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CHAPTER 4
PROBABILITY
Outline
• 4-1 Introduction
• 4-2 Sample Spaces and Probability
• 4-3 The Addition Rules for
Probability
• 4-4 The Multiplication Rules and
Conditional Probability
Objectives
• Determine the sample spaces and find
the probability of an event using classical
probability.
• Find the probability of an event using
empirical probability.
• Find the probability of compounds events
using the addition rules.
Objectives
• Find the probability of compounds
events using the multiplication rules.
• Find the conditional probability of an
event.
4-1 Introduction
• What is probability?
Probability is the chance of occurrence
of a particular event.
Probability is measured on a scale from 0 to 1.
0
Event can
never
happen
1
Event is
absolutely certain
to happen
4-2 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.
• Sample space is the set of all possible
outcomes of a probability experiment.
4-2 Sample Spaces Example
Experiment
Sample space
Toss a
coin
Head, tail
Roll a
die
1,2,3,4,5,6
4-2 Event
• Event consists of a set of outcomes of a
probability experiment.
• An event with one outcome is called a simple
event.
• Event that consists of more than one outcome
is called a compound event.
4-2 Simple Event
• Experiment: Select a day of a week and
getting Tuesday.
• Outcome: Tuesday (One outcome)
4-2 Compound Event
• Experiment: Roll a die and getting odd
number.
• Outcomes: 1, 3, 5 (Three outcomes)
4-2 Tree diagram
• A tree diagram can be used to find all
possible outcomes of a probability
experiment.
H
H
First Toss
T
Second Toss
T
H
T
4-2 Type of Probability
Classical
Probability
VS
Empirical
Probability
4-2 Formula for Classical
Probability
• Classical probability assumes that all
outcomes in the sample space are
equally likely to occur.
• Equally likely events are events that
have the same probability of occurring.
4-2 Formula for Classical
Probability
Probability of an event E
=
Number of outcomes in E
Total number of outcomes in the sample space
n(E)
P (E) =
n(S)
Answer a multiple choice question with four
choices (A, B, C and D).
The probability of occurrence for
each outcome is the same which is 1/4.
Deliver a baby. The gender for a baby can be
either male or female.
The probability of getting male or female
will always be 1/2.
4-2 Classical Probability
Example
• An ordinary die is thrown. Find the
probability that the number obtained
(a) is less than 5.
(b) is a multiple of 3.
(c) is 9.
4-2 Classical Probability
Solutions:
(a) The outcomes are 1, 2, 3, and 4.
P (less than 5) = 4/6 = 2/3
(b) The outcomes are 3 and 6.
P (multiple of 3) = 2/6 = 1/3
(c) It is impossible to get a 9 when a die is rolled.
P (9) = 0
4-2 Classical Probability
Question:
• A card is drawn at random from an ordinary
pack containing 52 playing cards. Find the
probability that the card drawn
(a) is the diamond
(b) is the four of spades
4-2 Empirical Probability
• Some of the outcomes are not equally
likely thus their probabilities need to be
determined through empirical method.
• Empirical probability estimate the
probability of an outcome based on the
actual experience, observation, or
experiment.
4-2 Formula for Empirical
Probability
Given a frequency distribution, the probability
of an event being a given class is
Frequency for the class
P (E) =
Total frequencies in the distribution
= f
n
4-2 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 type AB blood. Set
up a frequency distribution.
4-2 Empirical Probability
Type
A
B
AB
O
Frequency
22
5
2
21
n = 50
4-2 Empirical Probability
Find the following probabilities based on
the frequency distribution :
(a) A person has type O blood.
(b) A person has type A or type B blood.
(c) A person has neither type A nor type O
blood.
(d) A person does not have type AB blood.
4-2 Empirical Probability
Solutions:
(a) P(O) = 21/50
(b) P(A or B) = 22/50 + 5/50 = 27/50
(c) P(neither A nor O) = 5/50 + 2/50 = 7/50
(d) P(not AB) = 1 - 2/50 = 48/50
4-2 Empirical Probability
Question
A ball is drawn from a box containing 10 red,
15 white, 5 green, and 5 black. Find the
probability that the ball is
(a)black
(b)red or green
(c)not white
4-2 Complementary Events
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 (read as “E bar”).
4-2 Complementary Events
Venn Diagram
P(E)
P(S) = 1
(a) Simple probability
P(E)
P(E)
(b) P(E) = 1- P(E)
4-2
Rule for Complementary
Events
P (E) = 1 - P (E)
or
P (E) = 1 - P (E)
or
P (E) + P (E) = 1
4-2 Complementary Events
Examples:
Find the complement of each event.
(a)
Rolling a die and getting 4.
(b)
Selecting a month and getting a month
that begins with a J.
(c)
Selecting a day of the week and getting a
weekday.
4-2 Complementary Events
Solutions:
(a)
Getting 1, 2, 3, 5, or 6
(b)
Getting February, March, April, May,
August, September, October, November, or
December.
(c)
Getting Saturday or Sunday.
4-3 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).
4-3 Addition Rules 1
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)
4-3 The Addition Rules for
Probability
P (A)
P (B)
P (S) = 1
Mutually exclusive events
4-3 Addition Rules 1
Example 1:
A box contains 3 chocolate doughnuts, 4 jelly
doughnuts, and 5 strawberry doughnuts. If a
person select one doughnut randomly, find the
probability that it is either a chocolate doughnut
or a strawberry doughnuts.
P (A or B) = P (A) + P (B)
4-3 Addition Rules 1
Solution:
P (chocolate or strawberry)
= P (chocolate) + P (strawberry)
= 3/12 + 5/12
= 8/12
= 2/3
4-3 Addition Rules 1
Example 2:
A day of the week is selected at random.
Find the probability that it is a weekend.
P (A or B) = P (A) + P (B)
4-3 Addition Rules 1
Solution:
P (Saturday or Sunday)
= P (Saturday) + P (Sunday)
= 1/7 + 1/7
= 2/7
In a conference, there are 12 researchers,
10 scientists, and 8 educators. If an attendant
is selected, find the probability of getting a
researcher or an educator.
4-3 Addition Rules 2
When two events A and B are not
mutually exclusive, the probability that A
or
or B will occur is
P (A or B) = P (A) + P (B) - P (A and B)
4-3 Addition Rules 2
P (A and B)
P (A)
P (B)
P (S) = 1
Non-mutually exclusive events
4-3 Addition Rules 2
Example:
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.
4-3 Addition Rules 2
STAFF
FEMALES
MALES
TOTAL
NURSES
7
1
8
PHYSICIANS
3
2
5
TOTAL
10
3
13
4-3 Addition Rules 2
Solution:
P (nurse or male)
= P (nurse) + P (male) - P (male nurse)
= 8/13 + 3/13 - 1/13
= 10/13
In a group of 30 students all study at least one
of the subjects physics and biology, 20 attend
the physics class and 21 attend the biology class.
Find the probability that a student chosen at
random studies both physics and biology.
In a statistics class there are 18 juniors and 10
seniors; 6 of the seniors are females, and 12 of
the juniors are males. If a student is selected at
random, find the probability of selecting:
(a)A junior or a female
(b)A senior or a female
(c)A junior or a male
4-4 The Multiplication Rules and
Conditional Probability
• 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.
4-4 Multiplication Rules 1
When two events A and B are
independent, the probability
of both
or
occurring is
P (A and B) = P (A) • P (B)
4-4 Multiplication Rules 1
Example 1:
• A coin is flipped and a die is rolled.
Find the probability of getting a head on
the coin and a 4 on the die.
4-4 Multiplication Rules 1
Solution:
P (head and 4)
= P(head) • P(4)
= 1/2 • 1/6
= 1/12
or
4-4 Multiplication Rules 1
Example 2:
A Harris poll found that 46% of Americans
or
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 great stress at
least once a week.
4-4 Multiplication Rules 1
Solution:
Let S denote stress. Then
or
P (S and S and S) = P (S) • P (S) • P (S)
= (0.46) (0.46) (0.46)
= 0.097
4-4 Multiplication Rules 1
Example 3:
The probability that a specific medical
test will show positive is 0.32. If four
person are tested, find the probability
that all four will show positive.
4-4 Multiplication Rules 1
Solution:
Let T denote a positive test result. Then
or
P (T and T and T and T)
= P (T) • P (T) • P (T) • P (T)
= (0.32)4
= 0.010.
Two types of metal A and B, which have been
treated with a special coating of paint have
probabilities of 1/4 and 1/3 respectively of lasting
four years without rusting. Find the probability that
(a)Both last four years without rusting
(b)At least one of them lasts four years without
rusting.
At a local university 54% of incoming first-year
student have computers. If three students are
selected at random, find the probability that
(a)All have computers
(b) None have computers
(c) At least one has computer
4-4 The Multiplication Rules and
Conditional Probability
• When the outcome or occurrence of the
first event affects the outcome or
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.
4-4 The Multiplication Rules and
Conditional Probability
• The conditional probability of an event B in
relationship to an event A is the probability
r
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).
• Note: This does not mean B divided by A.
4-4 Multiplication Rules 2
When two events A and B are dependent,
the probability of both occurring is
or
P (A and B) = P (A) • P (B|A)
4-4 The Multiplication Rules
and Conditional Probability
Example 1:
In a shipment of 25 refrigerators, two are
defective. If two refrigerators are randomly
selected and tested, find the probability that
both are defective if the first one is not
replaced after it has been tested.
4-4 The Multiplication Rules
and Conditional Probability
Solution:
Let D denote defective. Since the events
or
are dependent,
P (D1 and D2) = P (D1) • P (D2|D1)
= (2/25) (1/24)
= 2/600
= 1/300
4-4 The Multiplication Rules
and Conditional Probability
Example 2:
A person owned a collection of 30 CDs, of
which 5 are country music. If 2 CDs are
selected at random, one by one without
replacement, find the probability that
both are country music.
4-4 The Multiplication Rules
and Conditional Probability
Solution:
Let C denote country music. Since the
or
events are dependent,
P (C1 and C2) = P (C1) • P (C2|C1)
= (5/30) (4/29)
= 20/870
= 2/87
4-4 The Multiplication Rules
and Conditional Probability
Example 3:
Box 1 contains 2 red balls and 1 blue ball.
or
Box 2 contains 3 blue balls and 1 red ball.
A coin is tossed. If it falls heads up, box 1
is selected and a ball is drawn. If it falls tail
up, box 2 is selected and a ball is drawn.
Find the probability of selecting a red ball.
4-4 The Multiplication Rules
and Conditional Probability
BALL
BOX
P(R|B1)
1/2
P(R|B2)
P(B2)
1/2
Red
1/3
Blue 1/2 • 1/3 = 1/6
1/4
Red
3/4
Blue 1/2 • 3/4= 3/8
1/2 • 2/3 = 1/3
Box 1
P(B|B1)
P(B1)
2/3
1/2 • 1/4 = 1/8
Box 2
P(R|B2)
4-4 The Multiplication Rules
and Conditional Probability
Solution:
Since red ball can be obtained from box 1
and box 2.
P(red) = 1/3 + 1/8
= 8/24 + 3/24
= 11/24
A box contains 6 red pens and 3 blue pens. A pen is
selected at random, the colour is noted and the
pen is retained. After this, a second pen is selected
and the colour is noted. Find the probabilities of
obtaining
i) both red pens
ii) two pens of different colours
4-4 Conditional Probability Formula
The probability that 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) =
P (A and B)
P (A)
4-4 Conditional` Probability
Example:
For married couples the probability that the husband
has passed his driving test is 7/10 and the probability
that the wife has passed her driving test is 1/2. The
probability that both of them passed the driving test
is 7/15. Find the probability that the husband has
passed, given that the wife has passed.
4-4 Conditional Probability
Solution
P(H) = 7/10; P(W) =1/2; P (H and W) = 7/15
P (H|W) = P (H and W) / P(W)
= 7/15 / 1/2
= 14/15