Probability Class 30 1

Homework Check Assignment: Chapter 7 – Exercise 7.20, 7.29, 7.47, 7.48, 7.53 and 7.57 Reading: Chapter 7 – p

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8 Independent and Dependent Events Two events are independent of each other if knowing that one will occur (or has occurred) does not change the probability that the other occurs. Two events are dependent if knowing that one will occur (or has occurred) changes the probability that the other occurs. The definitions can apply either … to events within the same random circumstance or to events from two separate random circumstances.

9 Example 7.10 Winning a Free Lunch Customers put business card in restaurant glass bowl. Drawing held once a week for free lunch. You and Vanessa put a card in two consecutive weeks. Event A = You win in week 1. Event B = Vanessa wins in week 1. Event C = Vanessa wins in week 2. Events A and B refer to the same random circumstance and are dependent. Events A and C refer to different random circumstances and are independent.

10 Example 7.11 Alicia Answering Event A = Alicia is selected to answer Question 1. Event B = Alicia is selected to answer Question 2. P(A) = 1/50. If event A occurs, her name is no longer in the bag, so P(B) = 0. If event A does not occur, there are 49 names in the bag (including Alicia’s name), so P(B) = 1/49. Events A and B refer to different random circumstances, but are A and B independent events? Knowing whether A occurred changes P(B). Thus, the events A and B are not independent.

11 Conditional Probabilities Conditional probability of the event B, given that the event A occurs, is the long-run relative frequency with which event B occurs when circumstances are such that A also occurs; written as P(B|A). P(B) = unconditional probability event B occurs. P(B|A) = “probability of B given A” = conditional probability event B occurs given that we know A has occurred or will occur.

12 Example 7.12 Probability That a Teenager Gambles Differs for Boys and Girls Notice dependence between “weekly gambling habit” and “sex.” Knowledge of a student’s sex changes the probability that s/he is a weekly gambler. Survey: 78,564 students (ninth- and twelfth-graders) The proportions of males and females admitting they gambled at least once a week during the previous year were reported. Results for ninth-grade: P(student is weekly gambler | student is boy) = P(student is weekly gambler | student is girl) = 0.045

Basic Rules for Finding Probabilities Rule 1 (for “not the event”): P(A C ) = 1 – P(A) Example 7.13 Probability a Stranger Does Not Share Your Birth Date P(next stranger you meet will share your birthday) = 1/365. P(next stranger you meet will not share your birthday) = 1 – 1/365 = 364/365 = Probability an Event Does Not Occur

14 Rule 2 (addition rule for “either/or/both”): Rule 2a (general): P(A or B) = P(A) + P(B) – P(A and B) Rule 2b (for mutually exclusive events): If A and B are mutually exclusive events, P(A or B) = P(A) + P(B) Probability That Either of Two Events Happen

15 Example 7.14 Roommate Compatibility Brett is off to college. There are 1000 male students. Brett hopes his roommate will not like to party and not snore. A = likes to partyP(A) = 250/1000 = 0.25 B = snoresP(B) = 350/1000 = 0.35 Probability Brett will be assigned a roommate who either likes to party or snores, or both is: P(A or B) = P(A) + P(B) – P(A and B) = – 0.15 = 0.45 So probability his roommate is acceptable is 1 – 0.45 = 0.55

16 Example 7.15 Probability of Either Two Boys or Two Girls in Two Births What is the probability that a woman who has two children has either two girls or two boys? Recall that the probability of a boy is and probability of a girl is Then we have (using Rule 3b): Event A = two girls P(A) = (0.488)(0.488) = Event B = two boys P(B) = (0.512)(0.512) = Probability woman has either two boys or two girls is: P(A or B) = P(A) + P(B) = = Note: Events A and B are mutually exclusive (disjoint).

17 Rule 3 (multiplication rule for “and”): Rule 3a (general): P(A and B) = P(A)P(B|A) Rule 3b (for independent events): If A and B are independent events, P(A and B) = P(A)P(B) Extension of Rule 3b (for > 2 indep events): For several independent events, P(A 1 and A 2 and … and A n ) = P(A 1 )P(A 2 )…P(A n ) Probability That Two or More Events Occur Together

18 Example 7.16 Probability of Male and Gambler About 11% of all ninth-graders are males and weekly gamblers. For ninth-graders, 22.9% of the boys and 4.5% of the girls admitted they gambled at least once a week during the previous year. The population consisted of 50.9% girls and 49.1% boys. P(male and gambler) = P(A and B) = P(A)P(B|A) = (0.491)(0.229) = Event A = male Event B = weekly gambler P(A) = 0.491P(B|A) = 0.229

19 Example 7.17 Probability Two Strangers Both Share Your Birth Month Note: The probability that 4 unrelated strangers all share your birth month would be (1/12) 4. Assume all 12 birth months are equally likely. What is the probability that the next two unrelated strangers you meet both share your birth month? P(both strangers share your birth month) = P(A and B) = P(A)P(B) = (1/12)(1/12) = Event A = 1 st stranger shares your birth month P(A) = 1/12 Event B = 2 nd stranger shares your birth month P(B) = 1/12 Note: Events A and B are independent.

20 Rule 4 (conditional probability): P(B|A) = P(A and B)/P(A) P(A|B) = P(A and B)/P(B) Determining a Conditional Probability

21 Example 7.22 Probability of an A grade on Final given an A grade on Mid-term A on midterm and A on final P(A and B) = 0.15 A on midterm P(B) = 0.25 A on final P(A) = 0.20 Given: the probability of getting an A on midterm is 0.25 the probability of getting an A on final is 0.20 the probability of getting an A on midterm an final is 0.15 What is the probability of getting an A on Final given an A grade on Midterm?

22 Example 7.22 Probability of an A grade on Final given an A grade on Mid-term P(A|B) = P(A and B)/P(B) = 0.15/0.25 = 0.6

23 In Summary … When two events are mutually exclusive and one happens, it turns the probability of the other one to 0. When two events are independent and one happens, it leaves the probability of the other one alone. Students sometimes confuse the definitions of independent and mutually exclusive events.

24 In Summary …

25 A sample is drawn with replacement if individuals are returned to the eligible pool for each selection. A sample is drawn without replacement if sampled individuals are not eligible for subsequent selection. Sampling with and without Replacement

Finding Complicated Probabilities Example 7.21 Winning the Lottery Event A = winning number is 956. What is P(A)? Method 1: With physical assumption that all 1000 possibilities are equally likely, P(A) = 1/1000. Method 2: Define three events, B 1 = 1 st digit is 9, B 2 = 2 nd digit is 5, B 3 = 3 rd digit is 6 Event A occurs if and only if all 3 of these events occur. Note: P(B 1 ) = P(B 2 ) = P(B 3 ) = 1/10. Since these events are all independent, we have P(A) = (1/10) 3 = 1/1000. * Can be more than one way to find a probability.

27 None and at Least One Example 7.23 Will Shaun’s Friends Be There for Him? Shawn asks three friends to call to wake him up. Probability each friend will call is 0.7. What is probability that none of them will call? Note: “at least one” is complement of “none” P(none call) = P(friend 1 does not call and friend 2 does not call and friend 3 does not call) = (.3)(.3)(.3) =.027 P(at least one friend calls) = 1 – =.973.

28 You know P(B|A) but want P(A|B): Use Rule 3a to find P(B) = P(A and B) + P(A C and B), then use Rule 4. Bayes’ Rule

29 Example 7.24 Alicia Is Probably Healthy Define the following events for her disease status: A = disease; A C = no disease P(A) = 1/1000 =.001; P(A C ) = Define the following events for her test results: B = test is positive, B C = test is negative P(B|A) = 0.95 (positive test given disease) P(B C |A) = 0.05 (negative test given disease) P(B|A C ) = 0.05 (positive test given no disease) P(B C |A C ) = 0.95 (negative test given no disease) What is the probability that Alicia has the disease given that the test was positive?

30 Example 7.24 Alicia Healthy? P(B) = P(B|A)P(A) + P(B|A C )P(A C ) = (.95)(.001) + (.05)(.999) = There is less than a 2% chance that Alicia has the disease, even though her test was positive.

Homework Assignment: Chapter 7 – Exercise 7.53 and 7,57 Reading: Chapter 7 – p