10.1. Randomness and Probability. Investigation: Flip a Coin EXAMPLE A CONDENSED LESSON

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1 CONDENSED LESSON 10.1 Randomness and Probability In this lesson you will simulate random processes find experimental probabilities based on the results of a large number of trials calculate theoretical probabilities by counting outcomes and by using an area model Rolling a die, flipping a coin, and drawing a card are examples of random processes. In a random process, no individual outcome is predictable even though the long-range pattern of many individual outcomes often is predictable. Investigation: Flip a Coin Read the investigation in your book, and then complete Steps 1. If possible, get your class s results for Steps 4 and 5 from another student, and examine them carefully. If you don t have access to your class s results, flip your coin to generate at least five more 10-flip sequences before completing Step 6. You can use a random process, such as rolling a die, to generate random numbers. Over the long run, each number is equally likely to occur and there is no pattern in any sequence of random numbers. You can use your calculator to generate lots of random numbers quickly. (See Calculator Note 1L to learn how to generate random numbers.) Example A in your book shows you how to use your calculator s random-number generator to simulate rolling two dice. The simulated rolls are then used to find the probability of rolling a sum of 6. Read that example carefully. Then, read the example below. EXAMPLE A Use a calculator s random-number generator to find the probability of spinning an odd product on these two (fair) spinners Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

2 Lesson 10.1 Randomness and Probability Solution Simulate 00 spins of the first spinner by generating 00 random integers from 1 to 4 and storing them in a column or list. Simulate 00 spins of the second spinner by generating 00 random integers from 1 to and storing them in a second list. Define a third list as the product of the first two lists. The histogram shows the 00 products. The possible odd products are 1,, and 9. Adding the bin heights for these products gives , or 99. So, out of 00 simulated spins, 99 have an odd product. Therefore, P (odd product) 99, or The results of a random process are called outcomes. An event consists of one or more outcomes. A simple event consists of only one outcome. Events that aren t simple are compound. The probability of an event is always between 0 and 1. The probability of an event that is certain to happen is 1, and the probability of an event that is impossible is 0. Probabilities that are based on trials and observations, such as the probabilities found in Example A in your book and in Example A above, are called experimental probabilities. Generally, the more trials an experimental probability is based on, the better it will predict behavior. Sometimes it is possible to find the theoretical probability of an event, without an experiment. To find the theoretical probability, you count the number of ways a desired event can happen and divide by the total number of equally likely outcomes. (Outcomes that are equally likely have the same chance of occurring.) The box on page 551 of your book gives the formulas for calculating experimental and theoretical probabilities. Read this box, the text immediately after it, and Example B, which shows how to find the theoretical probability of rolling a sum of 6 with a pair of dice. Then, read the example below. EXAMPLE B Solution Find the theoretical probability of spinning an odd product on the spinners pictured in Example A. The possible equally likely outcomes when you spin the spinners are represented by the 1 grid points in the diagram at right. A B The four possible outcomes with an odd product are labeled 1 C D A D. Point B, for example, represents an outcome of on the first spinner and on the second, for a product of First spinner The theoretical probability is the number of ways the event (in this case, an odd product) can occur, divided by the number of equally likely outcomes. So, P (odd product) , or %. This is a little more than the experimental probability found in Example A. Note that although the experimental probability can vary, the theoretical probability is a fixed value. Second spinner In Example C in your book, the outcomes are not limited to whole numbers. In this case, you can t simply count the possible outcomes. Instead, the solution uses an area model. Read that example carefully. A probability found by calculating a ratio of lengths or areas is called a geometric probability. 144 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

3 CONDENSED LESSON 10. Counting Outcomes and Tree Diagrams In this lesson you will use tree diagrams to count outcomes and to find probabilities compute probabilities of independent events and of dependent events learn the multiplication rule for finding the probability of a sequence of events When determining the theoretical probability of an event, it can be difficult to count outcomes. Example A in your book illustrates how making a tree diagram can help you organize information. Read that example, and make sure you understand it. Any path from left to right through a tree diagram is an outcome, or simple event. You can calculate probabilities for each stage of the process as well as for each complete path. Investigation: The Multiplication Rule Work through the investigation in your book, and then compare your results with those below. Step 1 Below is the tree diagram from Example A, part a, labeled with probabilities. The probability of each branch (getting a particular toy in a box) is 0.5. Because there are four equally likely paths, the probability of any one path is 4, or 0.5. Notice that the probability of a path is the product of the probabilities of its branches. 1st Box 0.5 Toy 1 nd Box 0.5 Toy 1 Toy 0.5 a b P (Toy 1 followed by Toy 1) 0.5 P (Toy 1 followed by Toy ) 0.5 Toy Toy 1 Toy 0.5 c d P (Toy followed by Toy 1) 0.5 P (Toy followed by Toy ) 0.5 Step The tree diagram from Example A, part b, labeled with probabilities, is shown on the next page. Note again that the probability of each path is the product of the probabilities of its branches. The sum of the probabilities of all the paths is 1. The sum of the probabilities of the six paths that contain all three toys is 6 7, or about 0.. Discovering Advanced Algebra Condensed Lessons CHAPTER

4 Lesson 10. Counting Outcomes and Tree Diagrams Toy 1 Toy Toy 1 Toy Toy 1 Toy Toy Toy 1 Toy Toy Toy Toy 1 P (Toy 1, Toy, Toy ) P (Toy 1, Toy, Toy ) P (Toy, Toy 1, Toy ) P (Toy, Toy, Toy 1) P (Toy, Toy 1, Toy ) P (Toy, Toy, Toy 1) Step a. If there were four different toys, equally distributed among the boxes, then the probability of any particular toy in a particular box would be 0.5. b. The probability that Talya finds a particular toy in one box does not influence the probability that she will find a particular toy in the next box. c. There are 4 4, or 56 different equally likely outcomes. The outcome (Toy, followed by Toy, followed by Toy 4, followed by Toy 1) is one of these outcomes, so its probability is 1 56, or about You can also figure this out by realizing that a complete tree diagram would have 56 paths and the probability along each branch is 0.5. The specified outcome is one path with four branches, so its probability is (0.5)(0.5)(0.5)(0.5), or about Step 4 To find the probability of a path, multiply the probabilities of its branches. 146 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

5 Lesson 10. Counting Outcomes and Tree Diagrams Step 5 There are 4 paths that include all four toys. (For a sequence to have four different toys, there are four possibilities for the first toy, then there are three possibilities for the second toy, then two possibilities for the third toy, then one possibility for the fourth. There are 4 1 such sequences.) So the probability of getting a complete set is 4, or about In Example B in your book, a tree diagram showing every possible outcome would be too much to draw. That example illustrates how, in such cases, you can often make a tree diagram with branches of different possibilities. Read that example carefully. Below is another example. EXAMPLE A Solution Zak is working as a disk jockey at a party. To start the party, he puts three CDs in his CD player. Disk 1 has 8 rock songs and 4 hip-hop songs. Disk has 5 rock songs and 5 hip-hop songs. Disk has rock songs and 1 hip-hop songs. If Zak randomly selects one song from Disk 1, then one from Disk, and then one from Disk, what is the probability he will play exactly two hip-hop songs? There are two branches for each CD, one for rock and one for hip-hop. Each branch is labeled with its probability. The highlighted paths include exactly two hip-hop songs. 1st CD nd CD rd CD R H _ R _ H R H _ R _ 5 H 4 5 R _ 5 H 4 5 R _ 5 H 4 5 R _ 5 H 4 5 Find the probability of each path by multiplying the probabilities of its branches. _ _ Now, add the probabilities of the three paths. P (exactly hip-hop songs) Discovering Advanced Algebra Condensed Lessons CHAPTER

6 Lesson 10. Counting Outcomes and Tree Diagrams In the preceding example, the probability that Zak plays a rock song from Disk is the same regardless of whether or not he plays a rock song from Disk 1. These events are called independent, meaning that the occurrence of one event has no influence on the probability of the other event. To find the probability of a sequence of independent events, you simply multiply the probabilities of the events. This is summarized in the box on page 56 of your book. In Example C in your book, the events are not independent. Read Example C carefully, and make sure you understand it. Here is a similar example. EXAMPLE B Solution In Example A, suppose the host of the party tells Zak not to play two rock songs in a row. What is the probability Zak will play a rock song from Disk? If Zak plays a rock song from Disk 1, then P (rock song from Disk ) 0 because he cannot play two rock songs in a row. If he plays a hip-hop song from Disk 1, then P (rock song from Disk ). There is a probability he will play a hip-hop song from Disk 1, so the probability he plays a rock song from Disk is, or 6. When the probability of an event depends on the occurrence of another event, the events are dependent. Independent and dependent events can be described using conditional probability. If event A depends on event B, then the probability of A occurring given that B occurred is different from the probability of A by itself. The probability of A given B is written P (A B). If A and B are dependent, then P (A B) P (A). If A and B are independent, then P (A B) P (A). In Example A, the events (rock on Disk ) and (hip-hop on Disk 1) are independent, so P (rock on Disk hip-hop on Disk 1) P (rock on Disk ) In Example B, the events (rock on Disk ) and (hip-hop on Disk 1) are dependent, so P (rock on Disk hip-hop on Disk 1) P (rock on Disk ) Page 564 of your book explains how you can use tree diagrams to find conditional probabilities. Read that text carefully, and then read the multiplication rule. 148 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

7 CONDENSED LESSON 10. Mutually Exclusive Events and Venn Diagrams In this lesson you will learn the addition rule for mutually exclusive events discover the general addition rule use a Venn diagram to break non-mutually exclusive events into mutually exclusive events Two events that cannot both happen are mutually exclusive. For example, passing the history midterm and failing the history midterm are mutually exclusive because you can t do both. In Lesson 10., you saw how a tree diagram allows you to break down a sequence of dependent events into a sequence of independent events. Similarly, you can use a Venn diagram to break down non-mutually exclusive events into mutually exclusive events. Here is an example. EXAMPLE Audrey asked a sample of students in her school if they like football and if they like golf. These events are not mutually exclusive because it is possible to like both sports. Based on her results, Audrey made this Venn diagram of probabilities: Likes football Likes golf a. What is the meaning of the region labeled 0.0? Of the region labeled 0.15? b. What is the probability that a randomly selected student likes golf? c. What is the probability that a randomly selected student likes football but does not like golf? Solution a. The region labeled 0.0 represents the probability that a student does not like football and does not like golf. The region labeled 0.15 represents the probability that a student likes both sports. b. Add the probabilities inside the Likes golf circle: c. This probability is represented by the region inside the Likes football circle but outside the Likes golf circle. The probability is Example A in your book gives a slightly more complicated example involving three events. Read that example carefully. Then, read the addition rule for mutually exclusive events on page 57. In the investigation you ll discover how the rule can be generalized to events that may not be mutually exclusive. Discovering Advanced Algebra Condensed Lessons CHAPTER

8 Lesson 10. Mutually Exclusive Events and Venn Diagrams Investigation: Addition Rule Work through Steps 1 5 of the investigation in your book. Then, compare your results with those below. Step 1 The events are not mutually exclusive because a student can take both math and science. Step Math 40 0 Science 0 Neither 10 Step Math Science 0. Neither 0.1 Step 4 The sum P (M) P (S) counts the intersection twice. Step 5 Because adding P (M) and P (S) counts the intersection twice, you need to subtract the intersection once. So, P (M or S) P (M) P (S) P (M and S) Step 6 Suppose two dice are rolled. Let A sum is 7 and B both dice. In the diagram below, the enlarged points represent rolls in event A and the points in the shaded region represent rolls in event B. 6 5 Second die First die 150 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

9 Lesson 10. Mutually Exclusive Events and Venn Diagrams This Venn diagram shows how the 6 possible rolls are distributed. A B Use this information to find the probabilities in parts a f of Step 6 in your book. Then, compare your results with those below. a. P (A) 6 6 b. P (B) c. P (A and B) d. P (A or B) e. P (not A and not B) f. P (A or B) P (A) P (B) P (A and B) 6 6 Step 7 P (A or B) P (A) P (B) P (A and B) The general addition rule is summarized in the box on page 574 of your book. To practice using the rule, solve the problem in Example B in your book. Two events that are mutually exclusive and that make up all possible outcomes are referred to as complements. In general, the complement of an event A is (not A), and P (A) P (not A) 1. Example C gives you practice with all the new ideas from this lesson. Try to solve both parts of the problem before you read the solution. Discovering Advanced Algebra Condensed Lessons CHAPTER

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11 CONDENSED LESSON 10.4 Random Variables and Expected Value In this lesson you will learn the meaning of random variable, geometric random variable, and discrete random variable calculate the expected value of a random variable At the end of basketball practice, the coach tells the players they can go home as soon as they make a free throw shot. Kate s free throw percentage is 78%. What is the probability Kate will have to shoot only one or two times before she can go home? The probability Kate will make her first shot is From the diagram, you see that the probability she will miss her first shot and make her second shot is (0.)(0.78), or To find the probability she will get to go home after only one or two shots, add the probabilities of the two mutually exclusive events: The probability is about 95%. Probabilities of success (in this case, making a basket) are often used to predict the number of independent trials before the first success is achieved. 1st Shot nd Shot Make 0.78 Make 0.78 Miss 0. Investigation: Dieing for a Four Complete the investigation on your own, and then compare your results with those below. If you don t have your class s data for Step 1, do 0 trials on your own and find the mean. Step 1 The results will vary, but the mean number of rolls should be about 6. Step Based on the experiment in Step 1, you would expect a 4 to come up on the sixth roll on average. Step In this perfect sequence, a 4 comes up every 6 rolls. Step 4 The probability of success on any one roll is 6, and the probability of failure is _ 5 6. So Miss 0. P (rolling first 4 on 1st roll) 6 _ P (rolling first 4 on nd roll) 5 6 _ 6 5 P (rolling first 4 on rd roll) _ P (rolling first 4 on 4th roll) _ Step 5 Using the pattern from Step 4, P (rolling first 4 on nth roll) 5 n 1 6 n. Step 6 The sum should be about 6. This is the average of the number of rolls it will take to roll a 4. Step 7 The sum you found in Step 6 should be close to your estimates in Steps and. Discovering Advanced Algebra Condensed Lessons CHAPTER 10 15

12 Lesson 10.4 Random Variables and Expected Value A random variable is a numerical variable whose value depends on the outcome of a chance experiment. In the investigation the random variable is the number of rolls before getting a 4 on a die. The average value you found in Steps,, and 6 is the expected value of this random variable. It is also known as the long-run value, or mean value. The random variable in the investigation is a discrete random variable because its values are integers. It is also called a geometric random variable because the probabilities form a geometric sequence. Geometric random variables occur when counting the number of independent trials before something happens (success or failure). In Example A in your book, the random variable is the sum of two dice. This random variable is not geometric. Read Example A carefully, and make sure you understand it. The solution to part b demonstrates that to find the expected value, you multiply each possible value of the random variable by the probability it will occur and then add the results. This is summarized in the Expected Value box on page 581 of your book. Note that the random variable in the investigation had infinitely many possible values (theoretically, the first 4 could occur on any roll), so the expected value is the sum of a series namely, n 5n 1 6. The value you found in Step 6 of n n 1 the investigation was an estimate of the expected value. (It was the sum of the first 100 terms.) Example B in your book demonstrates that even if the random variable is discrete and an integer, its expected value may not be an integer. Read that example, and then read the example below. EXAMPLE Solution When two fair dice are rolled, the product of the results varies. What is the expected value of the product? The random variable x has as its values all the possible products of two dice. Here are the possible values and the probability of each. (Make sure you understand how the probabilities were determined.) Outcome x Probability P (x) Outcome x Probability P (x) The expected value is CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

13 CONDENSED LESSON 10.5 Permutations and Probability In this lesson you will learn the counting principle for counting outcomes involving a sequence of choices solve probability and counting problems involving permutations use factorial notation to express the number of permutations of n objects taken r at a time Some probability situations involve a large number of possible outcomes. In this lesson, you will learn some shortcuts for counting outcomes. Investigation: Name That Tune Try to complete the investigation in your book on your own. If you get stuck, or if you want to check your answers, read the results below. Step 1 Consider each spot in the playlist as a slot to be filled by a song. If n 1 and r 1, then there is one song, A, and one slot to fill. In this case there is only one thing you can do put A in the slot. If n and r 1, then there are two songs, A and B, and one slot. In this case there are two things you can do put A in the slot or put B in the slot. If n and r, then there are two songs, A and B, and two slots. In this case there are two things you can do put A in slot 1 and B in slot or put B in slot 1 and A in slot. You can represent these two possibilities as AB and BA. Here are the possibilities for all the cases for n. You can use a similar listing method to find the number of possibilities for n 4 and n 5. n and r 1: A, B, C ( possibilities) n and r : AB, AC, BA, BC, CA, CB (6 possibilities) n and r : ABC, ACB, BAC, BCA, CAB, CBA (6 possibilities) Here is the completed table at right: Step Possible patterns: The numbers in the first row are consecutive integers. In the second row, the values increase by consecutive even numbers (4 then 6 then 8). In the third row, the numbers increase by multiples of 18. In every column, the results for r n and r n 1 are the same. This is because there is only one way to place the last song. Number of songs in playlist, r Number of songs in library, n n 1 n n n 4 n 5 r r r r r 5 10 Step There are 150 ways to fill the first slot, 149 ways to fill the second slot, and so on. So the number of ways to arrange a playlist of 10 songs out of a library of 150 songs is Discovering Advanced Algebra Condensed Lessons CHAPTER

14 Lesson 10.5 Permutations and Probability There is a shortcut for counting the possibilities in the investigation. For example, suppose n 4 and r. Then there are four songs and three slots to fill. There are four choices of songs to go in the first slot. After the first slot is filled, there are three choices left for the second slot. After the second slot is filled, there are two choices left for the third slot. So, the number of possibilities is 4 4. (If you have trouble understanding why you must multiply, read the text about choosing an outfit on page 587 of your book.) This shortcut, known as the counting principle, is stated on page 587. Read Example A in your book, which applies the counting principle. When a counting problem involves arrangements of objects in which each object can be used only once (as in the investigation), the arrangements are called arrangements without replacement. An arrangement of some or all of the objects from a set, without replacement, is called a permutation. The notation n P r is read the number of permutations of n things chosen r at a time. You calculate n P r by multiplying n(n 1)(n ) (n r 1). Example B in your book illustrates these ideas. Read that example, and then try the example below. EXAMPLE Solution Theo works as a dog walker. Today, he must walk Abby, Bruno, Coco, Denali, Emma, and Fargus. a. In how many different orders can he walk the dogs? b. Theo decides to choose the order at random. What is the probability he will walk Abby first and Fargus last? a. There are six choices for the first dog, five choices for the second dog, four choices for the third dog, and so on. The total number of possibilities is 6 P b. First, count how many orderings have Abby first and Fargus last. Draw six slots to represent the dogs. There is one possibility for the first slot (Abby) and one possibility for the last slot (Fargus). _1 _1 There are then four choices for the second slot, three choices for the third slot, two choices for the fourth slot, and one choice for the fifth slot. _1 _4 _1 _1 Using the counting principle, there are , or 4 orderings in which Abby is first and Fargus is last. Thus, the probability Theo will walk Abby first and Fargus last is 4 70, or 1 0. In part a of the example above, you can see that 6 P 6 is the product of all the integers from 6 down to 1. The product of integers from n down to 1 is called n factorial and is abbreviated n!. For example, 5! In general n P n n! n P r n! (n r)! To learn about this in more detail, read the remainder of the lesson in your book. 156 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

15 CONDENSED LESSON 10.6 Combinations and Probability In this lesson you will solve counting and probability problems involving combinations discover how the number of combinations of n objects taken r at a time is related to the number of permutations of n objects taken r at a time Suppose eight players have entered a tennis tournament. In the first round, each player must play each of the other players once. How many games will be played in the first round? You might consider using the counting principle: There are eight choices for the first player, and then there are seven choices left for her opponent, for a total of 8 7, or 56 possible pairings. However, notice that in this case the order of the players does not matter. In other words, the pairing Player A vs. Player B is the same as Player B vs. Player A. Therefore, the total number of games is actually 8 7, or 8. When you count collections of outcomes without regard to order, you are counting combinations. In the tennis-tournament example, you found the number of combinations of eight people taken two at a time. This is written 8 C. Although there are 8 P 56 permutations of eight people taken two at a time, there are only half that many combinations: 8 C 8P 56 8 Read the text up to Example B in your book. Notice that the situation in Example A is very similar to the tennis-tournament situation above. Example B in your book involves finding the number of combinations of four people taken three at a time. Read that example carefully, and then read the example below. EXAMPLE Solution Jason bought six new books. He wants to bring four of the books with him on his summer trip. How many different combinations of four books can he take? Because order doesn t matter, you want to find the number of combinations of six books taken four at a time. However, first think about the number of permutations of six books (call them A F) taken four at a time: 6 P This number counts each combination of four books 4!, or 4 times. For example, ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, and DCBA are counted separately because they are all different permutations. However, these 4 permutations represent only one combination. Because each combination is counted 4 times, you must divide the number of permutations by 4 to get the number of combinations. So 6 C 4 6 P 4 4! There are 15 different combinations of four books Jason can take on his trip. Discovering Advanced Algebra Condensed Lessons CHAPTER

16 Lesson 10.6 Combinations and Probability Read the text in the Combinations box on page 597 of your book. To make sure you understand the ideas, try to solve the problem in Example C before you read the solution. Investigation: Winning the Lottery Read Step 1 of the investigation in your book. If possible, talk to one of your classmates or to your teacher about what happened when your class simulated playing Lotto 47. Believe it or not, everyone in your class was probably seated after only three or four numbers were called! Step To find the probability that any one set of six numbers wins, first find the number of possible combinations. (Note that these are combinations because the order does not matter.) There are 47 possible numbers, so the number of possible combinations is 47 C 6 47! 10,77,57 6!(47 6)! 1 The probability that any one combination will win is 10,77,57, or about If possible, complete the investigation using data from your class s simulation. If not, use these made-up assumptions: There are six groups of four students in your class, each group generated a total of 100 combinations of six numbers, and there are 1000 students in your school. The results below are based on these assumptions. Steps 6 Each group member invested $5, your group invested $100, and your class invested $600. Assuming the 600 combinations are all different, the 600 probability that someone in your class wins is 10,77,57, or about Assuming each person in your school generated 5 combinations, for a total of 5,000 combinations, the probability that someone in your school wins is 5,000 or about Step 7 If each of the 10,77,57 combinations were written on a 1-inch chip and the chips were laid end to end, then the line of chips would be about 169 miles long (10,77,57 in. 894, ft 169 mi). 10,77,57, Step 8 Answers will vary. One simple answer is that the probability of winning Lotto 47 is the same as the probability of drawing your name randomly from a hat containing 10,77,57 different names (including yours, of course)! 158 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

17 CONDENSED LESSON 10.7 The Binomial Theorem and Pascal s Triangle In this lesson you will learn how the numbers of combinations are related to Pascal s triangle learn how the numbers of combinations are related to binomial expansions use binomial expansions to find probabilities in situations involving two outcomes use information from a sample to make predictions about the population At right are the first six rows of Pascal s triangle. The triangle contains many 1 different patterns that have been studied for centuries. For example, notice 1 1 that each number in the interior of the triangle is the sum of the two numbers above it. Take a few minutes to see what other patterns you 1 1 can find. 1 1 In Lesson 10.6, you studied numbers of combinations. These numbers can be found in Pascal s triangle. For example, the numbers 1, 5, 10, 10, 5, 1 in the sixth row are the values of 5 C r : 5 C C C 10 5 C 10 5 C C 5 1 In the investigation you will explore why the numbers of combinations appear in Pascal s triangle. Investigation: Pascal s Triangle and Combination Numbers Complete the investigation in your book, and then compare your results with those below. Step 1 5 C 10 Step If Leora is definitely at the table, then two more students can sit at the table. There are four students to choose the two students from, so the number of possible combinations is 4 C 6. Step If Leora is excluded, then three students must be chosen from the four remaining students. The number of possible combinations is 4 C 4. Step 4 If four students must be chosen from five students, then the number of combinations is 5 C 4 5. If Leora is definitely at the table, then the other three students must be chosen from the remaining four students. The number of combinations is then 4 C 4. If Leora is excluded, then the four students must be chosen from the four remaining students. The number of combinations is 1. 4 C 4 Step 5 5 C 4 C 4 C and 5 C 4 4 C 4 C 4. In general, n C r n 1 C r 1 n 1 C r. Step 6 In Pascal s triangle, each entry (except a 1) is the sum of the two entries above it. The r th entry in row n 1 is n C r, and the two entries above it are n 1 C r and n 1 C r 1. Therefore, n C r n 1 C r 1 n 1 C r, which is precisely the pattern you found in Step 5. Discovering Advanced Algebra Condensed Lessons CHAPTER

18 Lesson 10.7 The Binomial Theorem and Pascal s Triangle Pascal s triangle is also related to expansions of binomials. For example, the expansion of (x y) is 1x x y xy 1y. Notice that the coefficients of the expansion are the numbers in the fourth row of Pascal s triangle. Why are the numbers in Pascal s triangle equal to the coefficients of a binomial expansion? The numbers in Pascal s triangle are values of n C r, so you can restate this question as, Why are the coefficients of a binomial expansion equal to values of n C r? To explore this question, read Example A in your book and the text immediately following it. Then, read the statement of the Binomial Theorem on page 604. Example B in your book shows how you can use a binomial expansion to find the probabilities of outcomes that are not equally likely. Read that example, following along with a pencil and paper, and then read the example below. EXAMPLE A coin is bent so that the probability of getting a head on any one flip is 0.4. Suppose the coin is flipped five times. Find the probabilities of getting 0, 1,,, 4, and 5 heads. Solution Let P (x) be the probability of getting x heads in five flips. If p is the probability of getting a head on any one toss and q is the probability of not getting a head (that is, getting a tail), then P (x) is the term 5 C x p x q 5 x in the expansion of (p q) 5. Specifically, P (0) 5 C 0 p 0 q P (1) 5 C 1 p 1 q P () 5 C p q P () 5 C p q P (4) 5 C 4 p 4 q P (5) 5 C 5 p 5 q Read the text following Example B up to Example C in your book. Study the table of the probabilities that different numbers of birds survive on page 605. The histogram shows another way to display this information. The extended table shows other probabilities, such as the probability that at most two birds survive. (Because the values are rounded, your calculations may vary slightly from those shown.) The definition box summarizes the formulas for the probability of success in a binomial event. In Example B and in the text following it, you use probabilities for a population to estimate probabilities for a sample. In Example C, the situation is reversed. You use data from a sample to make predictions about a population. Work through Example C carefully. 160 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons