What is the marginal distribution of income? What proportion of the sample made over $50,000?

Stats: Modeling the World Chapter 6 From Randomness to Probability

Sample Space The set of all possible outcomes Try it! Albert/Barbara Albert/Carl Albert/Dianna Barbara/Carl Barbara/Dianna Carl/Dianna AND VICE VERSA!!!! P(Both Girls) = 2/12

Basic Probability Rules For any event A, 0 ≤ P(A) ≤ 1. The sum of the probabilities for all outcomes in the sample space must equal 1. P(S) = 1

Try it! Is this a possible sample space? Why or why not? d) {0, 0, 1, 0} e) {0.1, 0.2, 1.2, -1.5} Yes No

Addition Rule P(A or B) = P(A) + P(B) provided that A and B are disjoint. Disjoint (Mutually Exclusive) The two events, A and B, cannot occur at the same time.

Complement Rule The set of outcomes that are not in the event A is called the complement of A, denoted NOT A. The probability of an event occurring is 1 minus the probability that it doesn’t occur: P(A) = 1 – P(NOT A)

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Knowledge of event A does not affect the probability of event B Multiplication Rule P(A and B) = P(A) x P(B) provided that A and B are independent. Independent Events Knowledge of event A does not affect the probability of event B

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Example The Masterfoods company says that before the introduction of purple, yellow candies made up 20% of their plain M&Ms, red another 20%, and orange, blue, and green each made up 10%. The rest were brown. a) If you pick an M&M at random, what is the probability that: - it is brown? - it is yellow or orange? - it is not green? - it is striped? 30% 20%+10% = 30% 100% - 10% = 90% 0%

Example (.3)(.3)(.3) = 0.027 (.8)(.8)(.2) = 0.128 (.8)(.8)(.8) = 0.512 The Masterfoods company says that before the introduction of purple, yellow candies made up 20% of their plain M&Ms, red another 20%, and orange, blue, and green each made up 10%. The rest were brown. b) If you pick three M&Ms in a row, what is the probability that: - they are all brown? - the third one is the first one that’s red? - none are yellow? - at least one is green? (.3)(.3)(.3) = 0.027 (.8)(.8)(.2) = 0.128 (.8)(.8)(.8) = 0.512 1-No Greens = 1 - (.9)(.9)(.9) = 0.271

Example P(0) = 1 – 0.28 = 0.72 P(0 or 1) = 0.72 + 0.17 = 0.89 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. a) What is the probability that a car chosen at random will need: 1. no repairs? 2. no more than one repair? 3. some repairs? P(0) = 1 – 0.28 = 0.72 P(0 or 1) = 0.72 + 0.17 = 0.89 P(1 or more) = 1 – P(none) = 0.28

Example P(none and none) = (0.72)(0.72) = 0.5184 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. b) If you own two cars, what is the probability that: 1. neither will need repair? 2. both will need repair? 3. at least one will need repair? P(none and none) = (0.72)(0.72) = 0.5184 P(some and some) = (0.28)(0.28) = 0.0784 1 - P(none) = 1 – 0.5184 = 0.4816

Example P(no/no/no) = (0.3)(0.3)(0.3) = 0.027 A certain bowler can bowl a strike 70% of the time. What is the probability that she a) goes three consecutive frames without a strike? b) makes her first strike in the third frame? c) has at least one strike in the first three frames d) bowls a perfect game (12 consecutive strikes) P(no/no/no) = (0.3)(0.3)(0.3) = 0.027 P(no/no/yes) = (0.3)(0.3)(0.7) = 0.063 1 - P(no/no/no) = 1 – 0.027 = 0.973 P(12 strikes) = (0.7)^12 = 0.0138

Example P(RRR) = (0.29)(0.29)(0.29) = 0.024 Suppose that in your city 37% of voters are registered Democrats, 29% Republicans, and 11% other parties. Voters not aligned with any party are termed “independent”. You are conducting a poll by calling registered voters at random. In your first three calls, what is the probability you talk to: a. All Republicans? b. No Democrats? c. At least one Independent? P(RRR) = (0.29)(0.29)(0.29) = 0.024 P(ND/ND/ND) = (0.63)(0.63)(0.63) = 0.250 1 – P(no Indy)= 1 - (0.89)(0.89)(0.89) = 0.295

P(40 or 100) = 3/12 = 0.25 P(10/10) = (0.5)(0.5) = 0.25 P(20/20/20) = (0.25)(0.25)(0.25) = 0.0156

P(20 or less 4 times) = (0.75)(0.75)(0.75)(0.75) = 0.316 P(No/No/No/No/WIN) = (11/12)^4 * (1/12) = 0.0588 1 - P(no gold in 6) = 1 – (11/12)^6 = 0.407

Disjoint events CANNOT be independent!! Common Errors Don’t confuse disjoint and independent!! Disjoint events CANNOT be independent!!

Example In a previous WAP problem, we calculated the probabilities of getting various M&Ms. a) If you draw one M&M, are the events of getting a red one and getting an orange one disjoint or independent or neither? b) If you draw two M&Ms one after the other, are the events of getting a red on the first and a red on the second disjoint or independent or neither? c) Can disjoint events ever be independent? Explain.