Chapter 15
Probability Rules!
What I will know and be able to do
Use the rules of probability to find the probability of
an event given conditions and contingencies.
Assignment:
Read Chapter 15
The General Addition Rule
When two events
A
and
B
are disjoint, we can use
the addition rule for disjoint events from Chapter 13:
However, when our events are not disjoint, this
earlier addition rule will double count the probability
of both
A
and
B
occurring. Thus, we need the
General Addition Rule.
Let’s look at a picture…
The General Addition Rule (cont.)
General Addition Rule: For any two events A and B,
The following Venn diagram shows a situation in which we
would use the general addition rule:
Slide 15- 4
Example: Traffic Stops
Police report that 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests. What is the probability that a
randomly selected DWI suspect is given:
a. A test?
b. A blood test or a breath test, but not both?
c. Neither test?
Slide 15- 5
Example continued….
1.
Draw a picture (Venn Diagram)
2.
Figure out what you want to know in words.
3.Translate words to equations.
Example continued….
What is the probability that the suspect is given a test?
P(A U B)=P(A) + P(B) – P(A∩B) = .92
What is the probability that the suspect gets either a blood
test or a breath test but NOT both?
P(A U B) – P(A ∩ B) = .92 - .22 = .70
What is the probability that the suspect gets neither test?
P(neither test) = 1 – P(A or B)
= 1 – P(A or B)= 1 – .92 = 0.08
Blood test only?
Example: Goals
Two psychologists surveyed 478 children in grades 4, 5, and 6 in elementary
schools in Michigan. They stratified their sample, drawing one third each from rural, suburban, and urban schools. They asked the students whether their primary goal was to get good grades, to be popular, or to be good at sports. The results are shown in the table below.
1. What is the probability of randomly picking a girl?
2. What is the probability of randomly picking a student whose goal is to be popular?
3. What is the probability of picking a student who is a boy and wants to be good at sports? 4. What is the probability of picking a student who is a girl or wants to get good grades?
Grades Popular Sports Boy 117 50 60
It Depends…
Back in Chapter 3, we looked at contingency tables and
talked about conditional distributions.
When we want the probability of an event from a
conditional distribution, we write P(B|A) and pronounce it “the probability of B given A.”
A probability that takes into account a given condition is
It Depends… (cont.)
To find the probability of the event B given the event A, we
restrict our attention to the outcomes in A. We then find the fraction of those outcomes B that also occurred.
Note: P(A) cannot equal 0, since we know that A has
occurred.
P
(
B
|
A
)
=
P
(
A
Ç
B
)
Example: Girls and Sports
In a recent study it was found that the probability that a randomly selected student is a girl is .51 and is a girl and plays sports is .10. If the student is female, what is the probability that she plays sports?
(
)
.1
( |
)
.1961
( )
.51
P S
F
P S F
P F
Example: Boys and Sports
The probability that a randomly selected student plays
sports if they are male is .31. What is the probability that the student is male and plays sports if the probability that they are male is .49?
(
)
( |
)
.31
(
)
.49
.1519
P S
M
x
P S M
P M
x
Probabilities from two way tables
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Total 195 164 359
1) What is the probability that the driver is a student?
359
195
)
Probabilities from two way tables
2) What is the probability that the driver drives a
European car?
359
45
)
(
European
P
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Probabilities from two way tables
3) What is the probability that the driver drives an
American or Asian car?
Disjoint?
359
102
212
)
(
American
or
Asian
P
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Probabilities from two way tables
4) What is the probability that the driver is staff or drives an Asian car?
Disjoint?
)
164
359
102
47
(
Staff
or
Asian
P
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Probabilities from two way tables
5) What is the probability that the driver is staff and drives an Asian car?
359 47 )
(Staff and Asian
P
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Probabilities from two way tables
6) If the driver is a student, what is the probability that they
drive an American car?
Condition
195 107 )|
(American Student
P
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Probabilities from two way tables
7) What is the probability that the driver is a student if the
driver drives a European car?
Condition
45
33
)
|
(
Student
European
P
Student Staff Total American 107 105 212 European 33 12 45
Asian 55 47 102
Example: Conditional Probability
After School Activities
At George Washington HS, after school activities can be classified into three types: athletic, fine arts, and other. The following table gives the number of students participating in these types of activities by grade:
Example continued…
Is it true from the table that:
There are 160 10th graders participating in athletics?
The number of senior participating in fine arts activities is
125?
There are 435 students in fine arts activities?
GWHS has 410 juniors?
Example continued…
What is the probability that a randomly selected student is a
senior athlete? P(senior athlete)=
What is the probability that the selected student is an athlete,
given that the student is a senior? P(athlete|senior)= 09375 . 1600 150 students of number total athletes senior of number 3529 . 425 150 seniors of number total athletes senior of number
Example: GFI Switches
A GFI (ground fault interrupt) switch turns off power to a system in the event of an electrical malfunction. A spa manufacturer currently has 25 spas in stock, each equipped with a single GFI switch. Two different
companies supply the switches and some of the switches are defective as summarized in the table:
Nondefective Defective Total
Company 1 10 5 15
Company 2 8 2 10
Example continued…
Given:
E = event that GFI switch in selected spa is from
company 1
D = event that GFI switch in selected spa is defective
Find:
P(E)
P(D)
Example continued…
P(E)=15/25 = .6 P(D) = 7/25 = .28
P(E and D) = P(E∩D) = 5/25 = .2
Now suppose that testing reveals a defective switch. How likely is it that the switch came from the first company?
P(company 1 given defective switch)=P(E|D) =5/7 = .714
The General Multiplication Rule
When two events
A
and
B
are independent, we can
use the multiplication rule for independent events
from Chapter 14:
However, when our events are not independent, this
earlier multiplication rule does not work. Thus, we
need the
General Multiplication Rule
.
The General Multiplication Rule (cont.)
We encountered the general multiplication rule in
the form of conditional probability.
Rearranging the equation in the definition for
conditional probability, we get the
General
Multiplication Rule
:
For any two events
A
and
B
,
or
P
(
A
B
)
P
(
A
)
P
(
B
|
A
)
Independence
Independence of two events means that the outcome of
one event does not influence the probability of the other.
With our new notation for conditional probabilities, we can
now formalize this definition:
Events A and B are independent whenever P(B|A) =
Independent ≠ Disjoint
(Chapter 14 as well) Disjoint events cannot be independent! Well, why not?
Since we know that disjoint events have no outcomes in
common, knowing that one occurred means the other didn’t.
Thus, the probability of the second occurring changed
based on our knowledge that the first occurred.
It follows, then, that the two events are not independent.
A common error is to treat disjoint events as if they were
Depending on Independence
It’s much easier to think about independent events than to
deal with conditional probabilities.
It seems that most people’s natural intuition for
probabilities breaks down when it comes to conditional probabilities.
Don’t fall into this trap: whenever you see probabilities
Drawing Without Replacement
Sampling without replacement means that once one
individual is drawn it doesn’t go back into the pool.
We often sample without replacement, which doesn’t
matter too much when we are dealing with a large population.
However, when drawing from a small population, we need
to take note and adjust probabilities accordingly.
Drawing without replacement is just another instance of
Tree Diagrams
A tree diagram helps us think through conditional
probabilities by showing sequences of events as paths that look like branches of a tree.
Making a tree diagram for situations with conditional
Tree Diagrams (cont.)
This figure is a nice
example of a tree
diagram and shows how we multiply the
probabilities of the branches together.
All the final outcomes
are disjoint and must add up to one.
We can add the final
probabilities to find probabilities of
Example: Assembly
Management has determined that customers return 12% of the items assembled by inexperienced employees, whereas only 3% of the items assembled by experienced employees are returned. Due to turnover and
absenteeism at an assembly plant, inexperienced employees assemble 20% of the items. Construct a tree diagram or a chart for this data.
What is the probability that an item is returned?
If an item is returned, what is the probability that an inexperienced employee assembled it?
P(returned) = 4.8/100 = 0.048
P(inexperienced|returned) = 2.4/4.8 = 0.5
Returned Not
returned
Total
Experienced 2.4 77.6 80
Inexperienced 2.4 17.6 20
Reversing the Conditioning
Reversing the conditioning of two events is rarely intuitive.
Suppose we want to know P(A|B), and we know only P(A), P(B), and P(B|A).
We also know , since
From this information, we can find P(A|B):
P
(A|B)
=
P
(A
Ç
B)
P
(B)
P(A B) P(A) P(B | A)
What Can Go Wrong?
Don’t use a simple probability rule where a general rule is
appropriate:
Don’t assume that two events are independent or disjoint without
checking that they are.
Don’t find probabilities for samples drawn without replacement as if
they had been drawn with replacement.
Don’t reverse conditioning naively.
What have we learned?
The probability rules from Chapter 13 only work in special cases—
when events are disjoint or independent.
We now know the General Addition Rule and General Multiplication
Rule.
We also know about conditional probabilities and that reversing the
What have we learned? (cont.)
Venn diagrams, tables, and tree diagrams help organize our
thinking about probabilities.
We now know more about independence—a sound understanding
AP Tips
Read conditional probabilities carefully.
Most AP problems use data for probability problems. Become
skilled at finding probabilities from 2-way tables.
