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.

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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

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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…

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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:

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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?

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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.

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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?

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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

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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

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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

)

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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

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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

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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

)

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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

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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

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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

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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

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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

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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

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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:

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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?

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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

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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

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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)

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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

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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

.

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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

)

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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) =

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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

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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

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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

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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

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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

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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

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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(AB)  P(A)  P(B | A)

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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.

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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

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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

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AP Tips

 Read conditional probabilities carefully.

 Most AP problems use data for probability problems. Become

skilled at finding probabilities from 2-way tables.

Figure

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References

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