1. Copyright © 2010 Pearson Education, Inc. Chapter 6 Probability

2. Copyright © 2010 Pearson Education, Inc. Slide 14 - 2 Dealing with Random Phenomena A random phenomenon is a situation in which we know what outcomes could happen, but we don’t know which particular outcome did or will happen. Each occasion upon which we observe a random phenomenon is called a trial. Example: flipping a coin. At each trial, we note the value of the random phenomenon, and call it an outcome. Example: tails When we combine outcomes, the resulting combination is an event. Example: Flipping a coin twice and getting both tails. The collection of all possible outcomes is called the sample space. Example: List all the possible outcomes of flipping two coins.

3. Copyright © 2010 Pearson Education, Inc. Slide 14 - 3 Independent Events - the outcome of one trial doesn’t influence or change the outcome of another. Example: Are the following independent events? a. coin flips b. Rolling a die or a pair of dice. c. Having an IPod and have an ITunes gift card. d. Your grade in AP statistics and your grade in trigonometry.

4. Copyright © 2010 Pearson Education, Inc. Slide 14 - 4 The Law of Large Numbers The Law of Large Numbers (LLN) says that the long-run relative frequency of repeated independent events gets closer and closer to a single value. We call the single value the probability of the event.

5. Copyright © 2010 Pearson Education, Inc. Slide 14 - 5 The probability of an event is the number of outcomes in the event divided by the total number of possible outcomes. P(A) = Modeling Probability # of outcomes in A # of possible outcomes

6. Copyright © 2010 Pearson Education, Inc. Slide 14 - 6 Formal Probability 1.Two requirements for a probability: A probability is a number between 0 and 1. For any event A, 0 ≤ P(A) ≤ 1.

7. Copyright © 2010 Pearson Education, Inc. Slide 14 - 7 Formal Probability (cont.) 2.Probability Assignment Rule: The probability of the set of all possible outcomes of a trial must be 1. P(S) = 1 (S represents the set of all possible outcomes.)

8. Copyright © 2010 Pearson Education, Inc. Slide 14 - 8 Formal Probability (cont.) 3.Complement Rule:  The set of outcomes that are not in the event A is called the complement of A, denoted A C.  The probability of an event occurring is 1 minus the probability that it doesn’t occur: P(A) = 1 – P(A C )

9. Copyright © 2010 Pearson Education, Inc. Slide 14 - 9 Example: When we arrive at the intersection of Altama and Community Rd. the probability the light is green is about 35% of the time. If P(green) =.35, what is the probability the light isn’t green when you get to Altama and Community?

10. Copyright © 2010 Pearson Education, Inc. Slide 14 - 10 Formal Probability (cont.) 4.Addition Rule: Events that have no outcomes in common (and, thus, cannot occur together) are called disjoint (or mutually exclusive).

11. Copyright © 2010 Pearson Education, Inc. Slide 14 - 11 Formal Probability (cont.) 4.Addition Rule (cont.): 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), provided that A and B are disjoint.

12. Copyright © 2010 Pearson Education, Inc. Slide 14 - 12 Example: Knowing the P(green) =.35 and P(yellow) =.04, when we get to the light at the corner of Altama and Community: a. What is the probability the light is green or yellow? Written P(green yellow). b. When you get to the light are the events of the light being green and yellow disjoint or independent? c. What is the probability the light is red?

13. Copyright © 2010 Pearson Education, Inc. Slide 14 - 13 Formal Probability (cont.) 5.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), provided that A and B are independent.

14. Copyright © 2010 Pearson Education, Inc. Slide 14 - 14 Example: Opinion polling organizations contact their respondents by telephone. In 1990s this method could reach about 60% of US households. By 2003, the contact rate had risen to 76%. We can reasonably assume each household’s response to be independent of the others. What is the probability that … a. The interviewer successfully contacts the next household on the list? b. The interviewer successfully contacts both of the next two households on the list? c. The interviewer’s first successful contact is the third household on the list? d. The interviewer makes at least one successful contact among the next five households on the list?

15. Copyright © 2010 Pearson Education, Inc. Slide 14 - 15 Formal Probability (cont.) 5.Multiplication Rule (cont.): Two independent events A and B are not disjoint, provided the two events have probabilities greater than zero:

16. Copyright © 2010 Pearson Education, Inc. Slide 14 - 16 Formal Probability (cont.) 5.Multiplication Rule: Many Statistics methods require an Independence Assumption, but assuming independence doesn’t make it true. Always Think about whether that assumption is reasonable before using the Multiplication Rule.

17. Copyright © 2010 Pearson Education, Inc. Slide 14 - 17 Formal Probability - Notation In this text we use the notation P(A  B) and P(A  B). In other situations, you might see the following: P(A or B) instead of P(A  B) P(A and B) instead of P(A  B)

18. Copyright © 2010 Pearson Education, Inc. Slide 15 - 18 The General Addition Rule When two events A and B are disjoint, we can use the addition rule for disjoint events from Chapter 14: P(A  B) = P(A) + P(B) 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.

19. Copyright © 2010 Pearson Education, Inc. Slide 15 - 19 The General Addition Rule (cont.) General Addition Rule: For any two events A and B, P(A  B) = P(A) + P(B) – P(A  B) The following Venn diagram shows a situation in which we would use the general addition rule:

20. Copyright © 2010 Pearson Education, Inc. Slide 15 - 20 Example: A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program and 42% do both. What is the probability that a randomly selected student either lives or eats on campus?

21. Copyright © 2010 Pearson Education, Inc. Example: Draw a Venn Diagram for the previous example. What is the probability that a randomly selected student: a. Lives off campus and doesn’t have a meal program? b. Lives in a residence hall but doesn’t have a meal program? Slide 14 - 21

22. Copyright © 2010 Pearson Education, Inc. Slide 15 - 22 Example: Draw a Venn Diagram for the previous example. What is the probability that a randomly selected student: a. Lives off campus and doesn’t have a meal program? b. Lives in a residence hall but doesn’t have a meal program?

23. Copyright © 2010 Pearson Education, Inc. Disjoint No two events have any outcomes in common, then you can add the individual event probabilities Independent Knowing one event’s probability does not change the probability that another event occurs, then you can multiply the probabilities Flipping a coin Slide 14 - 23

24. Copyright © 2010 Pearson Education, Inc. Slide 15 - 24 Independent ≠ Disjoint Give an example of disjoint events: Give an example of independent events: Disjoint events cannot be independent. A common error is to treat disjoint events as if they were independent, and apply the Multiplication Rule for independent events—don’t make that mistake.

25. Copyright © 2010 Pearson Education, Inc. Example: It is reported that typically as few as 10% of random phone calls result in a completed interview. Reasons are varied, but some of the most common include no answer, refusal to cooperate and failure to complete the call. Which of the following events are independent, disjoint or neither independent nor disjoint? a. A = Your telephone number is randomly selected. B = You’re not at home at dinner time when they call. b. A = As a selected subject, you complete the interview. B = As a selected subject, you refuse to cooperate. c. A = You are not at home when they call at 11 am. B = You are employed full time. Slide 14 - 25

26. Copyright © 2010 Pearson Education, Inc. Slide 15 - 26 Example: It is reported that typically as few as 10% of random phone calls result in a completed interview. Reasons are varied, but some of the most common include no answer, refusal to cooperate and failure to complete the call. Which of the following events are independent, disjoint or neither independent nor disjoint? a. A = Your telephone number is randomly selected. B = You’re not at home at dinner time when they call. b. A = As a selected subject, you complete the interview. B = As a selected subject, you refuse to cooperate. c. A = You are not at home when they call at 11 am. B = You are employed full time.

27. Copyright © 2010 Pearson Education, Inc. Conditional Probability The probability of an event under the condition that some preceding event has occurred P(A l B)=P(A and B) P(B)

28. Copyright © 2010 Pearson Education, Inc. Slide 15 - 28 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 has already occurred.” A probability that takes into account a given condition is called a conditional probability.

29. Copyright © 2010 Pearson Education, Inc. Slide 15 - 29 The General Multiplication Rule When two events A and B are independent, we can use the multiplication rule for independent events from Chapter 14: P(A  B) = P(A) x P(B) However, when our events are not independent, this earlier multiplication rule does not work. Thus, we need the General Multiplication Rule. P(A  B) = P(A)  P(B|A) or P(A  B) = P(B)  P(A|B)

30. Copyright © 2010 Pearson Education, Inc. Conditional Probability You toss two coins. What is the probability that you toss two heads given that you have tossed at least 1 head? P(A) = The two coins come up heads P(B) = There is a least one head Different outcomes of two coins? (H,H)-(T,H)-(H,T)-(T,T) P(B)=3/4 P(A and B) = ¼ P(A l B) = P(A and B)/P(B) P(A l B) = (¼)/(3/4) P(A l B) = 1/3

31. Copyright © 2010 Pearson Education, Inc. Conditional Probability A neighborhood lets families have two pets. They can have two dogs, two cats, or one of each. What is the probability that the family will have exactly 2 cats if the second pet is a cat? P(A) = Two cats = 1/3 P(B) = At least one cat=2/3 P (A l B) = (1/3)/(2/3)=1/2

32. Copyright © 2010 Pearson Education, Inc. Conditional Probability Two number cubes are tossed. Find the probability that the numbers showing on the cubes match given that their sum is greater than five. P(A) = Cubes Match P(A) = 1/6 P(B) = Sum is greater than 5 P(B)=26/36 P(A and B) = 4/36 P(A l B) = (4/36)/(26/36) P(A l B) = 2/13

33. Copyright © 2010 Pearson Education, Inc. Conditional Probability One card is drawn from a standard deck of cards. What is the probability that it is a queen given that it is a face card? P(A) = Queen = 4/52 P(B) =Face card = 12/52 P(A and B) = 4/52 P(A l B) = (4/52)/(12/52) P(A l B) = 1/3

34. Copyright © 2010 Pearson Education, Inc. Practice Try problems6.54-6.55 on page 369 Slide 14 - 34

35. Copyright © 2010 Pearson Education, Inc. Slide 15 - 35 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.

36. Copyright © 2010 Pearson Education, Inc. Example: You draw two cards from a deck and place them in your hand. What is the probability both cards are black face cards? Slide 14 - 36

37. Copyright © 2010 Pearson Education, Inc. Slide 15 - 37 Example: You draw two cards from a deck and place them in your hand. What is the probability both cards are black face cards?

38. Copyright © 2010 Pearson Education, Inc. Slide 15 - 38 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 probabilities is consistent with our “make a picture” mantra.

39. Copyright © 2010 Pearson Education, Inc. Slide 15 - 39 Tree Diagrams (cont.) Figure 15.5 is an 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 compound events.

40. Copyright © 2010 Pearson Education, Inc. Slide 15 - 40 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 P(A  B), since P(A  B) = P(A) x P(B|A) From this information, we can find P(A|B):

41. Copyright © 2010 Pearson Education, Inc. Slide 15 - 41 Bayes’s Rule When we reverse the probability from the conditional probability that you’re originally given, you are actually using Bayes’s Rule.