Chapter 5
Probability
Created by Kathy Fritz
Can ultrasound accurately predict the
gender of a baby?
The paper “The Use of Three-Dimensional Ultrasound for
Fetal Gender Determination in the First Trimester” (The
British Journal of Radiology [2003]: 448-451) describes a
study of ultrasound gender prediction. An experienced
radiologist looked at 159 first trimester ultrasound images
and made a gender prediction for each one.
When each baby was born, the ultrasound gender prediction
was compared to the baby’s actual gender.
This table summarizes the resulting data:
Radiologist 1
Predicted Male
Predicted Female
Baby is Male
74
12
Baby is Female
14
59
Notice that
theof
gender
prediction
based can
on the
All
these
questions
be
ultrasound image is NOT always correct.
answered using the methods
introduced
in this
chapter.
The paper also
included gender
predictions
by a second
radiologist, who looked at 154 first trimester
ultrasound inmages.
Radiologist 2
Predicted Male
Predicted Female
Baby is Male
81
8
Baby is Female
7
58
Interpreting Probabilities
Probability
Relative Frequency
Law of Large Numbers
Basic Properties
Probability
We often find ourselves in situations where the
To is
quantify
the likelihood of an occurrence, a
outcome
uncertain:
number between 0 and 1 can be assigned to an
outcome.
When a ticketed passenger shows up at the airport, she
facesAtwo
possible outcomes:
she is able
to 1take
probability
is a number(1)between
0 and
thatthe
flight,reflects
or (2) she
islikelihood
denied a seat
as a resultof
ofsome
the
of occurrence
overbooking by the airline
and must take a later flight.
outcome.
Based on her past experience, the passenger believes
that the chance of being denied a seat is
small or unlikely.
Subjective Approach to Probability
The subjective interpretation of probability is when a
probability is interpreted as a personal measure of the
strength of the belief that an outcome will occur.
A probability of 1 represents a belief that the
outcome will certainly occur.
A probability of 0 represents a belief that the
outcome will certainly NOT occur.
All other probabilities fall between these
Because different people may have different
extremes.
subjective beliefs,two
they
may assign different
probabilities to the same outcome.
Relative Frequency Approach
In the relative frequency interpretation of
probability, the probability of an outcome, denoted
by P(outcome), is interpreted as the proportion of
the time that the outcome occurs in the long run.
Relative frequency can be computed by:
A probability of 1 corresponds to an outcome
that occurs 100% of the time.
number of times outcome occurs
A probability
𝑃 outcome
= of 0 corresponds to an outcome
that occurstotal
0% ofnumber
the time.of trials
A package delivery service promises 2-day delivery
between 2 cities in California but is often able to deliver
the packages in just 1 day. The company reports that the
probability of next-day delivery is 0.3.
Suppose
you trackthis
the probability
delivery of would
packages
shipped
One waythat
to interpret
be to
say
with that
this company.
each new
package
shipped,
in the longWith
run, about
30 out
of every
100 you
could compute
the relative
frequency
packages shipped
packages
shipped
arrive in of
1 day.
so far that have arrived in 1 day:
number of packages that arrived in 1 day
total number of packages shipped
Here is a graph displaying the relative frequencies
for each of the first 15 packages shipped.
Here is a graph
displaying the
As the number of packages
in the
relative frequencies
sequence increases, for
theeach
relative
of the
50 packages
frequency does not first
continue
to
shipped. settles
fluctuate wildly, but instead
down and approaches a specific value,
which is the probability of interest.
Here is a graph
displaying the
relative frequencies
for each of the
first 1000 packages
shipped.
Law of Large Numbers
As the number of observations increases,
the proportion of the time that an
outcome occurs gets close to the
probability of that outcome.
The Law of Large Numbers is the basis for
the relative frequency interpretation of
probabilities.
Some Basic Properties of Probability
1.
The probability of any outcome is a
number between 0 and 1.
2. If outcomes can’t occur at the same time,
then the probability that any one of them
will occur is the sum of their individual
probabilities.
A large auto center sells cars made by
many different manufacturers. Two
of these are Honda and Toyota.
Suppose: P(Honda) = 0.25 and P(Toyota) = 0.14
An interpretation
for
this two
value
Why don’t
these
Consider
the
make
ofabout
the next
car
sold.
Can the
outcomes
Honda
and
Toyota
is that
25
out
of
probabilities
have
a every
sumand
of 1?
happen
at the
time?
100 cars
soldsame
would
be Hondas.
What is the probability that the next car sold
is either a Honda or a Toyota?
P(Honda or Toyota) = 0.25 + 0.14 = 0.39
Some Basic Properties of Probability
3. The probability that an outcome will not
occur is equal to 1 minus the probability
that the outcome will occur.
Because a probability represents a long-run
relative
in situations
exact
Recall
the frequency,
car dealership
(P(Honda)where
= 0.25):
probabilities are not known, it is common to
estimate probabilities based on observation.
What is the probability that the next car sold
is not a Honda?
P(not Honda) = 1 - 0.25 = 0.75
Computing Probabilities
Chance Experiment
Sample Space
Event
Classical Approach to Probability
Chance Experiment
A chance experiment is any activity or situation
in which there is uncertainty about which of two
or more
outcomesof
will
result.experiments.
Thesepossible
are all examples
chance
These are the outcomes of chance experiments.
Suppose two six-sided dice are rolled and they
both land on sixes.
Or a coin is flipped and it lands on heads.
Or record the color of the next 20 cars to pass
an intersection.
Sample Space
The collection of all possible outcomes of a
chance experiment is the sample space for the
experiment.
Sample space = {MH, FH, MT, FT}
Consider a chance experiment to investigate whether men
or women are more likely to choose a hybrid engine over a
traditional internal combustion engine when purchasing a
This
an example
of a sample
Honda Civic
at aisparticular
dealership.
The typespace.
of vehicle
purchased (hybrid or traditional) will be determined and the
customer’s gender will be recorded.
A list of all possible outcomes are:
male, hybrid
female, hybrid
male, traditional
female, traditional
Chance Experiment
An event is any collection of outcomes from the
sample space of a chance experiment.
Recall
the can
situation
in which a by
person
purchases
Honda
An event
be represented
a name,
such asahybrid,
Civic: or by Sample
space letter,
= {MH, such
FH, MT,
an uppercase
as AFT}
, B, or C.
A simple
is an
event consisting
of
Eachevent
of these
4 outcomes
are simple events.
exactly on outcome.
Identify the following events:
traditional = {MT, FT}
female =
{FH, FT}
Classical Approach to Probability
When the outcomes in the sample space of a
chance experiment
are
equally likely,
the
The classical
approach
to probability
probability
of
an
event
E,
denoted
by
P
(
E
),
is
the
works well for chance experiments that
ratio ofhave
the number
favorable
to E
a finite of
setoutcomes
of outcomes
that are
to the total number equally
of outcomes
likely. in the sample
space:
number of outcomes favorable to E
𝑷 𝑬 =
number of outcomes in the sample space
Four students (Adam (A), Bettina (B), Carlos (C), and
Debra(D)) submitted correct solutions to a math contest
that had two prizes. The contest rules specify that if
more than two correct responses are submitted, the
winners will be selected at random from those submitting
correct responses.
What is the sample space for selecting the two winners
from the four correct responses?
Sample space = {AB, AC, AD, BC, BD, CD}
Because the winners are selected at random,
the six possible outcomes are equally likely.
Four students (Adam (A), Bettina (B), Carlos (C), and
Debra(D)) submitted correct solutions to a math contest
that had two prizes. The contest rules specify that if
more than two correct responses are submitted, the
winners will be selected at random from those submitting
correct responses.
Sample space = {AB, AC, AD, BC, BD, CD}
Let E be the event that both selected winners are the
same sex.
What is the probability of E?
2
𝑃 𝐸 = = 0.333
6
Four students (Adam (A), Bettina (B), Carlos (C), and
Debra(D)) submitted correct solutions to a math contest
that had two prizes. The contest rules specify that if
more than two correct responses are submitted, the
winners will be selected at random from those submitting
correct responses.
Sample space = {AB, AC, AD, BC, BD, CD}
Let F be the event that at least one of the selected
winners is female.
What is the probability of F?
5
𝑃 𝐸 = = 0.833
6
Relative Frequency Approach to Probability
The probability of an event E, denoted by P(E), is
When a chance experiment is performed,
defined to be the value approached by the
some events may be likely to occur,
relatively frequency of occurrence of E in a very
whereas others may not be as likely to
long series of observations from a chance
occur. In cases like these, the classical
experiment. If the number of observations is
approach is not appropriate.
large,
number of times E occurs
𝑃 𝐸 ≈
number of repetitions
Suppose that you perform a chance experiment
that consists of flipping a cap from a 20-ounce
bottle of soda and noting whether the cap lands
with the top up or down.
Do you think that the event U, the
cap landing
up, and
event D, the
You carry
out thistop
chance
experiment
by flipping
landing
areifequally
the capcap
1000
timestop
anddown,
record
it lands top up
likely?
Whylands
or Why
or top down.
The cap
top not?
up 694 times.
694
𝑃 𝑈𝑝 =
= 0.694
1000
Probabilities of More
Complex Events
Union
Intersection
Complement
Mutually Exclusive Events
Independents Events
Consider the chance experiment that consists of
selecting a student at random from those
enrolled at a particular college.
There are 9000 students enrolled at the college
Here are some possible events:
F = event that the selected student is female
O = event that the selected student is older than 30
A = event that the selected student favors the expansion
of the athletic program
S = event that the selected student is majoring is one of
the lab sciences
Complement
If E is an event, the complement of E, denoted
EC, is the event that E does not occur.
The probability of EC can be computed from the
probability of E as follows:
𝑃 𝐸 𝐶 = 1 − 𝑃(𝐸)
Suppose that
students
favor the
𝑃 4300
𝐴𝐶 of=the
1 9000
− 0.48
= 0.52
expansion of the athletic program.
4300
𝑃 𝐴 =
= 0.48
9000
What is the probability of event A not occurring?
Intersection
If E and F are events, the intersection of E and F
is denoted by 𝐸 ∩ 𝐹 and is the new event that
both E and F occur.
This is the symbol
for “intersection”.
Consider the events:
O = event that the selected student is older than 30
S = event that the selected student is majoring is one of
the lab science
This table summaries the occurrence of these events:
Intersection
If E and F are events, the intersection
ofin
Elab
and F
Majoring
Not majoring
in lab
science
science
AND
AND
is denoted by 𝐸 ∩ 𝐹 and is the new
event
that
Not
Over
over
3030
both E and
F
occur.
The numbers in red corresponds to the
S
SC
intersections
of the events.
(Majoring in
Lab Science)
(Not Majoring
in Lab Science)
Total
O (Over 30)
400
1700
2100
OC (Not over 30)
1100
5800
6900
Total
1500
7500
9000
What is the probability of a randomly selected student is
older than 30 AND is majoring in a lab science?
400
𝑃 𝑂∩𝑆 =
= 0.04
9000
Union
If E and F are events, the union is denoted by
𝐸 ∪ 𝐹 . The event 𝐸 ∪ 𝐹 is the new event that
E or F occur.
This is the symbol
Consider the events:
for “union”.
O = event that the selected student is older than 30
A = event that the selected student favors the expansion
of the athletic program
This table summaries the occurrence of these events:
Union
If E and F are events,
theA,union
denoted by
The event
favorsissale
𝐸 ∪The
𝐹 . The event 𝐸 ∪of
𝐹 alcohol
is the new event that
E or
F occur.
event
A
AC
O, over
30
(Favors
Expansion)
(Does Not Favor
Expansion)
Total
O (Over 30)
1600
500
2100
OC (Not over 30)
2700
4200
6900
Total
4300
4700
9000
What is the probability of a randomly selected
student is older than 30 OR favors the expansion of the
athletic program?
1600 + 500 + 2700
𝑃 𝑂∪𝐴 =
= 0.53
9000
Hypothetical 1000
You can use tables to compute the probability
of an intersection of two events and the
probability of a union of two events.
InInmany
situations, you may ONLY know the
the previous examples, this was possible because a
probabilities
some
events.
In this
it is
student was of
to be
selected
at random
andcase,
because
often
possible
to create
a “hypothetical
1000”
the number
of students
falling
into each of the
cells
the appropriate
tabletowere
given.
table andofthen
use the table
compute
probabilities.
The report “TV Drama/Comedy Viewers and Health
Information” (www.cdc.gov/healthmarketing)
describes a large survey that was conducted by the
Centers for Disease Control (CDC). The CDC believed that
the sample was representative of adult Americans.
Let’s investigate these events (taken from questions on
the survey):
L = event that a randomly selected adult American reports
learning something new about a health issue or disease
from a TV show in the previous 6 months.
F = event that a randomly selected adult American is
female.
Data from the survey were used to estimate the following
probabilities:
𝑃 𝐿 = 0.58 𝑃 𝐹 = 0.5 𝑃 𝐿 ∩ 𝐹 = 0.31
CDC study continued
𝑃 𝐿 = 0.58 𝑃 𝐹 = 0.5 𝑃 𝐿 ∩ 𝐹 = 0.31
F (female)
L (learned from TV)
Not L
310
190
Not F
270
230
Total
500
500
Total
580
420
1000
What is the probability that a randomly selected adult
P(
∩ Fhas
)you
tells
you58%
that
31%
of1000
the about
1000 people
are
American
learned
something
new
ashould
health
P(P(
L)FLtells
that
of
the
people
be
)
tells
you
that
the
F
row
is
(0.50)(1000)
=
500
Begin
by
rows
and
columns
ofprevious
thethe
table.
both
female
and
health
information
from
a TV
Fill
in labeling
the
remaining
cells
to in
complete
table.
issue
or
disease
from
a TV
show
the
6 Put
in the
Llearned
row:
(0.58)(1000)
=
580.
and
the
“hypothetical
1000”
in the bottom right cell.
show.
months or is female?
that the Not F row is310+270+190
1000 - 500 = 500
𝐿 ∪Not
𝐹 L=row have a sum
= of
0.770
The L row and𝑃 the
1000.
1000
The cell for L and F is (0.31)(1000) = 310.
Let’s look at the hypothetical table once more.
Suppose: P (A) = 0.6, P (B C) = 0.7, and P (A ∩ B) = 0.2
B
BC
Total
A
AC
Total
200
100
400
600
300
400
300
700
1000
It does not matter which
What is the probability
A oronBthe
happening?
eventofgoes
side or on
the top.
200 + 100 + 400
700
𝑃 𝐴∪𝐵 =
=
= 0.7
1000
1000
Mutually Exclusive Events
Two events E and F are mutually exclusive if they
can NOT occur at the same time.
Sometimes people call the emergency 9-1-1 number to
report situations that are not considered emergencies
(such as to report a lost dog). Let two events be:
M = event that the next call to 9-1-1 is for a medical
emergency
N= event that the next call to 9-1-1 is not considered an
emergency
Suppose that you know P(M) = 0.30 and P(N) = 0.20.
Events M and N are mutually exclusive because the next
call can’t be both a medical emergency and a call that is
not considered an emergency.
Mutually Exclusive Events
P(M) = 0.30 and P(N) = 0.20
𝑃 𝑀∩𝑁 =0
300 + 200
A𝑃
“hypothetical
𝑀 ∪ 𝑁 1000”
= table is shown
=below.
0.50 The
1000
uppermost cell must be 0
when the two events are
mutually exclusive.
N (Non-emergency)
Not N
Total
M (Medical Emergency)
Not M
0
300
300
200
500
700
Total
200
800
1000
Addition Rule for Mutually Exclusive Events
If E and F are mutually exclusive events, then
𝑃 𝐸∩𝐹 =0
and
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃(𝐹)
Independent Events
Two events are independent if the probability
that
Because
one the
event
twooccurs
components
is not
operate
affected
independently
by
of
each other,
learning that
monitor
has has
needed
knowledge
of whether
thethe
other
event
warranty service would not effect your assessment of
occurred.
the likelihood that the keyboard will need repair.
Suppose that you purchase a desktop computer system
with a separate monitor and keyboard. Two possible
events are:
Event 1: The monitor needs service while under
warranty.
Event 2: The keyboard needs service while under
warranty.
Dependent Events
Two events are dependent if knowing that one
event has occurred changes the probability that
the other event occurs.
Consider a university’s course registration process, which
divides students into 12 priority groups. Overall, only 10% of
all students receive all requested classes, but 75% of those in
the first priority group receive all requested classes.
You would say that the probability that a randomly selected
student at this university receives all requested class is 0.10.
However, if you know that the selected student is in the first
priority group, you revise the probability that the student
receives all requested classes to 0.75.
These two events are said to be dependent events.
Multiplication Rule for Two
Independent Events
If two events, E and F, independent, the
probability that both events occur is the product
of the individual event probabilities.
𝑃 𝐸 ∩ 𝐹 = 𝑃 𝐸 𝑃(𝐹)
More generally, if there are k independent events, the
probability that all the events occur is the product of all
individual event probabilities.
The Diablo Canyon nuclear power plant in California has a
warning system that includes a network of sirens. When
the system is tested, individual sirens sometimes fail.
The sirens operate independently of one another.
Imagine that you live near Diablo Canyon and that there
are two sirens that can be heard from your home. You
might be concerned about the probability that both Siren
1 and Siren 2 fail. (When the siren system is activated,
about 5% of the individual sirens fail.)
Using the multiplication rule for independent events:
𝑃 𝑆𝑖𝑟𝑒𝑛 1 𝑓𝑎𝑖𝑙𝑠 ∩ 𝑆𝑖𝑟𝑒𝑛 2 𝑓𝑎𝑖𝑙𝑠 = 0.05 0.05
= 0.0025
Conditional Probability
Sometimes the knowledge that one event has occurred
changes our assessment of the likelihood that another
event occurs.
Consider a population in which 0.1% of all the individuals
have a certain disease. The presence of the disease
cannot be discerned from appearances, but there is a
diagnostic test available. Unfortunately, the test is not
always correct.
Suppose that 80% of those with positive test results
actually have the disease and the other 20% of those
with positive test results actually do NOT have the
disease (false positive).
Disease example continued . . .
Consider the chance experiment in which an individual is
randomly selected from the population.
Let:
E = event that the individual has the disease
F =vertical
event that
individual's
The
line the
is read
“given”. diagnostic test is
This
is an example of conditional probability.
positive
P(E|F) denotes the probability that event
E (has disease) GIVEN that event F
(tested positive) occurs.
Conditional Probability
Conditional probability is a probability that
takes into account a given condition has
occurred.
P(A|B)
is read as
the probability of event A occurring GIVEN
event B has occurred.
Recall the example in the Chapter Preview section about
gender predictions based on ultrasounds performed
during the first trimester of pregnancy. The table
below summarizes the data for Radiologist 1.
Radiologist 1
Baby is Male
Predicted Male
Predicted
Female
Total
74
12
86
59
73
71
159
This question is about ALL
Baby is Female
14
159 ultrasound predictions.
Total
88
How likely is it that a predicted gender is correct?
74 + 59
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑔𝑒𝑛𝑑𝑒𝑟 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 =
= 0.836
159
Gender prediction example continued.
Radiologist 1
Predicted Male
Predicted
Female
Total
Baby is Male
74
12
86
Baby is Female
14
59
73
Total
88
71
159
Is a predicted gender more likely to be correct when
the baby is male than when the baby is female?
74
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑔𝑒𝑛𝑑𝑒𝑟 𝑖𝑠 𝑚𝑎𝑙𝑒|𝑏𝑎𝑏𝑦 𝑖𝑠 𝑚𝑎𝑙𝑒 =
= 0.86
86
This question is based on two conditions:
59
the 86
male babies
or the𝑏𝑎𝑏𝑦
73 female
babies. =
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑
𝑔𝑒𝑛𝑑𝑒𝑟
𝑖𝑠 𝑓𝑒𝑚𝑎𝑙𝑒
𝑖𝑠 𝑓𝑒𝑚𝑎𝑙𝑒
73
= 0.81
The appropriate row total or column total is used
Radiologist
1 is slightly in
more
to be correct
when the
as the denominator
the likely
probability
calculation.
baby is male than when the baby is female.
Gender prediction example continued.
Radiologist 1
Male
This is a Predicted
condition.
In thePredicted
probability
Female
statement, the condition follows the
Baby is Male
74
vertical
line “|”. 12
Total
86
Baby is Female
14
59
73
Total
88
71
159
If the predicted gender is female, should you paint the
nursery pink?
𝑃 𝑓𝑒𝑚𝑎𝑙𝑒|𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑓𝑒𝑚𝑎𝑙𝑒 =
59
71
= 0.83
For Radiologist 1, when the predicted gender is female,
about 83% of the time the baby is actually female.
So, if you painted the room pink, then the probability
that you would need to repaint is about 0.17 (1 – 0.83).
Let’s take the gender prediction example a little
further.
Suppose that two radiologists both work in the same
clinic; Radiologist 1 works part-time while Radiologist 2
(from the Chapter Preview section) works full-time.
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 1 = 0.30
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 2 = 0.70
Let’s answer these questions:
1. What is the probability that a gender prediction
based on a first-trimester ultrasound at this clinic is
correct?
2. If the first-trimester ultrasound gender prediction
is incorrect, what is the probability that the
prediction was made by Radiologist 2?
Gender prediction example continued.
From the data we know:
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 1 = 0.836
create
a “hypothetical
1000” 2 = 0.903
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛Let’s
𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡
𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛
𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡
𝑃 table
𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛
𝑖𝑠 𝑚𝑎𝑑𝑒 𝑏𝑦
𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡
1 = 0.30
to answer
the
two questions.
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 2 = 0.70
Prediction
Correct
Prediction
Incorrect
Total
Radiologist 1
251
49
300
Radiologist 2
632
68
700
833
117
1000
the probability
that the prediction
is correct
given that the prediction was made by Radiologist 1 is
0.836, then the value for this cell is:
You canSimilarly,
now fill inthe
thevalue
values
for
thecell
remaining cells.
this
(300)(0.836)
= for
250.8
≈ 251 is:
(700)(0.903) = 632.1 ≈ 632
(Cell values MUST be whole numbers since we are
counting how many are in each event.)
TotalSince
Gender prediction example continued.
From the data we know:
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 1 = 0.836
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 2 = 0.903
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 1 = 0.30
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 2 = 0.70
Prediction
Prediction
ofCorrect
the incorrectIncorrect
gender
Total
About 58.1%
predictions at
49
Radiologist 1 this clinic251
are made by Radiologist
2. 300
Radiologist
2
700 2
This seems
high 632
– but remember68that Radiologist
833 twice as many
117predictions1000
Total
does more than
as
Radiologist 1.
If
theisfirst-trimester
gender
prediction
is
What
the probabilityultrasound
that a gender
prediction
based
incorrect,
what is theultrasound
probability
prediction
on a first-trimester
atthat
this the
clinic
is correct?
was made by Radiologist 2?
833
68
𝑃 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛
𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡
=
==0.833 = 0.581
𝑃 𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡
2|𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛
1000
117
Calculating Probabilities –
A More Formal Approach
Probability Formulas
The Complement Rule
For any event E,
𝑃 𝐸 𝐶 = 1 − 𝑃(𝐸)
The Addition Rule
For any two events E and F,
𝑃 𝐸∪𝐹 =P E +P F −P E∩𝐹
For mutually exclusive events, this simplifies to
𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃(𝐹)
Probability Formulas Continued
The Multiplication Rule
For any two events E and F,
𝑃 𝐸∩𝐹 =𝑃 𝐸 𝐹 𝑃 𝐹
For independent events, this simplifies to
𝑃 𝐸 ∩ 𝐹 = 𝑃 𝐸 𝑃(𝐹)
Conditional Probabilities
For any two events E and F with P(F) ≠ 0,
𝑃(𝐸 ∩ 𝐹)
𝑃 𝐸𝐹 =
𝑃(𝐹)
Revisit CDC’s study . . .
Recall:
L = event that a randomly selected adult American reports
learning something new about a health issue or disease from
a TV show in the previous 6 months.
F = event that a randomly selected adult American is female.
Data from the survey were used to estimate the following
probabilities:
𝑃 𝐿 = 0.58 𝑃 𝐹 = 0.5 𝑃 𝐿 ∩ 𝐹 = 0.31
What is the probability that a randomly selected adult
American reports learning something new about a health
issue or disease from a TV show in the previous 6 months
or that a randomly selected adult American is female?
𝑃 𝐿∪𝐹 =𝑃 𝐿 +𝑃 𝐹 −𝑃 𝐿∩𝐹
= 0.58 + 0.5 − 0.31 = 0.77
The article “Chances Are You Know Someone with a
Tattoo, and He’s Not a Sailor” (Associated Press,
June 11, 2006) summarized data from a representative
sample of adults ages 18 to 50.
T = the event that a randomly selected person has a tattoo
A = the event that a randomly selected person is between
18 and 29 years old
The following probabilities were estimated based on data
from the sample:
Notice that the probability of
𝑃 𝑇 = 0.24,
𝐴 “T
= given
0.50 A”𝑃are
𝑇 ∩ 𝐴 = 0.18
“A given T”𝑃and
NOT the same!
𝑃 𝐴𝑇 =
𝑃(𝑇∩𝐴)
𝑃(𝑇)
0.18
0.24
=
= 0.75
𝑃 𝑇𝐴
𝑃(𝑇∩𝐴)
=
𝑃(𝐴)
0.18
=
0.50
= 0.36
Another Approach to Probability
A large electronics store sells two different portable DVD
players, Brand 1 and Brand 2. Based on past records, the
store manager reports that 70% of the DVD players sold
are Brand 1 and 30% are Brand 2.
The manager also reports that 20% of the people who buy
Brand 1 also purchase an extended warranty, and 40% of
the people who buy Brand 2 purchase an extended warranty.
Consider selecting a person at random from those who
purchased a DVD player from this store, what is the
probability that the person purchased extended warranty?
One way to do this problem would be to set
up a Hypothetical 1000 table.
DVD Players Continued
P(Brand 1) = 0.7
P(Brand 2) = 0.3
The manager also reports that 20% of the people who buy
Brand 1 also purchase an extended warranty, and 40% of
the people who buy Brand 2 purchase an extended warranty.
Brand 1
Brand 2
Total
Bought Extended
Warranty
140
120
260
Not Bought Extended
Warranty
560
180
740
Total
700
300
1000
Consider selecting a person at random from those who
purchased a DVD player from this store, what is the
probability that the person purchased extended warranty?
260
𝑃 Extended Warranty =
= 0.26
1000
DVD Players Continued
P(Brand 1) = 0.7
P(Brand 2) = 0.3
The manager also reports that 20% of the people who buy
Brand 1 also purchase an extended warranty, and 40% of
the people who buy Brand 2 purchase an extended warranty.
Consider selecting a person at random from those who
purchased a DVD player from this store, what is the
probability that the person purchased extended warranty?
and
B1 = 0.7
B2 = 0.3
and
(0.7)(0.2) to
= 0.14
Another approach
this problem
is to use a tree diagram.
EC = 0.8
E = 0.2
or
E = 0.4
This is an example of the Law of
(0.3)(0.4) = 0.12
Total
Probability!
EC = 0.6
P(E) = 0.14 + 0.12 = 0.26
The Law of Total Probability
If B1 and B2 are disjoint events with P(B1) + P(B2) = 1, then
for any event E
𝑃 𝐸 = 𝑃 𝐸 ∩ 𝐵1 + 𝑃 𝐸 ∩ 𝐵2
= 𝑃 𝐸 𝐵1 𝑃 𝐵1 + 𝑃 𝐸 𝐵2 𝑃(𝐵2 )
More generally, if B1, B2, ⋯, Bk are disjoint events with
P(B1) + P(B2) + ⋯ + P(Bk) = 1, then for any event E
𝑃 𝐸 = 𝑃 𝐸 ∩ 𝐵1 + 𝑃 𝐸 ∩ 𝐵2 + ⋯ + 𝑃 𝐸 ∩ 𝐵𝑘
= 𝑃 𝐸 𝐵1 𝑃 𝐵1 + 𝑃 𝐸 𝐵2 𝑃 𝐵2 + ⋯ + 𝑃 𝐸 𝐵𝑘 𝑃 𝐵𝑘
Let’s consider another type of problem . . .
Suppose the conditional probability of “a positive test
result given that the person has cancer” is known.
However, you would like to know the converse probability.
That is, you would like to know the probability of the
person having cancer given a positive test result.
This formula was discovered in the 1700’s by
the Reverend
Thomas probability
Bayes, an English
A converse
is the reversal
Presbyterian
minister. probability.
of a conditional
This converse probability can be computed using
Bayes’ Rule.
Bayes’ Rule
If B1 and B2 are disjoint events with P(B1) + P(B2) = 1, then
for any event E
𝑃 𝐸 𝐵1 𝑃(𝐵1 )
𝑃 𝐵1 𝐸 =
𝑃 𝐸 𝐵1 𝑃 𝐵1 + 𝑃 𝐸 𝐵2 𝑃(𝐵2 )
More generally, if B1, B2, ⋯, Bk are disjoint events with
Let’s look at an example.
P(B1) + P(B2) + ⋯ + P(Bk) = 1, then for any event E
𝑃 𝐵1 𝐸 =
𝑃 𝐸 𝐵1 𝑃 𝐵1
𝑃 𝐸 𝐵1 𝑃(𝐵1 )
+ 𝑃 𝐸 𝐵2 𝑃 𝐵2 + ⋯ + 𝑃 𝐸 𝐵𝑘 𝑃(𝐵𝑘 )
Internet addiction has been defined by researchers as a
disorder characterized by excessive time spent on the
Internet, impaired judgment and decision-making ability, social
withdrawal, and depression. In a study of adolescents, each
participant was assessed using the Chen Internet Addiction
Scale to determine if he or she suffered from Internet
addiction.
The following probabilities are based on survey results:
Although Bayes’
is not listed
AP® Statistics
P(F) Rule
= 0.518
P(M)in=the
0.482
course description,
are expected
be able to solve
P(I|F) = you
0.131
P(I|M)=to0.248
“Bayes’-like” problems. Besides using the formula, you can
What is the
probability
that
a randomly
selected
adolescent
also
solve using
tables
or tree
diagrams.
from the survey is female given that she has Internet addiction?
𝑃 𝐼 𝐹 𝑃(𝐹)
𝑃 𝐹𝐼 =
𝑃 𝐼 𝐹 𝑃 𝐹 + 𝑃 𝐼 𝑀 𝑃(𝑀)
(0.131)(0.518)
=
0.131 0.518 + (0.482)(0.248)
= 0.362
Probability as a Basis
for Making Decisions
Probability plays an important role in drawing
conclusions from data.
A professor planning to give a quiz that consists of 20
true-false questions is interested in knowing how
someone who answers by guessing would do on such a
quiz.
To investigate, he asks the 500 students in his
introductory psychology course to write the numbers
from 1 to 20 on a piece of paper and then to arbitrarily
write T or F next to each number.
This table summarizes the number of
The students are forced
to guess
at the
to each
correct
answers
onanswer
the quiz.
question, because they are not even told what the
questions are!
These answers are then collected and graded using the
key for the quiz.
Quiz example continued.
Number of
Correct
Responses
Number of
Students
Proportion of
Students
Number of
Correct
Responses
Number of
Students
Proportion of
Students
0
0
0.000
11
79
0.158
1
0
0.000
12
61
0.122
2
1
0.002
13
39
0.078
3
1
0.002
14
18
0.036
4
2
0.004
15
7
0.014
5
8
0.016
16
1
0.002
6
18
0.036
17
1
0.002
7
37
0.074
18
0
0.000
8
58
0.116
19
0
0.000
9
81
0.162
20
0
0.000
10
88
0.176
Would you be surprised if someone guessing on a 20question true-false quiz got only 3 correct?
Only about 2 in 1000 guessers would get exactly 3 correct.
Since this is so unlikely, this outcome is surprising!
Quiz example continued.
Number of
Correct
Responses
Number of
Students
Proportion of
Students
Number of
Correct
Responses
Number of
Students
Proportion of
Students
0
0
0.000
11
79
0.158
2
1
0.002
13
39
0.078
3
1
0.002
14
18
0.036
4
2
0.004
15
7
0.014
5
8
0.016
16
1
0.002
6
18
0.036
17
1
0.002
7
37
0.074
18
0
0.000
8
58
0.116
19
0
0.000
9
81
0.162
20
0
0.000
10
88
0.176
1
12
0.122
P(passing
quiz) 0≈ 0.014 +0.000
0.002 + 0.002
+ 0 +610 + 0 = 0.018
If a score of 15 or more correct is a passing grade on the
quiz, is it likely that someone who is guessing will pass?
It would be unlikely that a student who is guessing would be
able to pass.
Quiz example continued.
Number of
Correct
Responses
Number of
Students
Proportion of
Students
Number of
Correct
Responses
Number of
Students
Proportion of
Students
0
0
0.000
11
79
0.158
1
0
0.000
12
61
0.122
2
1
0.002
13
39
0.078
3
1
0.002
14
18
0.036
4
2
0.004
15
0.014
There
are two
possible
explanations
for 7a score of
16:
5
8
0.016
16
1
0.002
1)
The student
was
guessing 17and was REALLY
lucky
6
18
0.036
1
0.002
The student
wasstudent
not
guessing
7
37
0.074 the
18 just
0
0.000
Begin
by 2)
assuming
that
was
guessing
and
8
58
19
0
determine
whether
a0.116
score at least
as high
as 16 0.000
is a
9
81
0.162
20
0
0.000
Since the first
explanation
isoccurrence.
highly unlikely, you could
likely
or
an
likely
10
88
0.176
conclude
that
a student
with a score of 16 was not just
guessing.
The professor actually gives
the quiz, and a student scores 16
correct. Do you think that the student was just guessing?
P(scores 16 or higher) ≈ 0.002 + 0.002 + 0 + 0 + 0 = 0.004
Quiz example continued.
What score on the quiz would it take to convince you that a
student was not just guessing?
Score
Approximately Probability
20
0.000
19 or better
0.000 + 0.000 = 0.000
18 or better
0.000 + 0.000 + 0.000 = 0.000
17 or better
0.002 + 0.000 + 0.000 + 0.000 = 0.002
16 or better
0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.004
15 or better
0.014 + 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.018
14 or better
0.036 + 0.014 + 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.054
13 or better
0.078 + 0.036 + 0.014 + 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.132
You might say that a score of 14 or higher is reasonable
Consider
showing
approximate
probabilities
for
evidencethis
thattable
someone
is not
just guessing,
because the
a certain that
scorea or
higher.would score this
approximate probability
guesser
high is only 0.054.
Estimating Probabilities
Empirically and Using Simulation
Estimating Probabilities Empirically
It is fairly common practice to use observed longrun proportions to estimate probabilities.
The process used to estimate probabilities is
simple:
1. Observe a large number of chance outcomes
under controlled circumstances.
2. Interpreting probability as a long-run relative
frequency, estimate the probability of an event
by using the observed proportion of
occurrence.
To recruit a new faculty member, a university biology
department intends to advertise for someone with a Ph.D.
in biology and at least 10 years of college-level teaching
experience.
A
similar
university
just would
completed
a search
in which
The
biology
like to
determine
the
member
ofdepartment
the department
express
the belief
thatthere
was
no requirement
prior
experience.
probability
applicant
would
be excluded
becausewill
of the
requiring
atan
least
10for
years
of teaching
teaching
experience
However,
prior
teachingapplicants
experience
was
The
experience
requirement.
exclude most
potential
and
willrecorded.
exclude more
The probability
that
applicant
be excluded
resulting
data is summarized
in the would
following
table. due
female
applicants
than an
male
applicants.
to the requirement
of at
10 years experience is
Number
of least
Applicants
Less than 10 years 67.5%.10 or more years
Total
experience
experience
277
138
This is just178
a little more than two-thirds
of the
112
290
applicants. 21
Female
99
120
Male
Total
277
𝑃 𝑎𝑝𝑝𝑙𝑖𝑐𝑎𝑛𝑡 𝑖𝑠 𝑒𝑥𝑐𝑙𝑢𝑑𝑒𝑑 =
= 0.675
410
410
New faculty member example continued.
Now let’s determine if more females than males are
excluded due to the experience requirement.
Number of Applicants
It
Male
Less than 10 years
experience
10 or more years
experience
Total
appears that178
female applicants are
be
112more likely to
290
excluded due to99the experience requirement
than male
Female
21
120
applicants. 138
Total
277
410
About 61.4% 178
of the male applicants are excluded due
290
to the experience requirement.
99
About 82.5% 120
of the female applicants are excluded
due to the experience requirement.
Estimating Probabilities by Using Simulation
Simulation provides a way to estimate
probabilities when:
• You are unable to determine probabilities
analytically
• You do not have the time or resources to
determine probabilities
• It is impractical to estimate probabilities
empirically by observation
Simulations involves generating “observations” in a
situation that is similar to the real situation of
interest.
Using Simulation to Approximate a Probability
1.
Design a method that uses a random mechanism (such as
a random number generator or table, the selection of a
ball from a box, or the toss a coin) to represent an
observation. Be sure that the important characteristics
of the actual process are preserved.
2.
Generate an observation using the method in Step 1, and
determine if the event of interest has occurred.
3.
Repeat Step 2 a large number of times.
4.
Calculate the estimated probability by dividing the
number of observations for which the event of interest
occurred by the total number of observations generated.
Suppose that couples who wanted children were to continue
having children until a boy was born. Would this change the
proportion of boys in the population?
We will use simulation to estimate the proportion of boys in
the population if couples were to continue having children
until a boy was born.
1. You can use a single random digit to represent a child,
where odd digits represent a male birth and even digits
represent a female birth.
2. An observation is constructed by selecting a sequence
of random digits. If the first random number obtained
is odd (a boy), the observation is complete. If the first
random number obtained is even (a girl), another digit is
chosen. You would continue in this way until an odd digit
is obtained.
Baby Boy Simulation Continued . . .
Below are four rows from the random digit table.
Row
6 0 9 3 8 7 6 7 9 9 5 6 2 5 6 5 8 4 2 6 4
7 4 1 0 1 0 2 2 0 4 7 5 1 1 9 4 7 9 7 5 1
Notice that even with only 10 trials, the
8 6 4 7 3 6 3 4 5 1 2 3 1 1 8 0 0 4 8 2 0
proportion of boys is 10/22, which is
9 8 0 2 8 7 9 3 8 close
4 0 4to 20.5!
0 8 9 1 2 3 3 2
Trial 1: girl, boy
Trial 5: boy
Trial 9: girl, boy
Trial 2: boy
Trial 6: boy
Trial 3: girl, boy
Trial 7: boy
Trial 4: girl, boy
Trial 8: girl, girl, boy
Trial 10: girl, girl,
girl, girl, girl,
girl, boy