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. 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) 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) 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 12.1 Independence and the Multiplication Rule

Example/Poll: Blood Types ▸ Suppose two unrelated people are selected at random. Calculate the probability that neither have type O-negative blood. 12.1 Independence and the Multiplication Rule

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-) = (0.61)(0.61) (Independence) = 0.3721 12.1 Independence and the Multiplication Rule

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. 12.2 The General Addition Rule 0.462 0.063 0.4250.050.887 =.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)

Venn Diagrams and Probabilities ▸ Calculate the probability that a randomly selected person is right-hand dominant or female. 12.2 The General Addition Rule 0.462 0.063 0.4250.050.887 =.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)

Venn Diagrams and Probabilities ▸ Calculate the probability that a randomly selected person is right-hand dominant or female. ▸ 1 st Method  0.462 + 0.425 + 0.050 = 0.937 12.2 The General Addition Rule 0.462 0.063 0.4250.050.887 =.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)

Venn Diagrams and Probabilities ▸ Calculate the probability that a randomly selected person is right-hand dominant or female. ▸ 1 st Method  0.462 + 0.425 + 0.050 = 0.937 Let R = right-hand dominant (.887), and let F = female (.475). 12.2 The General Addition Rule 0.462 0.063 0.4250.050.887 =.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)

Venn Diagrams and Probabilities ▸ Calculate the probability that a randomly selected person is right-hand dominant or female. ▸ 1 st Method  0.462 + 0.425 + 0.050 = 0.937 Let R = right-hand dominant (.887), and let F = female (.475). ▸ 2 nd Method  0.887 + 0.475 – 0.425 = 0.937 This is actually = P(R) + P(F) – P(R and F) = P(R or F) 12.2 The General Addition Rule 0.462 0.063 0.4250.050.887 =.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)

The General Addition Rule ▸ 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? 12.2 The General Addition Rule

The General Addition Rule ▸ 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.” 12.2 The General Addition Rule

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? 12.2 The General Addition Rule

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) 12.2 The General Addition Rule

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) 12.2 The General Addition Rule

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) = P(A) + P(B) – 0 (Disjoint) 12.2 The General Addition Rule

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) = P(A) + P(B) – 0 (Disjoint) P(A or B) = P(A) + P(B) 12.2 The General Addition Rule

The Complement Rule and Addition Rule ▸ Calculate the probability that a randomly selected person is neither right-hand dominant nor female. 12.2 The General Addition Rule 0.462 0.063 0.4250.050

The Complement Rule and Addition Rule ▸ Calculate the probability that a randomly selected person is neither right-hand dominant nor female. P(neither R nor F) = 1 – P(R or F) 12.2 The General Addition Rule 0.462 0.063 0.4250.050

The Complement Rule and Addition Rule ▸ Calculate the probability that a randomly selected person is neither right-hand dominant nor female. P(neither R nor F) = 1 – P(R or F) = 1 – 0.937 = 0.063 12.2 The General Addition Rule 0.462 0.063 0.4250.050

The Complement Rule and Addition Rule ▸ Calculate the probability that a randomly selected person is neither right-hand dominant nor female. 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). 12.2 The General Addition Rule 0.462 0.063 0.4250.050

Conditional Probability 12.3 Conditional Probability

Conditional Probability ▸ 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. 12.3 Conditional Probability

General Multiplication Rule 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. 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. 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) 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)×P(I |C) 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)×P(I |C) = (0.81)(0.92) = 0.7452 12.4 General Multiplication Rule

Poll ▸ Two unrelated persons are in line to donate blood. The blood type of the 2 nd person is not impacted by the blood type of the first. a)Trueb)False 10.2 Randomness and Probability

Definition: Independence 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. 12.5 Independence Again Let S z = Zachary’s decision to smoke. Let S M = 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(S M and S Z ) 12.5 Independence Again Let S z = Zachary’s decision to smoke. Let S M = 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(S M and S Z )=P(S M )*P(S Z |S M )gen.rule of mult. 12.5 Independence Again Let S z = Zachary’s decision to smoke. Let S M = Megan’s decision to smoke.

Example: Smoking 12.5 Independence Again Let S z = Zachary’s decision to smoke. Let S M = Megan’s decision to smoke.

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 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 P(C)=.30 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 P(C)=.30 P(L) =.60 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) =.10P(2 nd | M) =.70 P(C)=.30 P(L) =.60 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) =.10P(2 nd | M) =.70 P(C)=.30P(2 nd | C) =.50 P(L) =.60 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) =.10P(2 nd | M) =.70 P(C)=.30P(2 nd | C) =.50 P(L) =.60P(2 nd | L) =.20 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) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 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 a) What percent of books are revised for a second edition? 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 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 a) What percent of books are revised for a second edition? 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07

Example: Textbooks a) What percent of books are revised for a second edition? 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07 P(C and 2 nd )=.15

Example: Textbooks a) What percent of books are revised for a second edition? 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07 P(C and 2 nd )=.15 P(L and 2 nd )=.12

Example: Textbooks a) What percent of books are revised for a second edition? 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07 P(C and 2 nd )=.15 P(L and 2 nd )=.12 P(2 nd )=.34

Example: Textbooks a) What percent of books are revised for a second edition? P(2 nd ) =.07 +.15 +.12 P(2 nd ) =.34 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07 P(C and 2 nd )=.15 P(L and 2 nd )=.12 P(2 nd )=.34

Example: Textbooks 12.6 Tree Diagrams P(M) =.10P(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07 P(C and 2 nd )=.15 P(L and 2 nd )=.12 P(2 nd )=.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(2 nd | M) =.70 P(not 2 nd | M) =.30 P(C)=.30P(2 nd | C) =.50P(not 2 nd | C) =.50P(L) =.60P(2 nd | L) =.20P(not 2 nd | L) =.80 Let M= textbooks that sell More than projected. Let C = textbooks that sell Close to projections. Let L= textbooks that sell Less than projected. P(M and 2 nd )=.07 P(C and 2 nd )=.15 P(L and 2 nd )=.12 P(2 nd )=.34

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. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Example: Global Warming Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Example: Global Warming ▸ “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? Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Example: Global Warming ▸ “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. Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Example: Global Warming ▸ “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. P(M or A) = P(M) + P(A) – P(M and A) Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Example: Global Warming ▸ “Conditional” question: What is the probability that a Midwesterner agrees that Global Warming increased temperatures during December 2011 and January 2012? Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Example: Global Warming ▸ “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. Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Probabilities and Two-Way Tables ▸ Independence question: Is Belief in Impact due to Global Warming (in terms of increase to December 2011 and January 2012 temperatures) independent of Region? Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

Probabilities and Two-Way Tables Belief that Global Warming Increased Temperatures During December 2011 and January 2012 Region of U.S. AgreeDisagreeTotal Northeast 14046186 Midwest 16851219 South 262112374 West 16268230 Total 7322771009 12.7 Probabilities and Two-Way Tables

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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