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Chapter 5 5.2 Probability Warm up Suppose I want to choose a simple random sample of size 6 from a group of 60 seniors and 30 juniors. To do this, I write each person’s name on an equally-sized piece of paper and mix the papers in a large grocery bag. Just as I am about to select the first name, a thoughtful student suggests that I should stratify by class. I agree, and we decide it would be appropriate to select 4 seniors and 2 juniors. However, because I have already mixed up the names, I don’t want to have to separate them all again. Instead, I will select names one at a time from the bag until I get 4 seniors and 2 juniors. Design and carry out a simulation using Table D to estimate the probability that you must draw 10 or more names to get 4 seniors and 2 juniors. State: What is the probability that it takes 10 or more selections to get 4 seniors and 2 juniors? Plan: Using pairs of digits from Table D, we’ll label the 60 seniors 01–60 and the 30 juniors 61–90. Numbers 00 and 91–99 will be skipped. Moving left to right across a row, we’ll look at pairs of digits until we have 4 different labels from 01–60 and 2 different labels from 61–90. Then we will count how many different labels from 01–90 we looked at. Do: Here is an example of one repetition, using line 101 from Table D: 19 (senior) 22 (senior) 39 (senior) 50 (senior) 34 (senior) 05 (senior) 75 Dot Plot (junior) 62 (junior) Collection 2 In this trial, it took exactly 8 selections to get at least 4 seniors and at least 2 juniors. Here are the results of 50 trials: 6 7 8 9 10 11 12 13 14 NumSel Conclude: In the simulation, 11 of the 50 trials required 10 or more selections to get 4 seniors and 2 juniors, so the probability that it takes 10 or more selections is approximately 0.22. 5.2 Probability Models • Sample space: set of all possible outcomes • Probability model: a mathematical description of a random phenomenon consisting of two parts: a sample space and a way of assigning probabilities • What is the sample space of rolling two die? – Display the outcomes – How many outcomes? – What are the possible sums of the die? Example: Roll the Dice Give a probability model for the chance process of rolling two fair, sixsided dice – one that’s red and one that’s green. Sample Space 36 Outcomes Since the dice are fair, each outcome is equally likely. Each outcome has probability 1/36. Probability models allow us to find the probability of any collection of outcomes. Definition: An event is any collection of outcomes from some chance process. That is, an event is a subset of the sample space. Events are usually designated by capital letters, like A, B, C, and so on. If A is any event, we write its probability as P(A). In the dice-rolling example, suppose we define event A as “sum is 5.” There are 4 outcomes that result in a sum of 5. Since each outcome has probability 1/36, P(A) = 4/36 = 1/9. Suppose event B is defined as “sum is not 5.” What is P(B)? P(B) = 1 – 4/36 = 32/36 = 8/9 Basic Rules of Probability • The probability of any event is a number between 0 and 1. 0 ≤ P(A) ≤ 1. • All possible outcomes together must have probabilities whose sum is exactly 1. P(S) = 1. • If all outcomes in the sample space are equally likely, the probability that event A occurs can be found using the formula number of outcomes corresponding to event A P(A) = total number of outcomes in sample space Complement Rule P(Ac) = 1- P(A) • The probability an event does not occur is 1 minus the probability it does occur • Ex: If the probability of you getting a 5 on the AP exam is 12%, what is the probability that you do not get a 5? Basic Rules of Probability Two events A and B are mutually exclusive (disjoint) if they have no outcomes in common and so can never occur together—that is, if P(A and B ) = 0. • If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. • Addition rule for mutually exclusive events: If A and B are mutually exclusive, P(A or B) = P(A) + P(B). Mutually exclusive (Disjoint) events • Getting an A and C in statistics the same 6 weeks. • Rolling a sum of 7 or 11 • Marital status of women ( divorced, single, married, widowed) • Suppose the probability that a randomly selected students is a sophomore is.20 and the probability that he or she is a junior is .3. What is the probability that the student is either a sophomore or a junior? Randomly select a student who took the 2013 AP® Statistics exam and record the student’s score. Here is the probability model: Score: 1 Probability: 0.235 2 0.188 3 0.249 4 0.202 5 0.126 Problem: (a) Show that this is a legitimate probability model. (a) All the probabilities are between 0 and 1 and the sum of the probabilities is 1, so this is a legitimate probability model. (b) Find the probability that the chosen student scored 3 or better. By the addition rule, P(3 or better) = 0.249 + 0.202 + 0.126 = 0.577. By the complement rule and addition rule, P(3 or better) = 1 – P(2 or less) = 1 – (0.235 + 0.188) = 1 – 0.423 = 0.577. Warm up 1. Mrs. R’s TV can pick up 120 stations. At 8pm, 42 stations are showing commercials, 18 sports, 25 movies, 7 news shows, 13 comedy programs, 12 drama programs, and 3 are sales programs (like QVC). A station is chosen at random. Find the following probabilities: a) Choosing a news show 7/120 b) Choosing a commercial 42/120 = 7/20 c) A comedy or drama 13/120 + 12/120 = 25/120 = 5/24 d) Sports, movie or news 18/120 + 25/120 +7/120 = 50/120 = 5/12 e) Not a sales program 1 – 3/120 = 117/120 = 39/40 f) Neither sports nor a movie 1 – (18/120 + 25/120) = 77/120 2. Choose an American adult at random. Define two events: A = the person has a cholesterol level of 240 mg/dl of blood or above (high cholesterol) B = the person has a cholesterol level of 200 to 239 mg/dl (borderline high cholesterol) According to the American Heart Association, P(A) = 0.16 and P(B) = 0.29 a) Explain why events A and B are mutually exclusive. A person cannot have a cholesterol level of both 240 or above and between 200 and 239 at the same time. b) Say in plain language what the event “A or B” is. What is P(A or B)? A person has either a cholesterol level of 240 or above or they have a cholesterol level between 200 and 239. P(A or B) = 0.16 + 0.29 = 0.45 c) If C is the vent that the person chosen has normal cholesterol (below 200 mg/dl), what’s P(C)? P(C) = 1 – 0.45 = 0.55 Two-Way Tables and Probability When finding probabilities involving two events, a two-way table can display the sample space in a way that makes probability calculations easier. Consider the example on page 309. Suppose we choose a student at random. Find the probability that the student (a) has pierced ears. (b) is a male with pierced ears. (c) is a male or has pierced ears. Define events A: is male and B: has pierced ears. (a) (b) Each (c) We want student to find is equally P(male likely or and pierced pierced to be ears), chosen. ears), that that 103 is,is, students P(A P(A orand B).have There B). pierced are ears. in So, theP(pierced classofand ears) 103 = individuals P(B) 103/178. with pierced ears.There Look90atmales the intersection the “Male” row=and “Yes” column. are 19 males with pierced ears. So, P(A and B) = 19/178. However, 19 males have pierced ears – don’t count them twice! P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178 Who owns a home? What is the relationship between educational achievement and home ownership? A random sample of 500 U.S. adults was selected. Each member of the sample was identified as a high school graduate (or not) and as a home owner (or not). The two-way table displays the data. Homeowner Not a homeowner Total High school graduate 221 89 310 Not a high school graduate 119 71 190 Total 340 160 500 Suppose we choose a member of the sample at random. Find the probability that the member: (a) is a high school graduate. Because 310 of the 500 members of the sample graduated from high school, P(G) = 310/500. (b) is a high school graduate and owns a home. Because 221 of the 500 members of the sample graduated from high school and own a home, P(G and H) = 221/500. (c) is a high school graduate or owns a home. Because there are 221 + 89 + 119 = 429 people who graduated from high school or own a home, P(G or H) is 429/500. Note that it is inappropriate to compute P(G) + P(H) to find this probability because the events G and H are not mutually exclusive—221 people both are high school graduates and own a home. If you did add these probabilities, the result would be 650/500, which is clearly wrong because it results in a probability that is greater than 1. General Addition Rule for Two Events We can’t use the addition rule for mutually exclusive events unless the events have no outcomes in common. General Addition Rule for Two Events If A and B are any two events resulting from some chance process, then P(A or B) = P(A) + P(B) – P(A and B) Who owns a home? Using the previous example, use the general addition rule to find the probability that a random person is a homeowner or graduate. P(G or H) = P(G) + P(H) – P(G and H) = 310/500 + 340/500 – 221/500 = 429/500. This matches our previous result. Venn Diagrams and Probability Because Venn diagrams have uses in other branches of mathematics, some standard vocabulary and notation have been developed. The complement AC contains exactly the outcomes that are not in A. The events A and B are mutually exclusive (disjoint) because they do not overlap. That is, they have no outcomes in common. Venn Diagrams and Probability The intersection of events A and B (A ∩ B) is the set of all outcomes in both events A and B. The union of events A and B (A ∪ B) is the set of all outcomes in either event A or B. Hint: To keep the symbols straight, remember ∪ for union and ∩ for intersection. Use the same two-way table from before summarizing the relationship between educational status and home ownership from the previous example. Our events of interest were G: is a high school graduate and H: is a homeowner. Make a Venn Diagram. The four distinct regions in the Venn diagram below correspond to the four cells in the two-way table as follows: Region in Venn diagram In the intersection of two circles Inside circle G, outside circle H Inside circle H, outside circle G Outside both circles In words In symbols Count Graduate and owns home G ∩ H 221 Graduate and doesn’t own home Not a graduate but owns home Not a graduate and doesn’t own home G ∩ HC 89 GC ∩ H 119 GC ∩ HC 71 Phone usage According to the National Center for Health Statistics (http://www.cdc.gov/nchs/data/nhis/earlyrelease/wireless201306.pdf), in December 2012, 60% of U.S. households had a traditional landline telephone, 89% of households had cell phones, and 51% had both. Suppose we randomly selected a household in December 2012. Problem: (a) Make a two-way table that displays the sample space of this chance process. (b) Construct a Venn diagram to represent the outcomes of this chance process. (c) Find the probability that the household has at least one of the two types of phones. (d) Find the probability that the household has a cell phone only. a) We will define events L: has a landline and C: has a cell phone. Cell phone 0.51 0.38 0.89 Landline No landline Total No cell phone 0.09 0.02 0.11 Total 0.60 0.40 1.00 C L .09 0.51 0.38 0.02 (c) To find the probability that the household has at least one of the two types of phones, we need to find the probability that the household has a landline, a cell phone, or both. P(L ᴜ C) = P(L) + P(C) - P(L ∩ C) = 0.60 + 0.89 – 0.51 = 0.98 There is a 0.98 probability that the household has at least one of the two types of phones. (d) P(cell phone only) = P(LC ∩ C) = 0.38. There is a 0.38 probability that the household is cell-phone only.