Probability Concepts and
Counting Rules
Chapter 4
Dr. Ateq Ahmed Al-Ghamedi
McGraw-Hill/Irwin
Department of Statistics
P O Box 80203
King Abdulaziz University
Jeddah 21589, Saudi Arabia
Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved.
[email protected]
GOALS
1.
2.
3.
4.
5.
6.
Define probability.
Describe the classical, empirical, and
subjective approaches to probability.
Explain
p
the terms experiment,
p
, event,,
outcome, permutations, and combinations.
Define the terms conditional probability and
joint probability.
Calculate probabilities using the rules of
addition and rules of multiplication.
Apply a tree diagram to organize and
compute probabilities.
5-2
Probability
PROBABILITY A value between zero and one,
inclusive, describing the relative possibility
(chance or likelihood) an event will occur.
5-3
1
Experiment, Outcome and Event
z
z
z
z
An experiment is a process that leads to the occurrence of one and only one of
several possible observations.
An outcome is the particular result of an experiment.
An event is the collection of one or more outcomes of an experiment.
An event with one outcome is called a simple event and with more than one
outcome is called compound event
event.
5-4
Tree Diagram
z
A tree diagram is a device used to list all possibilities
of a sequence of events in a systematic way.
H
H
First Toss
T
T
Second Toss
H
T
In class ex. Use a tree diagram to find the sample space
for the gender of the three children
5-5
Counting Rules – Multiplication
The multiplication formula indicates
that if there are m ways of doing
one thing and n ways of doing
another thing, there are m x n ways
of doing both.
Example: Mr. Ahmed has 10 shirts and 8
ties. How many shirt and tie outfits does
he have?
(10)(8) = 80
5-6
2
Example
z
z
Employees of a large corporation are to be issued
special coded identification cards. The card
consists of 4 letters of the alphabet. Each letter
can be used up
p to 4 times in the code. How many
y
different ID cards can be issued?
Solution: Since 4 letters are to be used, there are
4 spaces to fill ( _ _ _ _ ). Since there are 26
different letters to select from and each letter can
be used up to 4 times, then the total number of
identification cards that can be made is 26
× 26 × 26 × 26 = 456,976
5-7
Example
The digits 0, 1, 2, 3, and 4 are to be used
in a 4-digit ID card. How many different
cards are possible if repetitions are
permitted?
z Solution: Since there are four spaces to fill
and five choices for each space, the
solution is 5 × 5 × 5 × 5 = 54 = 625
z What if the repetitions were not permitted
in the previous example? (=120)!
z
5-8
Counting Rules - Permutation
A permutation is any arrangement of r
objects selected from n possible
objects. The order of arrangement is
important in permutations.
5-9
3
Permutation
Consider the possible arrangements of the
letters a, b, and c.
z The p
possible arrangements
g
are: abc
abc,, acb,
acb,
z
bac,, bca,
bac
bca, cab, cba.
cba.
z If the order of the arrangement is important
then we say that each arrangement is a
permutation of the three letters. Thus
there are six permutations of the three
letters
5-10
Example
z How
many different ways can a
chairperson and an assistant
chairperson be selected for a
research project if there are seven
scientists available?
z Solution: Number of ways
= 7P2 = 7! / ((7
7 – 2)! = 7!/5
!/5! = 42
42.
5-11
Example
z How
many different ways can four
books be arranged on a shelf if
they can be selected from nine
books?
z Solution: Number of ways
9 – 4)! = 9!/5
!/5! = 3024
3024.
=9P4 = 9! / ((9
5-12
4
Counting Rules- Combination
A combination is the number of
ways to choose r objects from a
group of n objects without regard to
order.
5-13
Combination
Consider the possible arrangements of the
letters a, b, and c.
z The p
possible arrangements
g
are: abc
abc,, acb,
acb,
z
z
bac,, bca,
bac
bca, cab, cba.
cba.
If the order of the arrangement is not
important then we say that each
arrangement is the same. We say there is
one combination of the three letters.
5-14
Example
In order to survey the opinions of
customers at local malls, a researcher
decides to select 5 malls from a total of 12
malls in a specific geographic area. How
many different ways can the selection be
made?
z Solution: Number of combinations:
z
12C5
= 12!
12! / [(
[(12
12 – 5)! 5!] = 12
12!/[
!/[7
7!5!] = 792
5-15
5
Example
z
z
In a work team there are 7 women and 5 men.
A committee of 3 women and 2 men is to be
chosen. How many different possibilities are
there?
Solution: Number of possibilities: (number of
ways of selecting 3 women from 7) × (number of
ways of selecting 2 men from 5) = 7C3 × 5C2 =
(35
35)(
)(10
10)) = 350
350..
5-16
Sample Space
Ex: Find the sample space for the gender of
the children if a family has three children.
Use B for boy and G for girl.
There are two gender and three children, so
there are 8 possibilities as shown here,
{BBB, BBG, BGB, GBB, GGG, GGB, GBG, BGG}
5-17
Mutually Exclusive Events
z
z
Events are mutually exclusive if the occurrence of
any one event means that none of the others can
occur at the same time.
Events are independent if the occurrence of one
eventt does
d
nott affect
ff t the
th occurrence off another.
th
z
Equally likely events are events that have the same
probability of occurring.
z
Venn diagrams are used to represent probabilities
pictorially.
5-18
6
Ways of Assigning Probability
There are three ways of assigning probability:
1. CLASSICAL PROBABILITY
Based on the assumption that the outcomes of an
experiment are equally likely
likely.
2. EMPIRICAL PROBABILITY
The probability of an event happening is the fraction
of the time similar events happened in the past.
3. SUBJECTIVE CONCEPT OF PROBABILITY
The likelihood (probability) of a particular event
happening that is assigned by an individual based on
whatever information is available.
5-19
Classical Probability
Consider an experiment of rolling a six-sided die. What is the
probability of the event “an
an even number of spots appear face up
up”?
?
The possible outcomes are:
There are three “favorable” outcomes (a two, a four, and a six) in the
collection of six equally likely possible outcomes. =(3/6)=(1/2)
5-20
Empirical Probability
The empirical approach to probability is based on what
is called the law of large numbers.
numbers The key to
establishing probabilities empirically is that more
observations will provide a more accurate estimate of
the probability.
5-21
7
Empirical Probability - Example
On February 1, 2003, the Space Shuttle Columbia
exploded. This was the second disaster in 113 space
missions for NASA. On the basis of this information,
what is the probability that a future mission is
successfully completed?
Number of successful flights
Total number of flights
111
=
= 0.98
113
Probability of a successful flight =
5-22
Subjective Probability - Example
z
If there is little or no past experience or information on which to
base a probability, it may be arrived at subjectively.
z
Illustrations of subjective probability are:
1. Estimating the likelihood the New England Patriots will play in the
Super Bowl next year.
2. Estimating the likelihood you will be married before the age of 30.
3. Estimating the likelihood the U.S. budget deficit will be reduced by
half in the next 10 years.
5-23
Summary of Types of Probability
5-24
8
The Complement Rule
The complement rule is used to determine the
probability of an event occurring by subtracting
the probability of the event not occurring from 1.
5-25
Examples
Find the complement of each event.
1-Rolling a die and getting a 4.
z Solution: Getting a 1, 2, 3, 5, or 6.
2 Selecting a letter of the alphabet and getting a vowel.
2-Selecting
Solution: Getting a consonant
3-Selecting a day of the week and getting a weekday.
z Solution: Getting a day of the weekend.
4-Selecting a one-child family and getting a boy.
z Solution: Getting a girl.
z
5-26
Rules for Computing Probabilities
Rules of Addition
z Special Rule of Addition - If two events
A and B are mutually exclusive, the
probability of one or the other event’s
g equals
q
the sum of their
occurring
probabilities.
P(A or B) = P(A) + P(B)
z
The General Rule of Addition - If A and
B are two events that are not mutually
exclusive, then P(A or B) is given by the
following formula:
P(A or B) = P(A) + P(B) - P(A and B)
5-27
9
Example
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.
13
(b) Since there is only one 6 of clubs, then P(
P(6
6 of
clubs) = 1/52.
52
5-28
(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(
P(3
3 or diamond) = 16
16//52 = 4/13.
13
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.
Type
A
B
AB
O
Frequency
22
5
2
21
50 = n
5-29
Example
Find the following probabilities for the previous example.
1-A person has type O blood.
z Solution
Solution:: P(O) = f /n = 21/50.
2 A person has
2h type A or type B bl
blood.
d
z Solution: P(A or B) = 22/50+ 5/50 = 27/50.
3- A person has neither type A nor type O blood.
z Solution: P(B or AB) = 5/50+2/50=7/50
4- A person does not have type AB blood.
z Solution: P(not AB) =1- P(AB) =1-2/50=48/50
5-30
10
Example
5-31
Determine which events are mutually exclusive and which are not,
when a single die is rolled.
a. Getting an odd number and getting an even number.
The events are mutually exclusive, since the first event can be 1, 3 or 5 and
the second event can be 2, 4 or 6.
b. Getting a 3 and getting an odd number.
The events are not mutually exclusive, since the first event is a 3 and then
second event can be 1, 3 or 5. Hence, 3 is contained in both events.
c. Getting an odd number and getting a number less than 4.
The events are not mutually exclusive, since the first event can be 1, 3 or 5
and the second event can be 1, 2 or 3. Hence, 1 and 3 are contained in
both events.
d. Getting a number greater than 4 and getting a number less than 4.
The events are mutually exclusive, since the first event can be 5 or 6 and
the second event can be 1, 2 or 3.
Example
In a hospital unit there are 8 nurses and 5 physicians; 7
nurses and 3 physicians are females. If a staff is selected,
find the probability that the subject is a nurse or a male.
The events are not mutually exclusive and the sample space
is
Staff
Female
Males
Total
Nurses
7
1
8
Physicians
3
2
5
Total
10
3
13
z
z
5-32
Solution: P(nurse or male)
= P(nurse)
+ P(male) – P(male nurse) = 8/13 + 3/13 – 1/13 =
10/13
Example
What is the probability that a card chosen at
random from a standard deck of cards will be
either a king or a heart?
P(A or B) = P(A) + P(B) - P(A and B)
= 4/52 + 13/52 - 1/52
= 16/52, or .3077
5-33
11
Example
The Venn Diagram shows the result of a survey of
200 tourists who visited Florida during the year.
The survey revealed that 120 went to Disney
World, 100 went to Busch Gardens and 60 visited
both.
What is the probability a selected person visited
either Disney World or Busch Gardens?
P(Disney or Busch) = P(Disney) + P(Busch) - P(both Disney and Busch)
= 120/200 + 100/200 – 60/200
= .60 + .50 – .80
5-34
Joint Probability – Venn Diagram
JOINT PROBABILITY A probability that
measures the likelihood two or more events
will happen concurrently.
5-35
Independent and Dependent Events
Two events A and B are independent if the
occurrence of one has no effect on the
probability of the occurrence of the other.
z This rule is written:
P(A and B) = P(A)P(B)
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
5-36
12
Example
5-37
1- 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.
z Solution: Because these two events are
independent (why?), P(queen and ace) =
(4/52)⋅(4/52) = 16/2704 = 1/169.
2- 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.
z Solution: Let T denote a positive test result. Then
P(T and T and T and T) = (0.32)4 = 0.010.
Conditional Probability
zA
conditional probability is the
probability of a particular event
occurring given that another event
occurring,
has occurred.
z The probability of the event A given
that the event B has occurred is
written P(A|B).
5-38
Example
z
z
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: Since the events are dependent,
P(D1 and D2) = P(D1)⋅P(D2| D1) = (2/25)(1/24)
= 2/600 = 1/300.
5-39
13
Example
z
z
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.
Solution: Since the events are dependent, P(H
and A) = P(H)⋅P(A|H) = (0.53)(0.27) = 0.1431.
5-40
Example
z
z
z
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.
Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 +
1/8 = 8/24 + 3/24 = 11/24
see the tree diagram
5-41
Tree Diagram
P(B1) 1/2
P(R|B1) 2/3
Red
Box 1
Bl
Blue
(1/2)(1/3)
P(B|B1) 1/3
(1/2)(2/3)
P(R|B2) 1/4
P(B2) 1/2
5-42
Box 2
P(B|B2) 3/4
Red
(1/2)(1/4)
Blue (1/2)(3/4)
14
Example
A golfer has 12 golf shirts in his closet.
Suppose 9 of these shirts are white and
the others blue. He gets dressed in the
dark, so he just grabs a shirt and puts it
on. He plays golf two days in a row and
does not do laundry.
What is the likelihood both shirts selected
are white?
z
The event that the first shirt selected is white is W 1. The probability is P(W 1)
= 9/12
The event that the second shirt (W 2 )selected is also white. The conditional
probability that the second shirt selected is white, given that the first shirt
selected is also white, is P(W 2 | W 1) = 8/11.
To determine the probability of 2 white shirts being selected we use formula:
z
P(AB) = P(A) P(B|A)
P(W 1 and W 2) = P(W 1)P(W 2 |W 1) = (9/12)(8/11) = 0.55
z
z
5-43
Contingency Tables
A CONTINGENCY TABLE is a table used to classify sample
observations according to two or more identifiable
characteristics
A recent survey asked 100 people if they thought women in the
armed forces should be permitted to participate in combat.
combat The
results are shown in the table
GG eennddeerr
M
M aal lee
FFeemm aal lee
TToottaal l
YY eess
NN oo
TToottaal l
3322
1188
5500
88
4422
5500
4400
6600
110000
5-44
Example
1- Find the probability that the respondent answered
“yes” given that the respondent was a female.
z Solution: Let M = respondent was a male; F = respondent
was a female; Y = respondent answered “yes”;
yes ; N =
respondent answered “no”.
P(Y|F) = [P( F and Y) ]/P(F) = [8/100]/[50/100] = 4/25.
5-45
2- Find the probability that the respondent was a
male, given that the respondent answered “no”.
z Solution: P(M|N) = [P(N and M)]/P(N) =
[18/100]/[60/100] = 3/10.
15
Using Tree Diagrams
Using a tree diagram is useful for portraying
conditional and joint probabilities. It is
particularly useful for analyzing business
decisions involving several stages.
5-46
Example
A box contains 24 transistors, 4 of which are defective. If 4 are
sold at random, find the following probabilities a. Exactly 2
are defective.
P (ex actly 2 def ectiv es ) =
=
( # o f selectin g 2 d efectives) × (# o f selectin g 2 n o nd efectives)
# o f selectin g 4 tran sisto rs
4
⎛ 4 ! ⎞ ⎛ 20 ! ⎞
⎜
⎟×⎜
⎟
C 2 × 20C 2
1140
190
2 !× 2 ! ⎠ ⎝ 1 8 !× 2 ! ⎠
= ⎝
=
=
24 !
C
10626 1771
24
4
2 0 !× 4 !
b. None is defective. P (no defective ) = # of selecting no defective =
# of selecting 4 transistors
c. All are defective.
C
# of selecting defective
1
P (all defective ) =
# of selecting 4 transistors
=
4
4
24
C4
=
C 4 1615
=
C 4 3542
20
24
10626
d. At least 1 is defective.
C4
1615 1927
P (at least 1 defective ) = 1 − P ( no defective ) = 1 −
= 1−
=
3542 3542
24C 4
20
5-47
16