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Making sense of the Common Core domain: “Conditional Probability and the Rules for Probability” David Spohn Hudson High School Hudson Ohio [email protected] Understand independence and conditional probability and use them to interpret data • S-CP.1. Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). • S-CP.2. Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. • S-CP.3. Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. Understand independence and conditional probability and use them to interpret data • S-CP.4. Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. • S-CP.5. Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. Use the rules of probability to compute probabilities of compound events in a uniform probability model • S-CP.6. Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model. • S-CP.7. Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. • S-CP.8. (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. • S-CP.9. (+) Use permutations and combinations to compute probabilities of compound events and solve problems. • The implementation of the CCSS will require many teachers to teach probability for the first time. It is important for students to learn probability in the spirit of the CCSS and be able to make sense of the underlying thought processes. This session will focus on conditional probability and the rules of probability; however we will start with activities that develop appropriate reasoning and understanding of probability concepts. Once these are developed we will attempt to make sense of the algebraic rules from the models we have created. Intended Outcome! “I expect you all to be independent, innovative, critical thinkers who will do exactly as I say!” The Definition of Probability P(A) = 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝐴/𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 Example 1: A fair die is tossed. What is the probability of a “5”? Outcome P(x) 1 2 3 4 5 1/6 1/6 1/6 1/6 1/6 6 1/6 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 Example 2: Two dice are tossed. What is the probability that the sum of the faces is 5? What is wrong with this solution? What was implied in the definition of probability that is not true in this situation? Outcome 2 3 4 5 6 7 8 9 10 11 12 P(x) 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 Example 2: Two dice are tossed. What is the probability that the sum of the faces is 5? The correct model demonstrates the probability. So the P(5) = 4/36 !! One must be convinced that all cells are equally likely! And that the dice are distinguishable! 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Read as “the probability of B given A.” • P(Ac) = 1 – P(A) • P(A or B) = P(A) + P(B) – P(A and B) • P(A and B) = P(A)*P(B│A) • P(B│A) = 𝑃(𝐴 𝑎𝑛𝑑 𝐵)/𝑃(𝐴) • Mutually Exclusive Events: P(A and B) = 0 • Independent Events: P(B│A) = P(B) The Rules! Given P(5) = P[(first die ok AND second die ok) = P(first die ok) * P(second ok | first ok) = 4/6 ∗ 1/6 = 4/36 1, 2, 3, or 4 Only one option will create a 5 given the first die If we follow the rules for two dice roll! • Many students will do better just ‘figuring out’ the probability from a model or thought. • The correct model is sometimes elusive at the beginning. • Ultimately, talented students need to be able to justify and use the rules for more abstract problems. • Some students will resist the models or resist the rules. Thoughts on Rules, Models, and Applications • I imagine that there are tremendously different levels of knowledge/confidence in probability • I will not be developing everything completely or in a linear fashion. • Do not worry! • Feel free to ask questions! • I actually believe that we need to develop these thoughts and learn as we go! • Learning comes from repeated exposure and necessity! Thoughts on the presentation Ice Cream! A school survey of 200 students found that 80 students liked vanilla ice cream, 95 liked chocolate, and 65 liked strawberry. 25 students liked both strawberry and chocolate (but not vanilla), 15 liked vanilla and strawberry (but not chocolate), 10 liked vanilla and chocolate (but not strawberry) and 20 students liked all three flavors. If a student is chosen at random, find the following: a) P(vanilla only) b) P(none of the flavors) c) P(vanilla and chocolate) d) P(chocolate or strawberry) e) P(strawberry │vanilla) f) P(vanilla │not chocolate) Ice Cream! Ice Cream Venn Diagram P(vanilla and chocolate) P(chocolate or strawberry) P(strawberry │vanilla) P(vanilla │not chocolate) Independence: • “Two events are independent if the outcome (knowledge) of one does not affect (change) the probability of the other” • P(B│A) = P(B) • This is the most prevalent topic in the standards for Probability!!1 Independence Using a two-way table I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 • P(B) • P(III) Find the following: • P(B and III) • P(B or III) • P(B | III) • P(III | B) Are (A, B, C) independent of (I, II, III, IV)? The Rows The Columns I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 P(B) = 360/720 P(III) = 288/720 I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 P(B and III) = 146/720 I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 P(B or III) = 64+120+146+20+110+32/720 = 502/720 I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 This is a good time to verify a rule: • P(B or III) = P(B) + P(III) – P(B and III) Notice that the 146 is counted for both the ROW B and COLUMN III totals. Therefore it has been included twice and needs to be subtracted out once. I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 P(B │III ) = 146/288 I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 P(III │ B) = 146/360 Are the rows and columns independent of each other? I II III IV A 45 60 110 25 240 B 64 120 146 30 360 C 35 36 32 17 120 144 216 288 72 720 • Notice that the elements in B are 50% of all elements • Therefore, if independent, B should represent 50% of each column (it does not) • Therefore the rows and columns are NOT independent By Rule: P(B│III) ≠ P(B) 146/288 360/720 I II III IV A 240 B 360 C 120 144 216 288 72 720 Fill in the table so that the rows and columns are independent. Titanic The chart below classifies each of the 2201 passengers on the Titanic by survival status and type of ticket. First Class tickets were the most expensive and Third Class tickets were the least expensive. First Second Third Titanic Class Class Class Ship's Survival Ticket Ticket Ticket Crew Total Alive 203 118 178 212 711 Dead 122 167 528 673 Total 325 285 706 885 2201 Find P(Alive) Find P(Alive │Third Class Ticket) Are ‘Survival Status’ and 1490 ‘Ticket Type’ Independent? Disease Control A test for a certain disease gives a falsepositive result for 5% of the people who do not have the disease. It gives a falsenegative result for (.3)% of the people who do have the disease. It is known that 2% of the population actually has the disease. • What is that probability that a randomly selected person tests positive? • Given that a person has tested positive, what is the probability that they have the disease? Adapted from 1997 AP Statistics Exam Understand independence and conditional probability and use them to interpret data • S-CP.1. Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). • S-CP.2. Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. • S-CP.3. Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. • S-CP.4. Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. • S-CP.5. Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. Use the rules of probability to compute probabilities of compound events in a uniform probability model • S-CP.6. Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model. • S-CP.7. Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. • S-CP.8. (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. • S-CP.9. (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Making sense of the Common Core domain: “Conditional Probability and the Rules for Probability” David Spohn Hudson High School Hudson Ohio [email protected]