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Transcript
35
Chapter 5
Probability
Section 5.1 Probability Rules
Objectives:
 Apply the rules of probabilities
 Compute and interpret probabilities using the empirical method
 Compute and interpret probabilities using the Classical method
5-2
Probability is a measure of the likelihood of a random
phenomenon or chance behavior.
The long-term proportion with which a certain
outcome is observed is the probability of that
outcome.
5-3
The Law of Large Numbers
As the number of repetitions of a probability
experiment increases, the proportion with which a
certain outcome is observed gets closer to the
probability of the outcome.
In probability, an experiment is any process that can
be repeated in which the results are uncertain.
Sample Space: the collection of all possible outcomes
of an experiment. It is denoted S
Event: a collection of outcomes from a probability
experiment (may consist of one outcome or more than
one outcome).
We will denote events with one outcome, sometimes
called simple events, ei. In general, events are denoted
using capital letters such as E.
EXAMPLE
Identifying Events and the Sample Space of a
Probability Experiment
Consider the probability experiment of having two
children.
(a) Identify the outcomes of the probability experiment.
(b) Determine the sample space.
(c) Define the event E = “have one boy”.
(a)
(b)
(c)
5-6
A probability model lists the possible outcomes of a probability
experiment and each outcome’s probability. A probability model
must satisfy rules 1 and 2 of the rules of probabilities.
5-8
If an event is impossible, the probability of the event is …...
If an event is a certainty, the probability of the event is …...
An unusual event is an event that has a
low probability of occurring.
5-9
EXAMPLE
A Probability Model
In a bag of peanut M&M milk chocolate
candies, the colors of the candies can be
brown, yellow, red, blue, orange, or green.
Suppose that a candy is randomly selected
from a bag. The table shows each color and
the probability of drawing that color. Verify
this is a probability model.
Rule1:
Rule2:
Color
Probability
Brown
0.12
Yellow
0.15
Red
0.12
Blue
0.23
Orange
0.23
Green
0.15
5-11
5-12
EXAMPLE Building a Probability Model
Pass the PigsTM is a Milton-Bradley
game in which pigs are used as dice.
Points are earned based on the way the
pig lands. There are six possible
outcomes when one pig is tossed. A
class of 52 students rolled pigs 3,939
times. The number of times each
outcome occurred is recorded in the
table at right.
(Source: http://www.members.tripod.com/~passpigs/prob.html)
Outcome
Frequency
Side with no dot
1344
Side with dot
1294
Razorback
767
Trotter
365
Snouter
137
Leaning Jowler
32
(a) Use the results of the experiment to build a probability model for the way the pig
lands.
(b) Estimate the probability that a thrown pig lands on the “side with dot”.
(c) Would it be unusual to throw a “Leaning Jowler”?
5-13
(a) Outcome
Side with no dot
Side with dot
Razorback
Trotter
Snouter
Leaning Jowler
(b)
(c)
Probability
5-15
The classical method of computing probabilities requires
equally likely outcomes.
An experiment is said to have equally likely outcomes when
each simple event has the same probability of occurring.
5-16
EXAMPLE
Computing Probabilities Using the Classical Method
Suppose a “fun size” bag of M&Ms contains 9 brown candies, 6 yellow
candies, 7 red candies, 4 orange candies, 2 blue candies, and 2 green
candies. Suppose that a candy is randomly selected.
(a) What is the probability that it is yellow?
(b) What is the probability that it is blue?
(c) Comment on the likelihood of the candy being yellow versus blue.
(a) There are a total of ….. candies, so N(S) =…...
(b) P(blue) =
(c)
5-17
Section 5.2 Probability Rules
5-18
5-19
Two events are disjoint if they have no
outcomes in common. Another name for
disjoint events is mutually exclusive events.
5-20
We often draw pictures of events using
Venn diagrams. These pictures
represent events as circles enclosed in
a rectangle. The rectangle represents
the sample space, and each circle
represents an event. For example,
suppose we randomly select a chip
from a bag where each chip in the bag
is labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Let E
represent the event “choose a number
less than or equal to 2,” and let F
represent the event “choose a number
greater than or equal to 8.” These
events are disjoint as shown in the
figure.
5-21
5-22
EXAMPLE
The Addition Rule for Disjoint Events
The probability model to
the right shows the
distribution of the
number of rooms in
housing units in the
United States.
(a) Verify that this is a probability
model.
Number of Rooms
in Housing Unit
Probability
One
0.010
Two
0.032
Three
0.093
Four
0.176
Five
0.219
Six
0.189
Seven
0.122
Eight
0.079
Nine or more
0.080
Source: American Community Survey,
U.S. Census Bureau
5-23
Number of Rooms
in Housing Unit
Probability
One
0.010
Two
0.032
Three
0.093
Four
0.176
Five
0.219
Six
0.189
Seven
0.122
Eight
0.079
Nine or more
0.080
(b) What is the probability a
randomly selected housing unit
has two or three rooms?
P(two or three)=
5-24
Number of Rooms
in Housing Unit
Probability
One
0.010
Two
0.032
Three
0.093
Four
0.176
Five
0.219
Six
0.189
Seven
0.122
Eight
0.079
Nine or more
0.080
(c) What is the probability a
randomly selected housing unit
has one or two or three rooms?
P(one or two or three)
5-25
5-26
5-27
EXAMPLE
Illustrating the General Addition Rule
Suppose that a pair of dice are thrown. Let E = “the first die is a
two” and let F = “the sum of the dice is less than or equal to 5”.
Find P(E or F) using the General Addition Rule.
5-28
N (E)
N (S )
6

36
1

6
P( E ) 
N (F )
N (S )
10

36
5

18
P( F ) 
N ( E and F )
N (S )
3

36
1

12
P( E and F ) 
P( E or F) = P(E) +P(F) – P(E and F)
5-29
5-30
Complement of an Event
Let S denote the sample space of a probability experiment
and let E denote an event. The complement of E, denoted EC,
is all outcomes in the sample space S that are not outcomes
in the event E.
5-31
Complement Rule
If E represents any event and EC represents the complement
of E, then
P(EC) = 1 – P(E)
5-32
EXAMPLE
Illustrating the Complement Rule
According to the American Veterinary Medical Association, 31.6%
of American households own a dog. What is the probability that
a randomly selected household does not own a dog?
P(do not own a dog) = 1 – P(own a dog)
5-33
EXAMPLE
Computing Probabilities Using Complements
The data to the right represent the
travel time to work for residents of
Hartford County, CT.
(a) What is the probability a randomly
selected resident has a travel time of
90 or more minutes?
There are a total of
24,358 + 39,112 + … + 4,895 = 393,186
residents in Hartford County, CT.
The probability a randomly selected
resident will have a commute time of “90
or more minutes” is
Source: United States Census Bureau
5-34
(b) Compute the probability that a randomly selected resident of
Hartford County, CT will have a commute time less than 90 minutes.
P(less than 90 minutes) = 1 – P(90 minutes or more)
5-35
Section 5.3 Independence and Multiplication Rule
5-36
5-37
Two events E and F are independent if the occurrence
of event E in a probability experiment does not affect
the probability of event F. Two events are dependent
if the occurrence of event E in a probability experiment
affects the probability of event F.
© 2010 Pearson Prentice Hall. All rights
reserved
5-38
EXAMPLE Independent or Not?
(a) Suppose you draw a card from a standard 52-card deck of
cards and then roll a die. The events “draw a heart” and
“roll an even number” are independent because the
results of choosing a card do not impact the results of the
die toss.
(b) Suppose two 40-year old women who live in the United
States are randomly selected. The events “woman 1
survives the year” and “woman 2 survives the year” are
independent.
(c) Suppose two 40-year old women live in the same
apartment complex. The events “woman 1 survives the
year” and “woman 2 survives the year” are dependent.
5-39
5-40
5-41
EXAMPLE
Computing Probabilities of Independent Events
The probability that a randomly selected female aged 60 years
old will survive the year is 99.186% according to the National
Vital Statistics Report, Vol. 47, No. 28. What is the probability
that two randomly selected 60 year old females will survive the
year?
The survival of the first female is independent of the survival of the second female.
We also have that P(survive) = 0.99186.
P  First survives and second survives   P  First survives   P  Second survives 
 (0.99186)(0.99186)
 0.9838
5-42
EXAMPLE
Computing Probabilities of Independent Events
A manufacturer of exercise equipment knows that 10% of their products are
defective. They also know that only 30% of their customers will actually use the
equipment in the first year after it is purchased. If there is a one-year warranty
on the equipment, what proportion of the customers will actually make a valid
warranty claim?
We assume that the defectiveness of the equipment is independent of the use
of the equipment. So,
P  defective and used   P  defective   P  used 
 (0.10)(0.30)
 0.03
5-43
5-44
EXAMPLE
Illustrating the Multiplication Principle for Independent Events
The probability that a randomly selected female aged 60 years old will survive
the year is 99.186% according to the National Vital Statistics Report, Vol. 47,
No. 28. What is the probability that four randomly selected 60 year old females
will survive the year?
P(all four survive)
= P (1st survives and 2nd survives and 3rd survives and 4th survives)
5-45
EXAMPLE
Computing “at least” Probabilities
The probability that a randomly selected female aged 60 years old will
survive the year is 99.186% according to the National Vital Statistics Report,
Vol. 47, No. 28. What is the probability that at least one of 500 randomly
selected 60 year old females will die during the course of the year?
P(at least one dies) = 1 – P(none die)
5-47
5-48
Section 5.4 Conditional Probability and the
General Multiplication Rule
5-49
5-50
Conditional Probability
The notation P(F | E) is read “the probability of
event F given event E”. It is the probability of
an event F given the occurrence of the event E.
5-51
EXAMPLE
An Introduction to Conditional Probability
Suppose that a single six-sided die is rolled. What is the probability that the
die comes up 4? Now suppose that the die is rolled a second time, but we are
told the outcome will be an even number. What is the probability that the die
comes up 4?
First roll:
Second roll:
S = {1, 2, 3, 4, 5, 6}
S = {2, 4, 6}
P(S ) 
1
6
P(S ) 
1
3
5-52
5-53
EXAMPLE
Conditional Probabilities on Belief about God and Region of the Country
A survey was conducted by the Gallup Organization conducted May 8 – 11, 2008 in
which 1,017 adult Americans were asked, “Which of the following statements comes
closest to your belief about God – you believe in God, you don’t believe in God, but you
do believe in a universal spirit or higher power, or you don’t believe in either?” The
results of the survey, by region of the country, are given in the table below.
Believe in
God
Believe in
universal spirit
Don’t believe
in either
East
204
36
15
Midwest
212
29
13
South
219
26
9
West
152
76
26
(a) What is the probability that a randomly selected adult American who lives in
the East believes in God?
(b) What is the probability that a randomly selected adult American who believes
in God lives in the East?
5-54
Believe in
God
Believe in
universal spirit
Don’t believe
in either
East
204
36
15
Midwest
212
29
13
South
219
26
9
West
152
76
26
(a) What is the probability that a randomly selected adult American who lives in
the East believes in God?
P  believes in God lives in the east  
N  believe in God and live in the east 
N  live in the east 
(b) What is the probability that a randomly selected adult American who believes
in God lives in the East?
P  lives in the east believes in God  
N  believe in God and live in the east 
N  believes in God 
5-55
EXAMPLE
Murder Victims
In 2005, 19.1% of all murder victims were between the ages of 20 and 24
years old. Also in 1998, 16.6% of all murder victims were 20 – 24 year old
males. What is the probability that a randomly selected murder victim in
2005 was male given that the victim is 20 - 24 years old?
P  male 20  24  
P  male and 20  24 
P  20  24 
=
5-56
Section 5.5
Counting Techniques
If a task consists of a sequence of choices in which
there are P selections for the first choice,
Q selections for the second choice, R selections for
the third choice, and so on,
then the task of making these selections can be
done in P·Q·R· … different ways.
5-58
Example: How many mix and match outfits can you make
from 5 pair of jeans and 3 shirts?
Example: How many license plates can be formed
containing 3 letters followed by 3 numbers?
(repeats allowed) ___ ____ ____ ____ ____ ____
(no repeats allowed) ___ ___ ___ ____ ___ _____
5-59
Factorials:
n! = n(n –1)(n – 2)(n – 3) · · · 3 · 2 · 1
Example:
5! =
(MATH, PRB, #4)
10! =
Factorial Rule – N items can be arranged in N! different ways.
Example: You are car shopping and there are 8 different Nissan dealerships
in the Dallas Metroplex.
If you want to visit all 8 dealerships, how many different routes are possible?
Example: How many different ways can 7 different video games be arranged
on a shelf?
A permutation is an ordered
arrangement in which r objects are
chosen from n distincts objects and
repetition is not allowed. The symbol nPr
represents the number of permutations
of r objects selected from n objects.
5-61
Number of Permutation of distinct Objects taken r at a time
The number of arrangements of r objects chosen from n
objects, in which
1. The n objects are distinct,
2. Repetition of objects is not allowed, and
3. Order is important,
Is given by the formula
Evaluate
7P3
; 6P6
nPr
= n! / (n-r)!
(MATH, PRB, #2)
Example: How many ways can a President, Vice-President,
Secretary, and Treasurer be chosen from
10 committee members?
Example: How many different 5-letter words can be formed from the
letters in the word “decagon”?
Example: Lisa has 15 framed pictures. She wants to pick 5 of them
to hang on the wall. How many
different ways can she hang 5 pictures on the wall?
A combination is an arrangement,
without regard to order, of n distinct
objects without repetitions. The symbol
nCr represents the number of
combinations of n distinct objects taken
r at a time, where r < n.
5-
Determine the value of 9C3.
(MATH, PRB, #3)
5-66
The United States Senate consists of 100 members. In how many
ways can 4 members be randomly selected to attend a luncheon at the
White House?
Example: How many ways can you pick 6 committee members from a
group of 20 people?
Example: How many different random samples of size 5 can be
obtained from a population whose
size is 30?
5-67
Computing Probabilities using Permutations
and Combinations:
Example: The Texas Lottery: Pick 6 numbers
from 1 – 54. (Order does not matter)
How many different number combinations are
there?
What is the probability of winning the Texas
Lottery?
Example: Outside a home, there is a keypad that
can be used to open the garage if the correct four-digit
code is entered. The numbers on the keypad are 0 – 9.
How many codes are possible?
What is the probability that a burglar would enter
the correct code on the first try?
Finding Probabilities Using “Categories”
Example: A committee consists of 8 Democrats
and 2 Republicans. Suppose you randomly select 4
people from the committee. Find the following
probabilities:
a) P (3 Democrats, 1 Republican)
b) P (all 4 are Democrats)
Example: Suppose you have received a shipment of 100 TVs.
You don’t know it, but 6 of the TVs are defective. To determine if you
are going to accept the shipment, you randomly select 5 TV and test
them.
If all 5 TVs work, you accept the shipment, otherwise, you reject it.
What is the probability that you accept the shipment?
Example: You pick 4 cards from a standard deck of cards.
What is the probability of getting all Aces?