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UNIT 4 PROBABILITY SCIENTIFIC INQUIRY AND ANALYSIS 1 OBJECTIVES UNIT 4 The student will be able to: • Describe events as subsets of a set of outcomes using characteristics or categories of the outcomes, or as unions, intersections, or complements of other events (“or”, “and”, “not”.) (CCSS.HSS.CP.A.1) • Find the number of ways a group of objects can be arranged in order (permutations with and without replacement). (CCSS.HSS.CP.B.9) • Find the number of ways to choose several objects from a group without regard to order (combinations with and without replacement). (CCSS.HSS.CP.B.9) SCIENTIFIC INQUIRY AND ANALYSIS 2 OBJECTIVES UNIT 4 • Find the probability of simple and compound events. (5.3.12.C1) • Determine if two events are independent by showing P(A given B) is the same as P(A) and that the P(B given A) is the same as P(B). (CCSS.HSS.CP.A.2, CCSS.HSS.CP.A.3) • Find conditional probabilities as probability of A given B as P(A and B)/P(B). (CCSS.HSS.CP.B.6) • Use the multiplication rule to find the probability of two events occurring in a sequence and to find conditional probabilities. (CCSS.HSS.CP. B.8) • Use the addition rule to find the probability of two events. (CCSS.HSS.CP.B.7) SCIENTIFIC INQUIRY AND ANALYSIS 3 PROBABILITY • Probability – the numerical measure of the likelihood of an event occurring. • Statistics – collection of methods for planning experiments, obtaining data, and then organizing, summarizing, presenting, analyzing, interpreting, and drawing conclusions from the data. SCIENTIFIC INQUIRY AND ANALYSIS 4 PROBABILITY • Sample Space – Before analyzing the probability that something can occur, it is necessary to understand the number of possible outcomes given a set of possible choices or selections. This is known as the sample space. SCIENTIFIC INQUIRY AND ANALYSIS 5 PROBABILITY • Fundamental Counting Principle – When determining the number of possible outcomes of an event, it is always not practical to count all of the outcomes. All of the possible outcomes is the sample space. • One method to use to find the sample space is to use the Tree – Method. • Another method to use to find the sample space is to us the Grid – Method. The grid – method is best if the number of selections is no more than 2. SCIENTIFIC INQUIRY AND ANALYSIS 6 PROBABILITY • Sample Space: (example using tree method) If you have a pair of blue pants, a pair of black pants, a white shirt and a red shirt, how many different ways can you wear one of your pants and one of your shirts? What are the events in this example? SCIENTIFIC INQUIRY AND ANALYSIS 7 PROBABILITY • Sample Space: (example using grid method) If you have a pair of blue pants, a pair of black pants, a white shirt and a blue shirt, how many different ways can you wear one of your pants and one of your shirts? White Shirt Blue Shirt Blue Pants White Shirt & Blue Pants Blue Shirt & Blue Pants Black Pants White Shirt & Black Pants Blue Shirt & Black Pants SCIENTIFIC INQUIRY AND ANALYSIS 8 PROBABILITY • Counting Example 1: – Kevin goes to a sandwich shop to order a sandwich. There is a choice of 5 different breads, 8 different meats, 5 different cheeses and 3 different condiments. • If Kevin chooses one of each, determine how many different sandwiches can be made through a tree diagram? SCIENTIFIC INQUIRY AND ANALYSIS 9 PROBABILITY AMERICAN CHEESE • Counting Example 1: MUSTARD MAYO OLIVE OIL SWISS CHEESE PROVOLONE CHEDDAR CHEESE BOILED HAM MUENSTER CHEESE HONEY HAM SALAMI PASTRAMI ROAST BEEF TURKEY CHICKEN WHOLE WHEAT BREAD BOLAGNA SCIENTIFIC INQUIRY AND ANALYSIS 10 PROBABILITY • Fundamental Counting Principle (Product Rule) – Instead of counting out every different occurrence through a tree diagram, the fundamental counting principle is used to determine the number of occurrences. – Principle: if one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is equal to m · n. (or m · n · o · p · q · r . . .) SCIENTIFIC INQUIRY AND ANALYSIS 11 PROBABILITY • Counting Example 1: – So, if we revisit this example, use the counting principle to answer the question. – Jonah goes to a sandwich shop to order a sandwich. There is a choice of 5 different breads, 8 different meats, 5 different cheeses and 3 different condiments. • If Jonah chooses one of each, how many different sandwiches can be made? SCIENTIFIC INQUIRY AND ANALYSIS 12 PROBABILITY • Fundamental Counting Principle without Repetition – Example: Conner goes to a sandwich shop the next day to order a sandwich. Again, there is a choice of 5 different breads, 8 different meats, 5 different cheeses and 3 different condiments. • Conner is feeling hungrier than Jonah. If Conner chooses three different meats and one each of everything else, how many different sandwiches can be made? So, for this problem, you can’t repeat the type of meat. SCIENTIFIC INQUIRY AND ANALYSIS 13 PROBABILITY AMERICAN CHEESE • Fundamental Counting Principle without Repetition WHOLE WHEAT BREAD BOILED HAM HONEY HAM SALAMI HONEY HAM SALAMI PASTRAMI SALAMI PASTRAMI ROAST BEEF PASTRAMI ROAST BEEF TURKEY ROAST BEEF TURKEY CHICKEN TURKEY CHICKEN BOLAGNA CHICKEN BOLAGNA MUSTARD MAYO OLIVE OIL SWISS CHEESE PROVOLONE CHEDDAR CHEESE MUENSTER CHEESE BOLAGNA SCIENTIFIC INQUIRY AND ANALYSIS 14 PROBABILITY • Fundamental Repetition Counting Principle without – If one event (choosing the first meat) can occur m ways, if that same event is done a second time (choosing the second meat) and if you don’t want the same outcome (the same meat twice, no repetitions), then the second event can occur m – 1 ways. Likewise, if the same event is done a third time (choosing a third meat) and if you don’t want the same outcome (no repetitions of the meat), then the third event can occur m – 2 ways. SCIENTIFIC INQUIRY AND ANALYSIS 15 PROBABILITY • Fundamental Repetition Counting Principle without – So, if we revisit this example, use the counting principle without repetitions to answer the question. – Connor goes to a sandwich shop the next day to order a sandwich. Again, there is a choice of 5 different breads, 8 different meats, 5 different cheeses and 3 different condiments. • Conner is feeling hungrier than Jonah. If Conner chooses three different meats and one each of everything else, how many different sandwiches can be made? So, for this problem, you can’t repeat the type of meat. SCIENTIFIC INQUIRY AND ANALYSIS 16 PROBABILITY • Fundamental Repetition Counting Principle with – Assuming the same example, Mike goes to a sandwich shop to order a sandwich and the choices are the same. • Again Mike is feeling hungrier than Jonah. If Mike chooses three meats, but for this example he can choose each meat more than once, how many different sandwiches can be made? SCIENTIFIC INQUIRY AND ANALYSIS 17 PROBABILITY • Counting Principle Example – In a high school, there are 273 freshmen, 291 sophomores, 252 juniors and 237 seniors. In how many different ways can a committee of 1 freshman, 1 sophomore, 1 junior and 1 senior be chosen? SCIENTIFIC INQUIRY AND ANALYSIS 18 PROBABILITY • Fundamental Counting Principle (Sum Rule) – Instead of counting out every different occurrence the fundamental counting principle is used to determine the number of occurrences. – Principle: if one event can be chosen from two disjoint groups, then the number of ways that the event can occur is equal to m + n. (or m + n + o + p + q + r . . .) SCIENTIFIC INQUIRY AND ANALYSIS 19 PROBABILITY • Fundamental Counting Principle (Sum Rule) – Example: Chris goes into a restaurant and can choose between 4 different chicken entrees or 3 different beef entrees or 5 different pasta entrees or 5 different seafood entrees. If he has one entrée, how many different entrees can he have? If he has two entrees, how many different ways can he order? SCIENTIFIC INQUIRY AND ANALYSIS 20 PROBABILITY • Factorials – Factorial (m!) is a mathematical operation that multiplies a number m by this method : 𝑚 × 𝑚 − 1 × (𝑚 − 2) ⋯ × 1 – Example: 5! = 5 × 4 × 3 × 2 × 1 = 120 – Note: 1! = 1 & 0! = 1 SCIENTIFIC INQUIRY AND ANALYSIS 21 PROBABILITY • Permutations – determines the number of ways n elements (objects) can be arranged in order r ways. • Permutation is calculated by figuring out the factorial. 𝑛! 𝑃 𝑛, 𝑟 = 𝑛𝑃𝑟 = 𝑛 − 𝑟 ! • P(n,r) is sometimes phrased: “Permutations of n taken r times” SCIENTIFIC INQUIRY AND ANALYSIS 22 PROBABILITY • Permutation properties 𝑊ℎ𝑒𝑛 𝑛 = 𝑟, 𝑡ℎ𝑒𝑛 𝑃 𝑛, 𝑟 = 𝑛𝑃𝑟 = 𝑛! 𝑊ℎ𝑒𝑛 𝑟 = 0, 𝑡ℎ𝑒𝑛 𝑃 𝑛, 0 = 𝑛𝑃0 = 1 SCIENTIFIC INQUIRY AND ANALYSIS 23 PROBABILITY • Permutation Example 1: – You are manager of a baseball team. You have 9 players on the team who can bat anywhere in the batting order. (i.e. 1st, 2nd, 3rd, 4th . . . 9th). How many different batting orders could you have? SCIENTIFIC INQUIRY AND ANALYSIS 24 PROBABILITY • Permutation Example 2: – You have 9 players that can bat in the first 3 positions of the batting order. How many different ways can the manager arrange the first 3 positions of the batting order? SCIENTIFIC INQUIRY AND ANALYSIS 25 PROBABILITY • Permutation Example 3: – Next year you are taking Calculus I, Intro to Engineering, Spanish II, Phys Ed, Physics I, American Literature, US History, Chemistry. Assume each class is offered during each of the 8 periods of the day. In how many different orders can you schedule your classes? SCIENTIFIC INQUIRY AND ANALYSIS 26 PROBABILITY • Permutations with Repetition: for the permutations taken so far, all of the objects were distinct. If they are not distinct, then the permutations are not indistinguishable. • For example: How many orders of three are there of the letters M – O – M? Which means how many permutations of the 3 letters take 3 ways: 3! 3! 3 × 2 × 1 𝑃 3,3 = 3𝑃3 = = = =6 3 − 3 ! 0! 1 • However some of the outcomes are not indistinguishable. MOM, OMM, MMO, MOM OMM, MMO • In this case only 3 outcomes are distinguishable. SCIENTIFIC INQUIRY AND ANALYSIS 27 PROBABILITY • Permutations with Repetition: the number of distinguishable permutations of n objects where one object is repeated q1 times, another is repeated q2 times, etc. 𝑛! 𝑞1 ! × 𝑞2 ! × ⋯ × 𝑞𝑘 ! SCIENTIFIC INQUIRY AND ANALYSIS 28 PROBABILITY • Permutation with Repetition Example 1: – Find the number of distinguishable permutations of the letters in REBECCA. SCIENTIFIC INQUIRY AND ANALYSIS 29 PROBABILITY • When not concerned with the order of an outcome, it is known at a combination. • Examples: – Being dealt 5 cards in a card game. Not concerned with the order. – Ordering an omelet with 1 cheese, 1 meat and 1 vegetable. SCIENTIFIC INQUIRY AND ANALYSIS 30 PROBABILITY • Combinations – how many different r objects out of n objects, no order. 𝑛! 𝐶 𝑛, 𝑟 = 𝑛𝐶𝑟 = 𝑛 − 𝑟 ! × 𝑟! • Given the same number of n objects and r objects, which is larger the combination or the permutation? SCIENTIFIC INQUIRY AND ANALYSIS 31 PROBABILITY • Combination Example 1: – You have 9 players that can bat in the first 3 positions of the batting order. How many different combinations of players can the manager utilize to fill in the first 3 positions of the batting order? – Remember: it is not important that the 3 players selected to bat in the 3 positions bat 1st, 2nd or 3rd. Compare this to permutation example 2. SCIENTIFIC INQUIRY AND ANALYSIS 32 PROBABILITY • Combination Example 2: – You are dealt 4 cards out of regular deck of 52. How many different combinations of the 4 cards out of the 52 can you get? SCIENTIFIC INQUIRY AND ANALYSIS 33 PROBABILITY • Multiple Combination Events – Sometimes there are multiple combination events which require you to calculate multiple times and then either multiply them or add them together. – For example you are dealt 7 cards out of 52. How many possible ways are there of getting 7 cards all of the same suit? OR – There are 5 field day events and you have to participate in at least 2 events. How many different combination of events can you participate? SCIENTIFIC INQUIRY AND ANALYSIS 34 PROBABILITY • Multiple Combination Events (Multiplication) – When finding the number of ways both one event (Event A) and another event (Event B) can occur, take the combination of both and multiply the two together. – Example: you are dealt 7 cards out of 52. How many are all of the same suit? Event A (chose 1 out 4 suits) Event B (chose 7 out of 13 cards in a suit) 4𝐶1 × 13𝐶7 SCIENTIFIC INQUIRY AND ANALYSIS 35 PROBABILITY • DO NOW Problem – Have out your homework. – Do the following problem: You are packing for a vacation. At home you have 10 shirts and 7 pairs of shorts. 1. 2. 3. In how many different ways can you choose 4 pairs of shorts to take on vacation? In how many different ways can you choose 2 shirts to wear on the 1st and 2nd days of vacation? If you bring 4 pairs of shorts and 6 shirts, how many different outfits can you make? SCIENTIFIC INQUIRY AND ANALYSIS 36 PROBABILITY • Probability – a number between 0 – 1 which indicates the likelihood the event will occur. – Experiment – a process by which an outcome is obtained. – Sample Space (S) – a set that is composed of a finite number of possible outcomes. – Event (E) – any subset of the sample space. SCIENTIFIC INQUIRY AND ANALYSIS 37 PROBABILITY • Probability – Probability of an outcome in E will occur is the ratio of the number of outcomes in E to the number of outcomes in Sample Space (S). OUTCOMES EVENT (E) SAMPLE SPACE (S) SCIENTIFIC INQUIRY AND ANALYSIS 38 PROBABILITY • Probability OUTCOMES EVENT (A) SAMPLE SPACE (S) 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝐴 𝑃 𝐴 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 SCIENTIFIC INQUIRY AND ANALYSIS 39 PROBABILITY • Example – Experiment – draw two cards out of a deck of 52 cards. What is the probability both will be a red face card? – Sample Space (S) = # of ways to select two cards out of deck of 52. – Event (E) = # of ways to get two red face cards. – Probability (P) = # 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑎 𝑟𝑒𝑑 𝑓𝑎𝑐𝑒 𝑐𝑎𝑟𝑑 # 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑠𝑒𝑙𝑒𝑐𝑡 2 𝑐𝑎𝑟𝑑𝑠 𝑜𝑢𝑡 𝑜𝑓 52 SCIENTIFIC INQUIRY AND ANALYSIS 40 PROBABILITY • Example using a Punnett Square SCIENTIFIC INQUIRY AND ANALYSIS 41 PROBABILITY • Example of Punnett Square – Experiment – genes that determine eye color are either Aa or Bb. Each parent contributes an A or a gene and a B or b gene. What is the probability that a baby whose parents are both AaBb will have green eyes? (3 lower case, 1 upper case) – Sample Space (S) = – Event (E) = – Probability (P) = SCIENTIFIC INQUIRY AND ANALYSIS 42 PROBABILITY • Combination/Probability Example: – The graph shows the % of 100 pizzas sold according to type. If 3 pizza were randomly chosen out of the 100, what is the probability that all 3 were thin crusts? – Sample Space (S) = combinations of 3 pizzas out of all 100 pizzas – Event (E) = combinations of 3 thin crust pizza – Probability (P) = Stuffed Crust Thick Crust 2% Pizzas Sold 23% SCIENTIFIC INQUIRY AND ANALYSIS Pan Pizza 22% Thin Crust 53% 43 PROBABILITY • Counting Principle Probability Example: – GP Clayton goes to a sandwich shop to order a sandwich. There is a choice of 5 different breads, 8 different meats, 5 different cheeses and 3 different condiments. • If GP Clayton chooses one of each, how many different sandwiches can be made? # of sandwiches = 5 x 8 x 5 x 3 = 600 SCIENTIFIC INQUIRY AND ANALYSIS 44 PROBABILITY • Counting Principle Probability Example (Continued): • Orest ran out of turkey. He is wondering what is the probability that GP Clayton will ask for a sandwich with turkey? – Sample Space (S) = the number of sandwiches that can be made. – Event (E) = the number of sandwiches with turkey – Probability (P) = SCIENTIFIC INQUIRY AND ANALYSIS 45 PROBABILITY • Counting Principle Probability Example (Continued): • Orest ran out of salami and Swiss cheese. He is wondering what is the probability that Kevin will ask for a sandwich with salami or Swiss? – Sample Space (S) = the number of sandwiches that can be made. – Event (E) = the number of sandwiches with salami or the number of sandwiches with Swiss – Probability (P) = SCIENTIFIC INQUIRY AND ANALYSIS 46 PROBABILITY • Geometric Probability: – Using a particular geometric shape to represent the event space and the whole geometric shape represents the sample space. – Example: given the blue square is 10’ x 10’ and the orange square is 7’ x 7’, what is the probability that a dart will fall in the orange square. SCIENTIFIC INQUIRY AND ANALYSIS 47 PROBABILITY • Empirical Probability - the ratio of the number of favorable outcomes to the total number of trials. • Law of Large Numbers – The theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed. SCIENTIFIC INQUIRY AND ANALYSIS 48 PROBABILITY • Example of Telephone Numbers – Experiment – What is the probability that given a telephone book page that a telephone number ends in 1?, 2?, 3?, 4?, 5?, 6?, 7?, 8?, 9?, 0? – Theoretical Probability (P) = – Sample Space (S) = SCIENTIFIC INQUIRY AND ANALYSIS 49 PROBABILITY • Example of Telephone Numbers EVENT NUMBER OF OUTCOMES FREQUENCY = # OF OUTCOMES/SAMPLE SPACE Telephone # ends in 1 Telephone # ends in 2 Telephone # ends in 3 Telephone # ends in 4 Telephone # ends in 5 Telephone # ends in 6 Telephone # ends in 7 Telephone # ends in 8 Telephone # ends in 9 Telephone # ends in 0 SCIENTIFIC INQUIRY AND ANALYSIS 50 PROBABILITY • DO NOW – As part of a monthly inspection at a hospital, the inspection team randomly selects reports from 8 of the 84 nurses who are on duty. What is the probability that none of the reports selected will be from the 10 most experienced nurses on duty? OR What is the probability that all of the reports selected will be from the 74 least experienced nurses on duty? SCIENTIFIC INQUIRY AND ANALYSIS 51 PROBABILITY • Unions & Intersections Review – A SET is a collection of elements. • For example Set A can be made up of the numbers {2, 3, 6, 8} Set B can be made up of the numbers {2, 6, 9, 12} – A UNION of sets A or B means that the elements belong to either SETS A or B. – An INTERSECTION of sets A and B means that the elements belong to SETS A and B ∪ 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑢𝑛𝑖𝑜𝑛 ∩ 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑓 𝐴 ∩ 𝐵 = 𝑒𝑚𝑝𝑡𝑦 𝑠𝑒𝑡, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 SCIENTIFIC INQUIRY AND ANALYSIS 52 PROBABILITY • Unions & Intersections Review 𝐴 ∪ 𝐵 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑢𝑛𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑡𝑠 𝐴 𝑜𝑟 𝐵 𝐴 ∩ 𝐵 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑡𝑠 𝐴 𝑎𝑛𝑑 𝐵 𝐼𝑓 𝐴 ∩ 𝐵 = 𝑒𝑚𝑝𝑡𝑦 𝑠𝑒𝑡, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 SCIENTIFIC INQUIRY AND ANALYSIS 53 PROBABILITY • Probabilities of Unions & Intersections – If A and B are events in the same sample space, then the probability of A or B occurring is: P(A U B) = P(A) + P(B) – P(A ∩ B) P(A U B) can also be written P(A or B) P(A ∩ B) can also be written P(A and B) SCIENTIFIC INQUIRY AND ANALYSIS 54 PROBABILITY • Probabilities of Unions & Intersections – If A and B are mutually exclusive events in the same sample space, then the probability of A or B occurring is: P(A U B) = P(A) + P(B) SCIENTIFIC INQUIRY AND ANALYSIS 55 PROBABILITY • Union Example 1: – One card is selected from a standard deck of 52 playing cards. What is the probability that the card is a heart or a face card? P(A or B) = P(A) + P(B) – P(A and B) SCIENTIFIC INQUIRY AND ANALYSIS 56 PROBABILITY • Union Example 2: – One card is selected from a standard deck of 52 playing cards. What is the probability that the card is a heart or a diamond card? P(A or B) = P(A) + P(B) – P(A and B) SCIENTIFIC INQUIRY AND ANALYSIS 57 PROBABILITY • Union Example 3: – In a poll of HS juniors, 6 out of 15 took a French class and 11 out 15 took a math class. Fourteen out of 15 students took French or Math. What is the probability that a student took both French and Math? P(A or B) = P(A) = P(B) = P(A and B) = SCIENTIFIC INQUIRY AND ANALYSIS 58 PROBABILITY • Complements A’ A – If A is an event in a sample space X, then the event consisting of all outcomes in X that are not in A is called the complement, A’: P(A’) = 1 - P(A) SCIENTIFIC INQUIRY AND ANALYSIS 59 PROBABILITY • Complement Example 1: – In throwing two 6 sided dice what is the probability that the sum is not 11? 𝑷 𝒔𝒖𝒎 𝒏𝒐𝒕 𝟏𝟏 = 𝟏 − 𝑷(𝒔𝒖𝒎 𝒊𝒔 𝟏𝟏) SCIENTIFIC INQUIRY AND ANALYSIS 60 PROBABILITY • Example 2: – The freezer case at a grocery store contains several frozen pies: 10 apple, 4 peach, 6 blueberry, 5 pumpkin, 3 peanut butter, and some pecan. You are shopping in a hurry and pick one of the pies from the freezer without looking at the type. The probability that the pie you picked is apple or pecan is 13/22. How many pecan pies are in the freezer case? SCIENTIFIC INQUIRY AND ANALYSIS 61 PROBABILITY • DO NOW – A deck of UNO cards is made up of 108 cards. (Twenty five each of red, yellow, blue and green, and 8 are wild cards). Each player is randomly dealt a 7 card hand. • What is the probability that a hand will have 2 wild cards? • What is the probability that a hand will have 2 wild cards, 2 red cards and 3 blue cards? • What is the probability that a hand will have at least 2 wild cards? SCIENTIFIC INQUIRY AND ANALYSIS 62 PROBABILITY • Independent Events – Two events are independent if the occurrence of one has no effect on the occurrence of the other. – Given A & B are independent events, the probability that A & B occur is: P(A ∩ B) = P(A) · P(B) Remember that P(A ∩ B) can be written P(A and B) SCIENTIFIC INQUIRY AND ANALYSIS 63 PROBABILITY • Independent Events – When calculating probability of independent events, the probability can be found on more than 2 independent events. – Given A & B & C are independent events, the probability that A & B & C occur is: P(A ∩ B ∩ C) = P(A) · P(B) · P(C) – Typically for independent events the sample space stays the same. SCIENTIFIC INQUIRY AND ANALYSIS 64 PROBABILITY • Independent Event Example 1: – In a survey, 9 out of 11 men and 4 out of 7 women said they were satisfied with a product. If the next 3 customers are 2 women and a man, what is the probability that they will all be satisfied? SCIENTIFIC INQUIRY AND ANALYSIS 65 PROBABILITY • Independent Event Example 2: – The following examples are better done using complements. – Given the previous survey of example 1, what is the probability that if men are the next 4 customers that at least 1 of them is not satisfied with the product? – Given the previous survey of example 1, what is the probability that if men are the next 4 customers that all of them are not satisfied with the product? SCIENTIFIC INQUIRY AND ANALYSIS 66 PROBABILITY • Dependent Events – Two events are dependent if the occurrence of one has an effect on the occurrence of the other. – Given A & B are dependent events, the probability that A & B occur is: P(A ∩ B) = P(A) · P(B|A) P(B|A) is known as a conditional probability that indicates the probability that B occurs given that A has occurred. SCIENTIFIC INQUIRY AND ANALYSIS 67 PROBABILITY • Dependent Events – When calculating probability of dependent events, the probability can be dependent on more than 2 events. – Given A & B & C are dependent events, the probability that A & B & C occur is: P(A ∩ B ∩ C) = P(A) · P(B|A) · P(C|A & B) SCIENTIFIC INQUIRY AND ANALYSIS 68 PROBABILITY • Dependent Event Example 1: – Three children have a choice of 12 summer camps that they can attend. If they each randomly choose which camp to attend what is the probability that they all attended different camps? SCIENTIFIC INQUIRY AND ANALYSIS 69 PROBABILITY • Independent vs. Dependent Events – A box contains the numbers 1 – 20. What is the probability of selecting 1 number less than 6, replacing the number and selecting a second number less than 6? – Is this process dependent or independent? – How can the above event be changed to the opposite type of event? SCIENTIFIC INQUIRY AND ANALYSIS 70 PROBABILITY • Example: – There are 25 pieces of paper, numbered 1 to 25, in a hat. You pick a piece of paper, replace it and then pick another piece of paper. What is the probability that each number is greater than 20 or less than 4? SCIENTIFIC INQUIRY AND ANALYSIS 71 PROBABILITY • Complements (Review) – When finding the probability of something occurring, it is sometimes easier to find the probability when something won’t occur (its complement) and then use the formula below. P(A’) = 1 - P(A) SCIENTIFIC INQUIRY AND ANALYSIS 72 PROBABILITY • Complement Example: – Suppose that 1% of the lights in a shipment of 10,000 lights are defective. If you make 100 random tests of the lights (each time choosing, testing and replacing the light), what is the probability that at least one of the tests reveals a defective light. – P(A’) = 9900 / 10,000 = 0.99 (selecting 1 good light) – P(A’)100 = 0.99100 ≈ 0.366 (selecting 100 good lights, 100 times) – P(A) = 1 - 0.99100 ≈ 0.634 (at least 1 bad after selecting 100) SCIENTIFIC INQUIRY AND ANALYSIS 73 PROBABILITY • DO NOW: In one town 95% of the students graduate from high school. Suppose a study showed that at age 25, 81% of the high school graduates held full-time jobs while only 63% of those who did not graduate held full-time jobs. What is the probability that a randomly selected student from the town will have a full time job at age 25? SCIENTIFIC INQUIRY AND ANALYSIS 74 HOMEWORK UNIT 4-1 ANSWERS • Homework Answers: – Pg. 705, 15 - 22. 15. 3 ways 16. 15 ways 17. 40 ways 18. 1512 ways 19. 17,576,000; 11,232,000 20. 45,697,600; 32,292,000 21. 6,760,000; 3,276,000 22. 118,813,760; 78,936,000 55. 480 56. 2772 57. 2,176,782,336; 1,402,410,240 SCIENTIFIC INQUIRY AND ANALYSIS 75 HOMEWORK UNIT 4-2 ANSWERS Pg. 705 23. 40,320 24. 120 25. 3,628,800 26. 362,880 31. 6 32. 20 33. 2 34. 5040 Pg. 706 60. 3,628,800 61.a. 720 b. 60,480 62. 210 63.12,612,600 Pg. 712 22. 495 23. 1 24. 3003 SCIENTIFIC INQUIRY AND ANALYSIS 25. 165 26. 792 27. 48 28. 778,320 Pg. 713 48. 1365 49. 315 52. 21,700 56. 210 76 HOMEWORK UNIT 4-3 ANSWERS Pg. 719 21. 0.5 22. 0.9231 23. 0.23 Pg. 720-721 30. 0.0218 31. 0.455 32. 0.477 33. 0.545 34. 0.262 35. 0.0385 36. 0.154 39. 5.6 x 10-8 40. 0.001 43. 0.0527 SCIENTIFIC INQUIRY AND ANALYSIS 77 HOMEWORK UNIT 4-4 ANSWERS Pg. 719-721 24. Exp = 21.7% Th = 16.7% 25. Exp = 30.8% Th = 33.3% 26. Exp = 49.2% Th = 50% 27. Exp = 50.8% Th = 50% 37. 26.2% 38. 1.285 x 10-10 41. a. P(tv) = 55.5% b. P(vid) = 3.8% 42. a. P(loss) = 23.5% b. P(gains) = 64.7% 46. 4.0% SCIENTIFIC INQUIRY AND ANALYSIS 78 PROBABILITY DO NOW: The Supreme Court of the US has 3 women and 6 men justices. On a certain case the justices voted 5 to 4 in favor of the defendant. 1. What is the probability that exactly 2 out of 3 women voted in favor of the defendant? 2. What is the probability that at least 2 out of 3 women voted in favor of the defendant? SCIENTIFIC INQUIRY AND ANALYSIS 79 HOMEWORK UNIT 4-5 ANSWERS Pg. 727 4. 0.5 6. 0.5 8. 0.6 10. 0.1 12. 0.5 14. 0.67 16. 0.25 17. 0.7 18. 0.45 (yes) 25. 0.66 27. 0.25 Pg. 728 42. 0.83 43. 0.75 46. 0.1 47. 0.1 SCIENTIFIC INQUIRY AND ANALYSIS 80 PROBABILITY DO NOW: Allergic reactions to poison ivy can be miserable. Plant oils cause the reaction. A study was performed to see the effects of washing the oil off within 5 minutes of exposure. A random sample of 1000 people with known allergies to poison ivy participated in the study. Oil from the poison ivy plant was rubbed on a patch of skin. For 500 of the subjects, the oil was washed off within 5 minutes. For the other 500 subjects, the oil was washed off after 5 minutes. The results are in the table: SCIENTIFIC INQUIRY AND ANALYSIS 81 HOMEWORK UNIT 4-6 ANSWERS Pg. 734 & 735 12. 0.0468 13. 0.0468 18.a. 0.0625 b. 0.0637 19.a. 0.0059 b. 0.0060 25. 0.00144 26. 0.606 33. 0.581 SCIENTIFIC INQUIRY AND ANALYSIS 82 PROBABILITY • DO NOW Problem – Have out your homework. – Do the following problem: Radio station call letters (i.e. WNBC) consists of four letters beginning with either K or W. How many different radio station call letters are possible if letters can be repeated? How many different radio station call letters are possible if letters cannot be repeated? SCIENTIFIC INQUIRY AND ANALYSIS 83 PROBABILITY Evaluate the following: a) b) c) d) e) f) 4! = 12! = 3P 3 = 5P 1 = 10P1 + 5P2 = 4P 3 x 3P 2 = 1. You are taking a Chemistry test and are given a list of 10 elements. You are to arrange the ten elements in order as they appear in the periodic table of elements. Suppose you have no idea of the correct order and simply guess. How many different ways can the ten elements be listed? 2. You are taking a Chemistry test and are given a list of 10 elements. You are told three of those elements are elements 5, 6 & 7 on the periodic table of elements. You are asked to pick elements 5, 6 & 7 and to arrange the three elements in order as they appear in the periodic table of elements. Suppose you have no idea of the correct elements or the correct order and simply guess. How many different ways can the three elements be listed? Suppose you know the three elements but you have no idea of the correct order and simply guess. How many SCIENTIFIC INQUIRY AND ANALYSIS different ways can the three elements be listed? 3. A student council is made up of 2 Freshmen, 2 Sophomores, 2 Juniors and 3 Seniors. There are 75 Freshman, 68 Sophomores, 71 Juniors and 82 Seniors. Selections for each class are done randomly where the first pick of the class is the active council member and subsequent picks are the alternates. The senior class has two alternates and all of the other classes have only one alternate. How many different student councils can be formed? 84 PROBABILITY 1. Evaluate the following: a) b) c) d) e) f) 2. 3. class were chosen, how many ways can they be selected? 3C3 = C 1000 25 = 4C3 x 6C2 = 5C3 x 2C2 = 5C1 + 8C2 = 5P1 + 8P2 = Junior Students 12 36 • 8 Seven cards are dealt from a regular 52 card deck. How many different ways can the seven cards be dealt? 14 To the right is an indication of how many students from the junior class are taking AP classes. Three students are4. picked from the junior class (order doesn’t matter). How many ways can those 3 students be chosen from the whole junior class? If only those students who took just 1 AP class were chosen, how many ways can they be selected? If only those students who took at least 1 AP Took 1 AP Class Took 2 AP Classes Took 3 AP Classes Took no AP Classes & 2nd places), but it does not matter in what order they finish. How many quinella bets are possible in this race? EXTRA: If you make 4 exacta bets in the race, what is the probability you select the correct one? If you make 4 quinella bets in the race, what is the probability you select the correct one? There are 10 horses in a race. One of the bets in a horse race is the “exacta” in which you select the 2 horses for “win and place” (1st & 2nd places) in that exact order. How many exacta bets are possible in this race? The “quinella” bet is similar to the “exacta” in which you select the 2 horses for “win and place” (1st SCIENTIFIC INQUIRY AND ANALYSIS 85