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+Chapter 5
Probability: What Are the Chances?
 5.1
Randomness, Probability, and Simulation
 5.2 Probability Rules
 5.3 Conditional Probability and Independence

Learning Objectives
After this section, you should be able to…

DEFINE conditional probability

COMPUTE conditional probabilities

DESCRIBE chance behavior with a tree diagram

DEFINE independent events

DETERMINE whether two events are independent

APPLY the general multiplication rule to solve probability questions
is Conditional Probability?
When we are trying to find the probability that one event will happen
under the condition that some other event is already known to have
occurred, we are trying to determine a conditional probability.
Definition:
The probability that one event happens given that another event
is already known to have happened is called a conditional
probability. Suppose we know that event A has happened.
Then the probability that event B happens given that event A
has happened is denoted by P(B | A).
Read | as “given that”
or “under the
condition that”
Conditional Probability and Independence
The probability we assign to an event can change if we know that some
other event has occurred. This idea is the key to many applications of
probability.
+
 What
Grade Distributions
+
 Example:
E: the grade comes from an EPS course, and
L: the grade is lower than a B.
Total
6300
1600
2100
Total 3392 2952
Find P(L)
P(L) = 3656 / 10000 = 0.3656
Find P(E | L)
P(E | L) = 800 / 3656 = 0.2188
Find P(L | E)
P(L| E) = 800 / 1600 = 0.5000
3656
10000
Conditional Probability and Independence
Consider the two-way table on page 314. Define events
Diagrams
Consider flipping a
coin twice.
What is the probability
of getting two heads?
Sample Space:
HH HT TH TT
So, P(two heads) = P(HH) = 1/4
Conditional Probability and Independence
We learned how to describe the sample space S of a chance
process in Section 5.2. Another way to model chance
behavior that involves a sequence of outcomes is to construct
a tree diagram.
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 Tree
Multiplication Rule
General Multiplication Rule
The probability that events A and B both occur can be
found using the general multiplication rule
P(A ∩ B) = P(A) • P(B | A)
where P(B | A) is the conditional probability that event
B occurs given that event A has already occurred.
Conditional Probability and Independence
The idea of multiplying along the branches in a tree diagram
leads to a general method for finding the probability P(A ∩ B)
that two events happen together.
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 General
For any two events A and B,
P(A and B) = P(A) ∙ P(B|A)
P(A and B) = P(B) ∙ P(A|B)
∙
The probability of A and B occurring jointly
is the probability that B occurs given that A has
already occurred multiplied by the probability that
A occurs
Teens with Online Profiles
+
 Example:
The Pew Internet and American Life Project finds that 93% of teenagers (ages
12 to 17) use the Internet, and that 55% of online teens have posted a profile
on a social-networking site.
P(online )  0.93
What percent of teens are online and have posted a profile?
P(profile | online )  0.55
P(A ∩ B) = P(A) • P(B | A)

P(online and have profile )  P(online
) P(profile | online )

 (0.93)(0.55)
P(online and have profile)  0.5115
51.15% of teens are online and have
posted a profile.

Who Visits YouTube?
See the example on page 320 regarding adult Internet users.
What percent of all adult Internet users visit video-sharing sites?
P(video yes ∩ 18 to 29) = 0.27 • 0.7
=0.1890
P(video yes ∩ 30 to 49) = 0.45 • 0.51
=0.2295
P(video yes ∩ 50 +) = 0.28 • 0.26
=0.0728
P(video yes) = 0.1890 + 0.2295 + 0.0728 = 0.4913
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 Example:
A Special Multiplication Rule
Definition:
Multiplication rule for independent events
If A and B are independent events, then the probability that A
and B both occur is
P(A ∩ B) = P(A) • P(B)
Example:
Following the Space Shuttle Challenger disaster, it was determined that the failure
of O-ring joints in the shuttle’s booster rockets was to blame. Under cold
conditions, it was estimated that the probability that an individual O-ring joint would
function properly was 0.977. Assuming O-ring joints succeed or fail independently,
what is the probability all six would function properly?
P(joint1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK)
=P(joint 1 OK) • P(joint 2 OK) • … • P(joint 6 OK)
=(0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87
Conditional Probability and Independence
When events A and B are independent, we can simplify the
general multiplication rule since P(B| A) = P(B).
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 Independence:
Two events A and B are independent if knowing that
one occurs does not change the probability that the
other occurs.
If A and B are independent, then P(A and B) = P(A) ∙ P(B)
If asked if two events are independent, you would prove this
by showing that this formula holds true. P(A and B) = P(A) ∙ P(B)
You can also show independence by showing that
P(A) = P(A|B)
Are disjoint events independent?
Disjoint events are never INDEPENDENT, for if one occurs,
then we know the other can not occur…so the outcome of
one will effect the outcome of the other.
Example:
Probability and Independence
Are the events “male” and “left-handed”
independent? Justify your answer.
P(A) = P(A|B)
P(left-handed) = P(left-handed | male)
7/50 = 3/23
0.14 = 0.13
These probabilities are not equal, therefore the
events “male” and “left-handed” are not independent.
P(A and B) = P(A) ∙ P(B)
P(left-handed and male) = P(left-handed) P(male)
3/50 =
7/50
23/50
0.06 =
0.14
0.13
0.06 =
0.0182
These probabilities are not equal, therefore the
events “male” and “left-handed” are not independent.
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 Conditional
Conditional Probabilities
General Multiplication Rule
P(A ∩ B) = P(A) • P(B | A)
Conditional Probability Formula
To find the conditional probability P(B | A), use the formula
=
Conditional Probability and Independence
If we rearrange the terms in the general multiplication rule, we
can get a formula for the conditional probability P(B | A).
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 Calculating
When P(A) > 0, the conditional probability of B
given A is:
The probability of B occurring, given that A
occurs, is the probability of A & B jointly
occurring divided by the probability of the stated
condition (A) occurring
Who Reads the Newspaper?
What is the probability that a randomly selected resident who reads USA
Today also reads the New York Times?
P(A  B)
P(B | A) 
P(A)


P(A  B)  0.05
P(A)  0.40
0.05
P(B | A) 
 0.125
0.40
There is a 12.5% chance that a randomly selected resident who reads USA
Today also reads the New York Times.
Conditional Probability and Independence
In Section 5.2, we noted that residents of a large apartment complex can be
classified based on the events A: reads USA Today and B: reads the New
York Times. The Venn Diagram below describes the residents.
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 Example:
+ Section 5.3
Conditional Probability and Independence
Summary
In this section, we learned that…

If one event has happened, the chance that another event will happen is a
conditional probability. P(B|A) represents the probability that event B
occurs given that event A has occurred.

Events A and B are independent if the chance that event B occurs is not
affected by whether event A occurs. If two events are mutually exclusive
(disjoint), they cannot be independent.

When chance behavior involves a sequence of outcomes, a tree diagram
can be used to describe the sample space.

The general multiplication rule states that the probability of events A
and B occurring together is P(A ∩ B)=P(A) • P(B|A)

In the special case of independent events, P(A ∩ B)=P(A) • P(B)

The conditional probability formula states P(B|A) = P(A ∩ B) / P(A)
+
Homework: #63-70, 77, 83, 84, 87
In the next Chapter…
We’ll learn how to describe chance processes using the
concept of a random variable.
We’ll learn about
 Discrete and Continuous Random Variables
 Transforming and Combining Random Variables
 Binomial and Geometric Random Variables
Conditional Probability – Example #1
Using a deck of playing cards: Initiating a game and
starting to deal out cards, what is the likelihood a
person is dealt a Heart?
If the person is dealt a heart, what is NOW the
likelihood the person is again dealt a heart (i.e.,
without replacement)?
Does the probability going from 25% to 23.5% make sense?
Conditional Probability – Example #2
Suppose that for a certain Caribbean island in any 3-year period:
the probability of a major hurricane is .25, the probability of water
damage is .44, and the probability of both a hurricane and water
damage is .22.
What is the probability of water damage
given that there is a hurricane?
P  water damage|hurricane  
P  water damage and hurricane 
.22
.25
 .88

P  hurricane 
Conditional Probability – Example #3
If P(A) = 0.45, P(B) = 0.35, and A & B are
independent, find P(A or B).
Is A & B disjoint?
NO, independent events cannot be disjoint
If A &
are –disjoint,
are they
P(A or B) = P(A)
+B
P(B)
P(A & B)
Disjoint
events are
independent?
Disjoint events
doIf not happen at the same
dependent!
How can you
independent,
time.
find the
P(A or B) =So,
.45if +A multiply
.35
.45(.35)
=
0.6425
occurs, can B occur?
probability of
A & B?
Conditional Probability – Example #4
Suppose I will pick two cards from a standard deck without
replacement. What is the probability that I select two spades?
Are the cards independent? NO
P(A & B) = P(A) · P(B|A)
Read “probability of B
given that A occurs”
P(Spade & Spade) = 1/4 · 12/51 = 1/17
The probability of getting a spade given
that a spade has already been drawn.
Conditional Probability – Example #5
In a recent study it was found that the
probability that a randomly selected student
is a girl is .51 and a girl that plays sports is
.10. If the student is female, what is the
probability that she plays sports?
P(S  F) .1
P(S | F) 

 .1961
P(F)
.51
Conditional Probability TABLES – Example #6
A survey was given to 200 college students. The data breakdown of
gender and pierced ears is listed below, with defined events:
Using the table, calculate the following probabilities:
a) Probability of Male or Pierced Ears: b) Probability of Male and Pierced Ears:
Pierced Ears?
Gender
Yes
No
Total
Male
24
76
100
Female
96
4
100
Total
120
80
200
Conditional Probability Tables – Example #6 cont.
c) Find the probability the student is Male,
given the student has pierced ears.
d) Find the probability the student is Female,
given that the student has pierced ears.
Gender
Male
Female
Total
Pierced Ears?
Yes
No
24
76
96
4
120
80
Total
100
100
200
e) Find the probability the student
is Female, given that the student
does not have pierced ears.
Conditional Probability TREE DIAGRAMS– Example #7
Leah is flying from Boston to Denver with a connection in Chicago. The
probability her first flight leaves on time is .15. If the flight is on time, the
probability that her luggage will make the connecting flight in Chicago is
.95, but if the first flight is delayed, the probability the luggage will make
it is only .65.
a) Are the first flight leaving in time and the luggage making the connection
independent events?
b) What is the probability that her luggage arrives in Denver with her?
c)
Suppose you pick up Leah in Denver and her luggage is not there. What is the
probabilty that Leah’s first flight is delayed?
a)
b)
c)
Are the first flight leaving in time and the luggage making the connection
P (on time and make) = P(on time) · P(make)
independent events?
.1425 = .15 · (.1425+.5525)
What is the probability that her luggage
.1425 = .10425
P (makes) = .1425 + .5525
arrives in Denver with her?
= .695
Not Independent
Suppose you pick up Leah in Denver and her luggage is not there. What is the
probability that Leah’s first flight is delayed?
.15
.85
.95
On Time
.05
.65
Delayed
.35
Luggage makes connection
On time and makes (.15)(.95) = .1425
On time and not make (.15)(.05) = .0075
Luggage does not make connection
Luggage makes connection
Delayed and makes (.85)(.65) = .5525
delayed and not make (.85)(.35) = .2975
Luggage does not make connection
All should add up to 1
Homework: #63-70, 77, 83, 84, 87
In the next Chapter…
We’ll learn how to describe chance processes using the
concept of a random variable.
We’ll learn about
 Discrete and Continuous Random Variables
 Transforming and Combining Random Variables
 Binomial and Geometric Random Variables