chapter 12: general rules of probability stat 1450

Chapter 12: General Rules of Probability

STAT 1450

Connecting Chapter 12 to our Current Knowledge of Statistics

▸ Probability theory leads us from data collection to inference.

▸ The introduction to probability from Chapter 10 will now be fortified by additional rules to allow us to consider multiple types of events. The rules of probability will help us develop models so that we can generalize from our (properly collected) sample to our population of interest.

12.0 General Rules of Probability

Independence

▸ Two events A and B are independent if knowing that one occurs does

not change the probability that the other occurs.

Thus, if A and B are independent,

P(A and B) = P(A)*P(B)

12.1 Independence and the Multiplication Rule

Example: Blood Types

▸ Someone with type O-negative blood is considered to be a “universal donor.” According to the American Association of Blood Banks, 39% of people are type O-negative. Two unrelated people are selected at random.

Calculate the probability that both have type O-negative blood.





▸ We can apply the concept of independence. P(both type O-negative) = P(first type O-) P(second type O-negative)






= (0.39)(0.39) (Independence)







= 0.1521


Example/Poll: Blood Types

▸ Suppose two unrelated people are selected at random. Calculate the probability that neither have type O-

negative blood.


Example/Poll: Blood Types

▸ Suppose two unrelated people are selected at random. Calculate the probability that neither have type O-

negative blood.

▸ We can again apply the concept of independence. P(neither type is O-) = P(not O-) P(not O-)


= 0.3721


The General Addition Rule

▸ Two-Way tables are helpful ways to picture two events.

▸ Venn diagrams are an alternative means of displaying multiple events.

▸ Both can be used to answer many questions involving probabilities.

12.2 The General Addition Rule

Venn Diagrams and Probabilities

▸ Example: In a sample of 1000 people, 88.7% of them were right-hand

dominant, 47.5% of them were female, and 42.5% of them were

female and right-hand dominant. Draw a Venn diagram for this

situation.


0.462

0.063

0.425 0.050

.887 = .462 + .425 P(R) =P(M and R) + P(F and R)

.475 = .425 + .050 P(F) = P(F and R) + P(F and L)


▸ Calculate the probability that a randomly selected person is right-hand

dominant or female.


0.462

0.063

0.425 0.050

.887 = .462 + .425 P(R) =P(M and R) + P(F and R)

.475 = .425 + .050 P(F) = P(F and R) + P(F and L)



dominant or female.▸ 1st Method 0.462 + 0.425 + 0.050 = 0.937


0.462

0.063

0.425 0.050

.887 = .462 + .425 P(R) =P(M and R) + P(F and R)

.475 = .425 + .050 P(F) = P(F and R) + P(F and L)




Let R = right-hand dominant (.887), and let F = female (.475).


0.462

0.063

0.425 0.050

.887 = .462 + .425 P(R) =P(M and R) + P(F and R)

.475 = .425 + .050 P(F) = P(F and R) + P(F and L)




Let R = right-hand dominant (.887), and let F = female (.475).

▸ 2nd Method 0.887 + 0.475 – 0.425 = 0.937 This is actually = P(R) + P(F) – P(R and F) = P(R or

F)


0.462

0.063

0.425 0.050

.887 = .462 + .425 P(R) =P(M and R) + P(F and R)

.475 = .425 + .050 P(F) = P(F and R) + P(F and L)


▸ We just used the general addition rule:

▸ For any two events A and B,

P(A or B) = P(A) + P(B) – P(A and B).

▸ Question: Where did we see this concept previously?



▸ We just used the general addition rule:

▸ For any two events A and B,

P(A or B) = P(A) + P(B) – P(A and B).

▸ Question: Where did we see this concept previously?

“OR” questions from Chapter 6 where two events

“overlapped.”


The Addition Rule for Disjoint Events

▸ What if there is no overlap of the events A and B?

▸ Events A and B are disjoint if they have no outcomes in common.

▸ Question: What is P(A or B) when A and B are disjoint?






P(A and B) = 0 (no outcomes in common)







P(A or B) = P(A) + P(B) – P(A and B) (General

Addition Rule)








Addition Rule)

= P(A) + P(B) – 0

(Disjoint)








Addition Rule)

= P(A) + P(B) – 0

(Disjoint)

P(A or B) = P(A) + P(B)


The Complement Rule and Addition Rule

▸ Calculate the probability that a randomly selected person is neither

right-hand dominant nor female.


0.462

0.063

0.425 0.050




P(neither R nor F) = 1 – P(R or F)


0.462

0.063

0.425 0.050




P(neither R nor F) = 1 – P(R or F) = 1 – 0.937 = 0.063


0.462

0.063

0.425 0.050




P(neither R nor F) = 1 – P(R or F) = 1 – 0.937 = 0.063

▸ We just used the complement rule:

For any event A, P(A does not occur) = P(not A) = 1 – P(A).


0.462

0.063

0.425 0.050

Conditional Probability

▸ When P(A) > 0, the conditional probability of B given A is

▸ Note that since P(B|A) assumes that the event A has occurred,

P(A) will be greater than 0.

12.3 Conditional Probability


▸ Note: Wording for conditional probabilities can often be subtle, so be sure to read carefully.

Occasionally you will come across problems that embed conditional probabilities into the question. Consider the following example:

▸ Example: Calculate the probability that a

randomly selected female is right-hand dominant.

=







= =







= = = 0.8947



▸ The | means “given.” The event behind the | is the conditioning event.

▸ The idea of a conditional probability P(B|A) of one event B given that

another event A occurs is the proportion of all occurrences of A for

which B also occurs.


General Multiplication Rule

▸ The probability that both of two events A and B happen together can

be found by

▸ Here P(B|A) is the conditional probability that B occurs given the

information that A occurs.

12.4 General Multiplication Rule

Example: Internet Access

▸ An A.C. Nielsen study found that 81% of households in the United States have computers. Of those 81%, 92% have Internet access. Calculate the probability that a randomly selected U.S. household has a computer and has Internet access.




▸ Let C =household has a computer. Let I = household has Internet access.





▸ P(C and I)





▸ P(C and I) = P(C)×P(I |C)





▸ P(C and I) = P(C)×P(I |C) = (0.81)(0.92) = 0.7452


Poll

▸ Two unrelated persons are in line to donate blood.

The blood type of the 2nd person is not impacted by the blood type of the first.

a) True b) False

10.2 Randomness and Probability

Definition: Independence

▸ Two events A and B that both have positive probability are

independent if

▸ The fact that A has occurred does not impact B’s probability of

occurrence.

12.5 Independence Again

Example: Smoking

▸ Smoking has been a “hot” topic in recent months. Twenty percent of people (in a city with a population in of about 1 million) are smokers. Two people, Megan and Zach, are selected at random. Confirm mathematically and practically, that Megan’s decision to smoke is independent of Zach’s.


Let Sz= Zachary’s decision to smoke.Let SM = Megan’s decision to smoke.

Example: Smoking

▸ Smoking has been a “hot” topic in recent months. Twenty percent of people (in a city with a population in of about 1 million) are smokers. Two people, Megan and Zach, are selected at random. Confirm mathematically and practically, that Megan’s decision to smoke is independent of Zach’s. Mathematically, P(SM and SZ )



Example: Smoking

▸ Smoking has been a “hot” topic in recent months. Twenty percent of people (in a city with a population in of about 1 million) are smokers. Two people, Megan and Zach, are selected at random. Confirm mathematically and practically, that Megan’s decision to smoke is independent of Zach’s. Mathematically, P(SM and SZ ) =P(SM )*P(SZ |SM) gen.rule of mult.



Example: Smoking


=



Example: Smoking


= P(SM and SZ ) =



Example: Smoking


= P(SM and SZ ) =

P(SM and SZ ) =.04



Example: Smoking


= P(SM and SZ ) =

P(SM and SZ ) =.04 = .20*.20



Example: Smoking


= P(SM and SZ ) =

P(SM and SZ ) =.04 = .20*.20 = P(SM )*P(SZ) (independence)



Example: Smoking


= P(SM and SZ ) =


Practically, Zach’s decision to smoke is not influenced by a randomly selected person from the same city (Megan, in this case).



Example: Smoking


= P(SM and SZ ) =


Practically, Zach’s decision to smoke is not influenced by a randomly selected person from the same city (Megan, in this case). Therefore, it makes sense that P(SZ |SM) = P(SZ ).



Tree Diagrams

▸ Tree diagrams can be helpful when we have several stages of a probability model. The graph begins with line segments (branches) that correspond to probabilities for specific mutually exclusive events.

▸ Subsequent sets of branches represent probabilities at each stage conditional on the outcomes of earlier stages. Take a look at Example 12.10 on page 319 of the text to see how to work with tree diagrams.

12.6 Tree Diagrams

Example: Textbooks

Textbook editors, must estimate the sales of new (first-edition) books. The records of one major publishing company indicate that 10% of all new books sell more than projected, 30% sell close to projected, and 60% sell less than projected. Of those that sell more than projected, 70% are revised for a second edition, as are 50% of those that sell close to projected, and 20% of those that sell less than projected.

What percent of books are revised for a second edition?

12.6 Tree Diagrams

Example: Textbooks

12.6 Tree Diagrams

Let M= textbooks that sell More than projected.Let C = textbooks that sell Close to projections.

Let L= textbooks that sell Less than projected.

Example: Textbooks

12.6 Tree Diagrams

P(M) = .10



Example: Textbooks

12.6 Tree Diagrams

P(M) = .10

P(C)=.30



Example: Textbooks

12.6 Tree Diagrams

P(M) = .10

P(C)=.30

P(L) = .60



Example: Textbooks

12.6 Tree Diagrams

P(M) = .10P(2nd | M) = .70

P(C)=.30

P(L) = .60



Example: Textbooks

12.6 Tree Diagrams

P(M) = .10P(2nd | M) = .70

P(C)=.30P(2nd | C) = .50

P(L) = .60



Example: Textbooks

12.6 Tree Diagrams

P(M) = .10P(2nd | M) = .70

P(C)=.30P(2nd | C) = .50

P(L) = .60P(2nd | L) = .20



Example: Textbooks

12.6 Tree Diagrams

P(M) = .10P(2nd | M) = .70

P(not 2nd | M) =.30

P(C)=.30P(2nd | C) = .50

P(not 2nd | C) = .50

P(L) = .60P(2nd | L) = .20

P(not 2nd | L) = .80



P(M and 2nd)=.07

P(C and 2nd)=.15

P(L and 2nd)=.12

P(2nd)=.34

b) You noticed that one of your textbooks is in its second edition. What’s the probability that the first edition sold more than expected? Does this probability surprise you? Why or Why not?

Example: Textbooks

12.6 Tree Diagrams

P(M) = .10P(2nd | M) = .70

P(not 2nd | M) =.30

P(C)=.30P(2nd | C) = .50

P(not 2nd | C) = .50

P(L) = .60P(2nd | L) = .20

P(not 2nd | L) = .80



P(M and 2nd)=.07

P(C and 2nd)=.15

P(L and 2nd)=.12

P(2nd)=.34


P(M | 2nd) = = = .2059 Not too surprising since the textbooks that sell more than expected are the smallest group in the first and final stages of the Tree Diagram.

Probabilities and Two-Way Tables

▸ Our work with Two-Way Tables is rooted in probability rules.

12.7 Probabilities and Two-Way Tables

Example: Global Warming

▸ “And” question: What is the probability that a randomly selected respondent is a Midwesterner who agrees that Global Warming increased temperatures during December 2011 and January 2012?

▸ Let M = Midwesterner, and let A = Agrees that Global Warming Increased Temperatures

Belief that Global Warming Increased Temperatures During December 2011 and January 2012

Region of U.S. Agree Disagree Total Northeast 140 46 186Midwest 168 51 219South 262 112 374West 162 68 230Total 732 277 1009



▸ “And” question: What is the probability that a randomly selected respondent is a Midwesterner who agrees that Global Warming increased temperatures during December 2011 and January 2012?

▸ Let M = Midwesterner, and let A = Agrees that Global Warming Increased Temperatures

▸ P(M and A) = = 0.1665 Belief that Global Warming Increased Temperatures During December 2011 and January 2012




▸ “Or” question: What is the probability that a randomly selected respondent is a Midwesterner or someone who agrees that Global Warming increased temperatures during December 2011 and January 2012?






▸ Use the general addition rule.






▸ Use the general addition rule. P(M or A) = P(M) + P(A) – P(M and A)






▸ Use the general addition rule. P(M or A) = P(M) + P(A) – P(M and A) = + - =







= 0.217 + 0.7255 – 0.1665







= 0.217 + 0.7255 – 0.1665 P(M or A) = 0.777







= 0.217 + 0.7255 – 0.1665 P(M or A) = 0.777

Or some would add = .051 to .726 = .777





▸ “Conditional” question: What is the probability that a Midwesterner

agrees that Global Warming increased temperatures during December

2011 and January 2012?








Use the Conditional Prob. Rule.








Use the Conditional Prob. Rule.

=





▸ Independence question: Is Belief in Impact due to Global Warming (in terms of increase to December 2011 and January 2012 temperatures) independent of Region?






P(A) = 732/1009=0.725






P(A) = 732/1009=0.725Because , the events are not independent.







Alternatively, via multiplication rule.







Alternatively, via multiplication rule.

The events are not independent.




Five-Minute Summary

▸ List at least 3 concepts that had the most impact on your knowledge of

general rules of probability.

___________ _____________

________________

chapter 12: general rules of probability stat 1450

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