53)(0.27)
= 0.1431.
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5-4 The Multiplication Rules and
Conditional Probability - Example
5
Box 1 contains two red balls and one
blue ball. Box 2 contains three blue
balls and one red ball. A coin is tossed.
If it falls heads up, box 1 is selected
and a ball is drawn. If it falls tails up,
box 2 is selected and a ball is drawn.
Find the probability of selecting a red
ball.
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5-4 Tree Diagram for Example
P(R|B1) 2/3
P(B1) 1/2
P(B2) 1/2
Box 1
Blue (1/2)(1/3)
P(B|B1) 1/3
P(R|B2) 1/4
Box 2
Red (1/2)(1/4)
P(B|B2) 3/4 Blue (1/2)(3/4)
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5-4 The Multiplication Rules and
Conditional Probability - Example
Solution: P(red) = (1/2)(2/3) +
(1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24
= 11/24.
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5-4 Conditional Probability Formula
The probability that the second event B occurs
given that the first event A has occurred can be
found by dividing the probability that both events
occurred by the probability that the first event has
occurred . The formula is
P( B| A) =
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Red (1/2)(2/3)
P( A and B)
.
P( A)
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5-4 Conditional Probability Example
The probability that Sam parks in a noparking zone and gets a parking ticket is
0.06, and the probability that Sam cannot
find a legal parking space and has to park
in the no-parking zone is 0.2. On Tuesday,
Sam arrives at school and has to park in a
no-parking zone. Find the probability that
he will get a ticket.
5-4 Conditional Probability Example
Solution: Let N = parking in a noparking zone and T = getting a
ticket.
Then P(T|N) = [P(N and T) ]/P(N) =
0.06/0.2 = 0.30.
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5-4 Conditional Probability Example
5-4 Conditional Probability Example
A recent survey asked 100 people
if they thought women in the
armed forces should be permitted
to participate in combat. The
results are shown in the table on
the next slide.
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Find the probability that the respondent
answered “yes” given that the respondent
was a female.
Solution: Let M = respondent was a male;
F = respondent was a female;
Y = respondent answered “yes”;
N = respondent answered “no”.
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Yes
Yes
No
No
Total
Total
Male
Male
32
32
18
18
50
50
Female
Female
88
42
42
50
50
Total
Total
40
40
60
60
100
100
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5-4 Conditional Probability Example
Gender
Gender
5-4 Conditional Probability Example
P(Y|F) = [P( F and Y) ]/P(F) =
[8/100]/[50/100] = 4/25.
Find the probability that the respondent
was a male, given that the respondent
answered “no”.
Solution: P(M|N) = [P(N and M)]/P(N) =
[18/100]/[60/100] = 3/10.
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