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Thinking Mathematically Chapter 11: Counting Methods and Probability Theory Thinking Mathematically Section 1: The Fundamental Counting Principle The Fundamental Counting Principle If you can choose one item from a group of M items and a second item from a group of N items, then the total number of two-item choices is M N. You MULTIPLY the numbers! The Fundamental Counting Principle At breakfast, you can have eggs, pancakes or cereal. You get a free juice with your meal: either OJ or apple juice. How many different breakfasts are possible? eggs pancakes cereal OJ apple OJ apple OJ apple 1 2 3 4 5 6 Example: Applying the Fundamental Counting Principle • The Greasy Spoon Restaurant offers 6 appetizers and 14 main courses. How many different meals can be created by selecting one appetizer and one main course? • Using the fundamental counting principle, there are 14 6 = 84 different ways a person can order a two-course meal. Example: Applying the Fundamental Counting Principle • This is the semester that you decide to take your required psychology and social science courses. • Because you decide to register early, there are 15 sections of psychology from which you can choose. Furthermore, there are 9 sections of social science that are available at times that do not conflict with those for psychology. In how many ways can you create two-course schedules that satisfy the psychology-social science requirement? Solution The number of ways that you can satisfy the requirement is found by multiplying the number of choices for each course. You can choose your psychology course from 15 sections and your social science course from 9 sections. For both courses you have: 15 9, or 135 choices. The Fundamental Counting Principle The number of ways a series of successive things can occur is found by multiplying the number of ways in which each thing can occur. Example: Options in Planning a Course Schedule Next semester you are planning to take three courses - math, English, and humanities. Based on time blocks and highly recommended professors, there are 8 sections of math, 5 of English, and 4 of humanities that you find suitable. Assuming no scheduling conflicts, there are: 8 5 4 = 160 different three course schedules. Example Car manufacturers are now experimenting with lightweight three-wheeled cars, designed for a driver and one passenger, and considered ideal for city driving. Suppose you could order such a car with a choice of 9 possible colors, with or without air-conditioning, with or without a removable roof, and with or without an onboard computer. In how many ways can this car be ordered in terms of options? Solution This situation involves making choices with four groups of items. color - air-conditioning - removable roof - computer 9 2 2 2 = 72 Thus the car can be ordered in 72 different ways. Example: A Multiple Choice Test You are taking a multiple-choice test that has ten questions. Each of the questions has four choices, with one correct choice per question. If you select one of these options per question and leave nothing blank, in how many ways can you answer the questions? Solution We DON’T blindly multiply the first two numbers we see. The answer is not 10 4 = 40. We use the Fundamental Counting Principle to determine the number of ways you can answer the test. Multiply the number of choices, 4, for each of the ten questions 4444444444 =1,048,576 Example: Telephone Numbers in the United States Telephone numbers in the United States begin with three-digit area codes followed by seven-digit local telephone numbers. Area codes and local telephone numbers cannot begin with 0 or 1. How many different telephone numbers are possible? Solution We use the Fundamental Counting Principle to determine the number of different telephone numbers that are possible. 8 10 10 8 10 10 10 10 10 10 =6,400,000,000 Thinking Mathematically Section 2: Permutations Permutations • A permutation is an arrangement of objects. – No item is used more than once. – The order of arrangement makes a difference. Example: Counting Permutations Based on their long-standing contribution to rock music, you decide that the Rolling Stones should be the last group to perform at the four-group Offspring, Pink Floyd, Sublime, Rolling Stones concert. Given this decision, in how many ways can you put together the concert? Solution We use the Fundamental Counting Principle to find the number of ways you can put together the concert. Multiply the choices: 3211=6 3 choices 2 choices 1 choice 1 choice offspring whichever only one pink floyd of the two remaining stones sublime remaining Thus, there are six different ways to arrange the concert if the Rolling Stones are the final group to perform. Example: Counting Permutations You need to arrange seven of your favorite books along a small shelf. How many different ways can you arrange the books, assuming that the order of the books makes a difference to you? Solution You may choose any of the seven books for the first position on the shelf. This leaves six choices for second position. After the first two positions are filled, there are five books to choose from for the third position, four choices left for the fourth position, three choices left for the fifth position, then two choices for the sixth position, and only one choice left for the last position. 7 6 5 4 3 2 1 = 5040 There are 5040 different possible permutations. Factorial Notation If n is a positive integer, the notation n! is the product of all positive integers from n down through 1. n! = n(n-1)(n-2)…(3)(2)(1) note that 0!, by definition, is 1. 0!=1 Permutations of n Things Taken r at a Time The number of permutations possible if r items are taken from n items: n! nPr = (n – r)! = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) n! = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) (n - r) (n - r - 1) . . . (2)(1) (n – r)! = (n - r) (n - r - 1) . . . (2)(1) Permutations of n Things Taken r at a Time The number of permutations possible if r items are taken from n items: nPr: starting at n, write down r numbers going down by one: nPr = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) 1 2 3 4 r Problem A math club has eight members, and it must choose 5 officers --- president, vice-president, secretary, treasurer and student government representative. Assuming that each office is to be held by one person and no person can hold more than one office, in how many ways can those five positions be filled? We are arranging 5 out of 8 people into the five distinct offices. Any of the eight can be president. Once selected, any of the remaining seven can be vice-president. Clearly this is an arrangement, or permutation, problem. 8P5 = 8!/(8-5)! = 8!/3! = 8 · 7 · 6 · 5 · 4 = 6720 Permutations with duplicates. • In how many ways can you arrange the letters of the word minty? • That's 5 letters that have to be arranged, so the answer is 5P5 = 5! = 120 • But how many ways can you arrange the letters of the word messes? • You would think 6!, but you'd be wrong! messes here are six permutations of messes me s s e s 1 well, all 3! arrangements of the s's me s s e s 2 look the same to me!!!! me s s e s This is true for any arrangement 3 of the six letters in messes, so me s s e s 4 every six permutations should count only once. me s s e s 5 The same applies for the 2! arrangement of the e's 6 Permutations with duplicates. • How many ways can you arrange the letters of the word messes? • The problem is that there are three s's and 2 e's. It doesn't matter in which order the s's are placed, because they all look the same! • This is called permutations with duplicates. Permutations with duplicates. • Since there are 3! = 6 ways to arrange the s's, there are 6 permutations that should count as one. Same with the e's. There are 2! = 2 permutations of them that should count as 1. • So we divide 6! by 3! and also by 2! • There are 6!/3!2! = 720/12 = 60 ways to arrange the word messes. Permutations with duplicates. • In general if we want to arrange n items, of which m1, m2, .... are identical, the number of permutations is n! m1!m2!m3! Problem A signal can be formed by running different colored flags up a pole, one above the other. Find the number of different signals consisting of 6 flags that can be made if 3 of the flags are white, 2 are red, and 1 is blue 6!/3!2!1! = 720/(6)(2)(1) = 720/12 = 60 Thinking Mathematically Section 3: Combinations Combination: definition A combination of items occurs when: • The item are selected from the same group. • No item is used more than once. • The order of the items makes no difference. How to know when the problem is a permutation problem or a combination problem • Permutation: – arrangement, arrange – order matters • Combination – selection, select – order does not matter. Example: Distinguishing between Permutations and Combinations • For each of the following problems, explain if the problem is one involving permutations or combinations. • Six students are running for student government president, vice-president, and treasurer. The student with the greatest number of votes becomes the president, the second biggest vote-getter becomes vice-president, and the student who gets the third largest number of votes will be student government treasurer. How many different outcomes are possible for these three positions? Solution • Students are choosing three student government officers from six candidates. The order in which the officers are chosen makes a difference because each of the offices (president, vice-president, treasurer) is different. Order matters. This is a problem involving permutations. Example: Distinguishing between Permutations and Combinations • Six people are on the volunteer board of supervisors for your neighborhood park. A three-person committee is needed to study the possibility of expanding the park. How many different committees could be formed from the six people on the board of supervisors? Solution • A three-person committee is to be formed from the six-person board of supervisors. The order in which the three people are selected does not matter because they are not filling different roles on the committee. Because order makes no difference, this is a problem involving combinations. Example: Distinguishing between Permutations and Combinations • Baskin-Robbins offers 31 different flavors of ice cream. One of their items is a bowl consisting of three scoops of ice cream, each a different flavor. How many such bowls are possible? Solution • A three-scoop bowl of three different flavors is to be formed from Baskin-Robbin’s 31 flavors. The order in which the three scoops of ice cream are put into the bowl is irrelevant. A bowl with chocolate, vanilla, and strawberry is exactly the same as a bowl with vanilla, strawberry, and chocolate. Different orderings do not change things, and so this problem is combinations. Combinations of n Things Taken r at a Time n n! = nCr = r r!(n – r)! Note that the sum of the two numbers on the bottom (denominator) should add up to the number on the top (numerator). Computing Combinations • Suppose we need to compute 9C3 9! 9! 9 C3 9 3!(9 3)! 3!6! 3!6! • r = 3, n – r = 6 • The denominator is the factorial of smaller of the two: 3! Computing Combinations • Suppose we need to compute 9C3 9! 9! 9 C3 9 3!(9 3)! 3!6! 3!6! • r = 3, n – r = 6 • In the numerator write (the product of) all the numbers from 9 down to n - r + 1 = 6 + 1 = 7: • There should be the same number of terms in the numerator and denominator: 9 8 7 Computing Combinations • If called upon, there's a fairly easy way to compute combinations. – Given nCr , decide which is bigger: r or n – r. – Take the smaller of the two and write out the factorial (of the number you picked) as a product. – Draw a line over the expression you just wrote. Computing Combinations • If called upon, there's a fairly easy way to compute combinations. – Now, put n directly above the line and directly above the leftmost number below. – Eliminate common factors in the numerator and denominator. – Do the remaining multiplications. – You're done! Computing Combinations • Suppose we need to compute 9C3 . – n – r = 6, and the smaller of 3 and 6 is 3. 3 4 987 = 3 4 7 = 84 321 1 1 Finding Probabilities from Odds • If the odds in favor of an event E are a to b, then the probability of the event is given by P( E ) a ab Finding Probabilities from Odds • If the odds against an event E are a to b, then the probability of the event is given by P( E ) b ab Finding Probabilities from Odds • Example: – Suppose Bluebell is listed as 7:1 in the third race at the Meadowlands. – The odds listed on a horse are odds against that horse winning, that is, losing. – The probability of him losing is 7 / (7+1) = 7/8. – The probability of him winning is 1/8. Finding Probabilities from Odds • Example: – Suppose Bluebell is listed as 7:1 in the third race at the Meadowlands. (a:b against) – The odds listed on a horse are odds against that horse winning, that is, losing. – The probability of him losing is a 7 / (7+1) = 7/8. a b b – The probability of him winning is 1/8. a b ab Thinking Mathematically Section 7: Events Involving And; Conditional Probability Independent Events • Two events are independent events if the occurrence of either of them has no effect on the probability of the other. • For example, if you roll a pair of dice two times, then the two events are independent. What gets rolled on the second throw is not affected by what happened on the first throw. And Probabilities with Independent Events • If A and B are independent events, then P(A and B) = P(A) P(B) • The example of choosing from four pairs of socks and then choosing from three pairs of shoes (= 12 possible combinations) is an example of two independent events. Dependent Events • Two events are dependent events if the occurrence of one of them does have an effect on the probability of the other. • Selecting two Kings from a deck of cards by selecting one card, putting it aside, and then selecting a second card, is an example of two dependent events. • The probability of picking a King on the second selection changes because the deck now contains only 51, not 52, cards. And Probabilities with Dependent Events • If A and B are dependent events, then • P(A and B) = P(A) P(B given that A has occurred) • written as P(A) P(B|A) Conditional Probability • The conditional probability of B, given A, written P(B|A), is the probability that event B will occur computed on the assumption that event A has occurred. • Notice that when the two events are independent, P(B|A) = P(B). Conditional Probability • Example: – Suppose you are picking two cards from a deck of cards. What is the probability you will pick a King and then another face card? – The probability of an King is 4 = 1 . 52 13 – Once the King is selected, there are 11 face cards left in a deck holding 51 cards. – P(A) = 1 . P(B|A) = 11 13 51 – The probability in question is 1 11 13 51 Applying Conditional Probability to Real-World Data P(B|A) = observed number of times B and A occur together observed number of times A occurs Review P(not E) 1 – P(E) P(A or B) P(A) + P(B) mutually – P(A and B) exclusive: P(A) + P(B) P(A and B) P(A) P(B|A) independent: P(A) P(B) Odds in favor - P(E) / P(not E) probability is a:b a/(a+b) Odds against - P(not E) / P(E) probability is a:b b/(a+b) Thinking Mathematically Section 8: Expected Value Expected Value • Expected value is a mathematical way to use probabilities to determine what to expect in various situations over the long run. • For example, we can use expected value to find the outcomes of the roll of a fair dice. • The outcomes are 1, 2, 3, 4, 5, and 6, each with a probability of 1 . The expected value, E, is computed 6 by multiplying each outcome by its probability and then adding these products. • E = 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 1 6 6 6 6 6 6 = (1+2+3+4+5+6)/6 = 21 = 3.5 6 Expected Value E = 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 1 6 6 6 6 6 6 = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21 = 3.5 6 Of course, you can't roll a 3½ . But the average value of a roll of a die over a long period of time will be around 3½. Example Expected Value and Roulette A roulette wheel has 38 different "numbers." • One way to bet in roulette is to place $1 on a single number. • If the ball lands on that number, you are awarded $35 and get to keep the $1 that you paid to play the game. • If the ball lands on any one of the other 37 slots, you are awarded nothing and the $1 you bet is collected. Example Expected Value and Roulette • 38 different numbers. • If the ball lands on your number, you win awarded $35 and you keep the $1 you paid to play the game. • If the ball lands on any of the other 37 slots, you are awarded nothing and you lose the $1 you bet. • Find the expected value for playing roulette if you bet $1 on number 11 every time. Describe what this means. Solution Outcome Gain/Loss Probability 1 11 $35 38 $1 37 Not 11 38 E = $35( 1 ) + (-$1)( 37) 38 38 35 37 2 = $ - $ = -$ ≈ -$0.05 38 38 38 This means that in the long run, a player can expect to lose about 5 cents for each game played. Expected Value • A real estate agent is selling a house. She gets a 4- month listing. There are 3 possibilities: – she sells the house: (30% chance) earns $25,000 – another agent sells the house: (20% chance) earns $10,000 – house not sold: (50% chance) loses $5,000 • What is the expected profit (or loss)? • If the expected profit is at least $6000 she would consider it a good deal. Expected Value Outcome Probability Profit or product loss she sells 0.3 +$25,000 +$7,500 other sells 0.2 +$10,000 +$2,000 doesn't sell 0.5 -$5,000 -$2,500 +$7,000 The realtor can expect to make $7,000. Make the deal!!!!

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