Recognize when events are disjoint and when events are independent.

Recognize when events are disjoint and when events are independent.
AP Statistics Objectives Ch14 Understand that the relative frequency of an outcome of a random phenomenon settles down as we gather more random outcomes. Recognize when events are disjoint and when events are independent.

Know how and when to apply the Addition Rule.
AP Statistics Objectives Ch14 Know how and when to apply the Addition Rule. Know how and when to apply the Multiplication Rule. Know how to use the Complement Rule to make calculating probability simpler. (“.at least….”)

Random phenomenon Law of Large Numbers Probability Event Independence
Vocabulary Random phenomenon Law of Large Numbers Probability Event Independence Complement Rule Disjoint

Chapter 14 Practice Questions
Chapter 14 Notes Chapter 14 Practice Questions Chapter 14 Assignment

1. Probability the long-run proportion (relative frequency) of times an event would occur each attempt to see the event occur must be independent of all other attempts

2. Rules of Probability Probability will always be between 0 and 1
0 ≤ P(event) ≤ 1 Probability of 0 means that it never occurs. P(Coal in my stocking) = 0 Probability of 1 means that it always occurs. P(Seeing Christmas Lights) = 1

Law of Large Numbers (LLN)
the long-run relative frequency of repeated independent events settles down to the true probability as the number of trials increases. LLN does not apply to short-run behavior ”Law of Averages” does not exist!!!

What is the probability of a tossed coin
Coin Toss Lab What is the probability of a tossed coin landing on “Heads”? P(Heads) = 0.5 = 50%

If you toss it two times, how many
Coin Toss Lab If you toss it two times, how many “Heads” will you get? Could get two heads or …. just one or …. no heads.

Coin Toss Lab Teacher tosses 10 pennies and fills in the table below with the students. Next each student will toss ten pennies five times. The class will then complete the table.

Coin Toss Lab This is an example of the Law of Large Numbers.

8. Sample Space the collection of all possible outcomes of an event
EXAMPLES Roll a die S={1,2,3,4,5,6} Toss a single coin S={H,T}

Sample Space Toss two coins S={HH,HT,TH,TT} EXAMPLES HEADS TAILS HH TH

Sample Space Toss three coins S={HHH,HHT,HTH,HTT, THH,THT,TTH,TTT}
EXAMPLES Toss three coins S={HHH,HHT,HTH,HTT, THH,THT,TTH,TTT} H H H T T H T T HEADS HHH HHT HTH HTT TAILS THH THT TTH TTT

9. Event Rolling a prime number on a die E={2, 3, 5}
any collection of outcomes from the sample space. EXAMPLES Rolling a prime number on a die E={2, 3, 5} Rolling a prime or even # E={2, 3, 4, 5, 6}

Event Rolling two dice and getting a seven
EXAMPLES Rolling two dice and getting a seven E= {1&6, 2&5, 3&4, 4&3, 5&2, 6&1} Rolling two dice and getting a 3 or 4 E={1&2, 1&3, 2&1, 2&2, 3&1} Drawing an Ace from a deck of cards E={♠Ace,♣Ace,♥Ace,♦Ace}

10. Venn Diagrams Used to display relationships between events
Helpful in calculating probabilities

Venn Diagrams 11. Draw a Venn Diagram for each of the following:
a. Number of students wearing a belt, ring, or shoe laces. Now you and your neighbor must create two probability questions from this Venn Diagram. Make sure that you write it in sentence and notation format. Make sure that you calculate the correct answers as well. Belt Shoe Laces Ring How many students are in the class? How many are not wearing any of these? How many are wearing all of these? Count the combinations of two next. Finish the Venn Diagram. P(Belt & Ring & Shoe Laces) = P(Wearing None of them) =

12. Complement Consists of all outcomes that are not in the event
Not rolling an even # E(Rolling an even #)C={1,3,5} EXAMPLES

-Not rolling a three on a single die E(Rolling a 3)C={1,2,4,5,6}
Complement EXAMPLES -Not rolling a three on a single die E(Rolling a 3)C={1,2,4,5,6} -Rolling two dice and not getting a total higher than three. E(Rolling higher than a 3)C={1&1, 1&2, 2&1} -The Probability of NOT wearing a ring P(Ring)C= 1 – P(Ring) =

Complement Complement of Event A A B 𝑨 𝑪

Note that “2” is in both sets (ie “2” is both prime & even)
13. Union the event A OR B happening consists of all outcomes that are in at least one of the two events (can be in both) EXAMPLE Rolling a prime # or even number Note that “2” is in both sets (ie “2” is both prime & even) 𝐸 𝑷𝒓𝒊𝒎𝒆 𝑶𝑹 𝑬𝒗𝒆𝒏 ={𝟐, 𝟑, 𝟒, 𝟓, 𝟔}

Union A OR B A B 𝑨 ∪𝑩

14. Intersection the event A and B happening
consists of all outcomes that are in both events EXAMPLE - Drawing from a deck and getting red and a “2” E(Red and 2)={♥2,♦2} Note that “2 of Spades” is not included Even though it is a “2”, it isn’t red.

Intersection A and B A B 𝑨 ∩𝑩

15. Must be able to know symbols.
D ∩ I tersection ∪ nion OR

16. Mutually Exclusive (disjoint)
two events have no outcomes in common

Mutually Exclusive (disjoint)
17. Sketch a Venn Diagram of the following two Events: A = Rolling an even number on a single die B=Rolling an odd number on a single die A B 1 2 4 3 6 5

18. More on Probability a. The probability of the set of all possible outcomes must be 1. b. The probability of an event occurring is 1 minus the probability that it doesn’t occur. P(A) = 1 – P(Ac) 60% A 100 – 60 = 40 40%

Statistics? 19. Stat Cal Comp Sci

Computer Science? 20. Stat Cal Comp Sci

Not Calculus? 21. Stat Cal Comp Sci Complement of Calculus

22. Statistics AND Computer Science AND not Calculus? Stat Cal
Comp Sci

23. Cal Stat Comp Sci Calculus OR Computer Science

24. (Statistics OR Computer Science) AND not Calculus Stat Cal
Comp Sci (Statistics OR Computer Science) AND not Calculus

25. Statistics and not (Computer Science or Calculus) Cal Stat
Comp Sci Statistics and not (Computer Science or Calculus)

Caution: Must be disjoint
26. Addition Rule For two disjoint events A and B, the probability that one OR the other occurs is the sum of the probabilities of the two events. P(A∪B) = P(A) + P(B) 15% A B 35% Caution: Must be disjoint Here: Events A and B Are disjoint P(A∪B) = = 50%

Classes Taken by Students
27. Stat Cal 105 30 20 10 10 20 80 Comp Sci Stat : _____ Calc : _____ Comp Sci : _____ = 70 = 155 = 120

27. Stat Cal Total Number of Students 105 30 20 30 + 20 + 10 + 80 +105 10 10 20 80 Comp Sci 275 Stat : _____ Calc : _____ Comp Sci : _____ 70 Why is this NOT the sum of the number of students taking Stats + Calc + Comp Sci? 155 The events are not DISJOINT, so they cannot be added. 120

27. Stat Cal Total Number of Students 105 30 20 275 10 10 20 Stat : _____ Calc : _____ Comp Sci : _____ 70 155 80 120 Comp Sci P(Stat) = _____ P(Calc) = _____ P(Comp Sci) = _____ 70 / 275 = 25.5% The events are not DISJOINT, so they DON’T add to 100%. 155 / 275 = 56.4% 120 / 275 = 43.6%

28. Stat Cal 105 30 20 10 10 20 80 Comp Sci How many Statistics or Computer Science? 170 =

28. Classes Taken by Students Stat Cal Total Number of Students 105 30 20 275 10 10 20 80 Comp Sci What is the probability that a randomly selected student is taking Statistics or Computer Science? 𝟏𝟕𝟎 𝟐𝟕𝟓 = 61.8%

29. Classes Taken by Students Stat Cal 105 30 20 10 10 20 80 Comp Sci How many Statistics and Computer Science? 20 =

29. Classes Taken by Students Stat Cal Total Number of Students 105 30 20 275 10 10 20 80 Comp Sci What is the probability that a randomly selected student is taking Statistics and Computer Science? 𝟐𝟎 𝟐𝟕𝟓 = 7.3%

30. Independent Two events are independent of each other if the outcome of one does not influence the outcome of the other EXAMPLE: Flipping a coin once, doesn’t affect the result of the next flip of the coin.

31. If two events are disjoint (mutually exclusive), then they cannot be independent of each other. 𝑬𝑿𝑨𝑴𝑷𝑳𝑬 ∗ : A = {you get an A in this class} B = {you get a B in this class} Events A and B are disjoint , because you can’t get both an A and a B in this class. If you find out that A is true, does that change the probability of B? YES! The probability of a B is now 0%, because you already have an A.

Caution: Must be independent
32. Multiplication Rule For two INDEPENDENT events A and B, the probability that both A AND B occur is the product of the probabilities of the two events. P(A∩B) = P(A) * P(B) Caution: Must be independent What is the probability of flipping a coin 2 times and getting heads both times? Here: Each coin toss is independent. P(Heads ∩ Heads) = .50 * .50 = .25

33. Classes Taken by Students How many Statistics and Computer Science? Stat Cal 27 3 3 2 2 18 4 18 27 Comp Sci Total Number of Students? 100

33. Classes Taken by Students What is the probability that a randomly selected student is taking both Stats and Comp Sci? Stat Cal 27 3 3 2 2 18 18 𝟒 𝟏𝟎𝟎 = 4% 27 Comp Sci Since being in Stats is independent of being in Comp Sci (proven later)… P(Stat ∩ Comp Sci) = P(Stat) * P(Comp Sci) 4% = .10 * .40 = .04 =

90 How many Statistics or (Computer Science and Calculus)? Stat Cal 54
3 3 2 2 36 Comp Sci How many Statistics or (Computer Science and Calculus)? 90

50 How many (Statistics or Computer Science) and Calculus? Stat Cal 54
3 3 2 2 36 Comp Sci How many (Statistics or Computer Science) and Calculus? 50

7. Insurance companies collect annual payments from homeowners in exchange for paying to rebuild houses that burn down. Why should you be reluctant to accept a $1000 payment from your neighbor to replace his house should it burn down during the coming year? Why can the insurance company make that offer? It would be foolish to insure your neighbor’s house for $1000. Although you would probably simply collect $1000, there is a chance you could end up paying much more than $1000. That risk is not worth $1000. The insurance company insures many people. The overwhelming majority of customers pay the insurance and never have a claim. The few customers who do have a claim are offset by the many who simply send their premiums without a claim. The relative risk to the insurance company is low.

= 1 – [P(1 repair) + P(2 repairs) + P(3+ repairs)]
11. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need a) no repair? Since all of the events listed are disjoint, the addition rule can be used. P(no repairs) = 1 – P(some repairs) = 1 – [P(1 repair) + P(2 repairs) + P(3+ repairs)] = 1 – [ ] = 1 – 0.28 = 0.72

11. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need a) no repair? Number of Car Repairs 1 2 1 – [ ] = 1 – 0.28 = 0.72 17% 7% 4% 72% 3+

P(no more than one repair) = P(one repair or no repairs)
11. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need b) no more than one repair? Since all of the events listed are disjoint, the addition rule can be used. P(no more than one repair) = P(one repair or no repairs) = P(1 repair) + P(0 repairs) = = 0.89

11. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need b) no more than one repair? Number of Car Repairs = 0.89 1 2 17% 7% 4% 72% 3+

= P(1 repair) + P(2 repairs) + P(3+ repairs) = 0.17 + 0.07 + 0.04
11. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need c) some repairs? Since all of the events listed are disjoint, the addition rule can be used. P(some repairs) = P(1 repair) + P(2 repairs) + P(3+ repairs) = = 0.28

11. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need c) some repairs? Since all of the events listed are disjoint, the addition rule can be used. Number of Car Repairs 1 2 = = 0.28 17% 7% 4% 72% 3+

P(neither will need repair)
13. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own two cars, what is the probability that a) neither will need repair? First we must assume that the repairs on the two cars are independent from one another in order to use the multiplication rule. (Use results from #11.) P(neither will need repair) = P(1st car no repairs) *P(2nd car no repairs) = (0.72) (0.72) =

a) neither will need repair?
13. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own two cars, what is the probability that a) neither will need repair? First we must assume that the repairs on the two cars are independent from one another in order to use the multiplication rule. (Use results from #11.) 2nd car Some repair 2nd car No repair 0.28 0.72 1st car Some repair 1st car No repair 0.72 0.28 P(None need repair) = (0.72)(0.72) =

P(both will need repair)
13. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own two cars, what is the probability that b) both will need repair? First we must assume that the repairs on the two cars are independent from one another in order to use the multiplication rule. (Use results from #11.) P(both will need repair) = P(1st car some repair) *P(2nd car some repair) = (0.28) (0.28) =

13. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own two cars, what is the probability that b) both will need repair? First we must assume that the repairs on the two cars are independent from one another in order to use the multiplication rule. (Use results from #11.) P(Both need repair) = (0.28)(0.28) = 2nd car Some repair 0.28 1st car Some repair 2nd car No repair 0.28 0.72 2nd car Some repair 1st car No repair 0.28 0.72 2nd car No repair 0.72

“AT LEAST ONE” is the COMPLEMENT of “None”
13. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own two cars, what is the probability that c) at least one car will need repair? First we must assume that the repairs on the two cars are independent from one another in order to use the multiplication rule. (Use results from #11.) “AT LEAST ONE” is the COMPLEMENT of “None” P(at least one will need repair) = 1 – P(Neither needs repair) = 1 – =

13. A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own two cars, what is the probability that c) at least one car will need repair? First we must assume that the repairs on the two cars are independent from one another in order to use the multiplication rule. (Use results from #11.) 2nd car Some repair 0.0784 0.28 1st car Some repair 2nd car No repair + 0.28 0.72 0.2016 2nd car Some repair 1st car No repair 0.28 + 0.72 0.2016 0.4816 2nd car No repair 0.72

15. You used the Multiplication Rule to calculate repair probabilities for your cars.
What must be true about your cars in order to make that approach valid? Do you think this assumption is reasonable? Explain. The repairs on the two cars must be independent from one another in order to use the multiplication rule.

Response Number More production 332 More conservation 563 Both 80
17. A Gallup poll in March 2001 asked 1005 U.S. adults how the United States should deal with the current energy situation: by more production, more conservation, or both? Here are the results: If we select a person at random from this sample of 1005 adults, a) What is the probability that the person responded “More production”? Response Number More production 332 More conservation 563 Both 80 No opinion 30 TOTAL 1005

17. A Gallup poll in March 2001 asked 1005 U.S. adults how the United States should deal with the current energy situation: by more production, more conservation, or both? Here are the results: If we select a person at random from this sample of 1005 adults, b) what is the probability that the person responded “Both” or had no opinion? Response Number More production 332 More conservation 563 Both 80 No opinion 30 TOTAL 1005 110

19. A Gallup poll in March 2001 asked 1005 U.S. adults how the United States should deal with the current energy situation: by more production, more conservation, or both? Here are the results: Suppose we select three people at random from this sample. a) what is the probability that all three responded “More conservation”? Response Number More production 332 More conservation 563 Both 80 No opinion 30 TOTAL 1005

19. A Gallup poll in March 2001 asked 1005 U.S. adults how the United States should deal with the current energy situation: by more production, more conservation, or both? Here are the results: Suppose we select three people at random from this sample. b) What is the probability that none responded “Both”? Response Number More production 332 More conservation 563 Both 80 No opinion 30 TOTAL 1005 1005 -80 925

19. A Gallup poll in March 2001 asked 1005 U.S. adults how the United States should deal with the current energy situation: by more production, more conservation, or both? Here are the results: Suppose we select three people at random from this sample. c) What assumption did you make in computing these probabilities? Response Number More production 332 More conservation 563 Both 80 No opinion 30 TOTAL 1005

19. A Gallup poll in March 2001 asked 1005 U.S. adults how the United States should deal with the current energy situation: by more production, more conservation, or both? Here are the results: Suppose we select three people at random from this sample. d) Explain why you think that assumption is reasonable. Response Number More production 332 More conservation 563 Both 80 No opinion 30 TOTAL 1005

21. Opinion-polling organizations contact their respondents by sampling random telephone numbers. Although interviewers now can reach about 76% of U.S. households, the percentage of those contacted who agree to cooperate with the survey has fallen from 58% in 1997 to only 38% in 2003 (Pew Research Center for the People and the Press). Each household, of course, is independent of the others. a) What is the probability that the next household on the list will be contacted, but will refuse to cooperate?

= 0.4712 Agree to cooperate Refuse to cooperate 0.38 0.62 Household
21. Opinion-polling organizations contact their respondents by sampling random telephone numbers. Although interviewers now can reach about 76% of U.S. households, the percentage of those contacted who agree to cooperate with the survey has fallen from 58% in 1997 to only 38% in 2003 (Pew Research Center for the People and the Press). Each household, of course, is independent of the others. a) What is the probability that the next household on the list will be contacted, but will refuse to cooperate? Agree to cooperate Refuse to cooperate 0.38 0.62 Household contacted Not Contacted 0.24 0.76 P(Will be contacted, but refuse to cooperate) = (0.76)(0.62) =

21. Opinion-polling organizations contact their respondents by sampling random telephone numbers. Although interviewers now can reach about 76% of U.S. households, the percentage of those contacted who agree to cooperate with the survey has fallen from 58% in 1997 to only 38% in 2003 (Pew Research Center for the People and the Press). Each household, of course, is independent of the others. b) What is the probability (in 2003) of failing to contact a household or of contacting the household but not getting them to agree to interview?

21. Opinion-polling organizations contact their respondents by sampling random telephone numbers. Although interviewers now can reach about 76% of U.S. households, the percentage of those contacted who agree to cooperate with the survey has fallen from 58% in 1997 to only 38% in 2003 (Pew Research Center for the People and the Press). Each household, of course, is independent of the others. b) What is the probability (in 2003) of failing to contact a household OR of contacting the household but not getting them to agree to interview? Agree to cooperate Refuse to cooperate 0.38 0.62 Household contacted Not Contacted 0.24 0.76 (0.76)(0.62) = 0.24 =

23. The Masterfood company says that before the introduction of purple, yellow candies made up 20% of their plain M&M’s, red another 20%, and orange, blue, and green each made up 10%. The rest were brown. If you pick an M&M at random, what is the probability that It is brown? It is yellow or orange?

23. The Masterfood company says that before the introduction of purple, yellow candies made up 20% of their plain M&M’s, red another 20%, and orange, blue, and green each made up 10%. The rest were brown. If you pick an M&M at random, what is the probability that 3) It is not green? 4) It is striped?

23. The Masterfood company says that before the introduction of purple, yellow candies made up 20% of their plain M&M’s, red another 20%, and orange, blue, and green each made up 10%. The rest were brown. b) If you pick three M&M’s in a row, what is the probability that 1) They are all brown? 2) The third one is the first one that’s red?

23. The Masterfood company says that before the introduction of purple, yellow candies made up 20% of their plain M&M’s, red another 20%, and orange, blue, and green each made up 10%. The rest were brown. b) If you pick three M&M’s in a row, what is the probability that 3) None are yellow? 4) At least one is green? “AT LEAST ONE” is the COMPLEMENT of “None”

27. You roll a fair die three times. What is the probability that
You roll all 6’s? You roll all odd numbers?

27. You roll a fair die three times. What is the probability that c) None of your rolls gets a number divisible by 3? d) You roll at least one 5? “AT LEAST ONE” is the COMPLEMENT of “None”

27. You roll a fair die three times. What is the probability that
e) The numbers you roll are not all 5’s?

28. A slot machine has three wheels that spin independently
28. A slot machine has three wheels that spin independently. Each has 10 equally likely symbols: 4 bars, 3 lemons, 2 cherries, and a bell. If you play, what is the probability You get 3 lemons? You get no fruit symbols?

28. A slot machine has three wheels that spin independently. Each has 10 equally likely symbols: 4 bars, 3 lemons, 2 cherries, and a bell. If you play, what is the probability c) You get 3 bells (the jackpot)? d) You get no bells?

28. A slot machine has three wheels that spin independently. Each has 10 equally likely symbols: 4 bars, 3 lemons, 2 cherries, and a bell. If you play, what is the probability e) You get at least one bar (an automatic loser)?

Chapter 14 Assignment Part III Part I pp. 341-342 #24, 26 p. 339 #6, 8
Part IV pp , 32, 36 Part I p. 339 #6, 8 Part II pp &14, 18&20

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Recognize when events are disjoint and when events are independent.

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