12.1. Randomness and Probability

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1 CONDENSED LESSON. Randomness and Probability In this lesson, you Simulate random processes with your calculator 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. 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 0-flip sequences before completing Step. 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 L 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. 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. 4 Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

2 Lesson. Randomness and Probability Solution Simulate 00 spins of the first spinner by generating 00 random integers from to 4 and storing them in list L. Simulate 00 spins of the second spinner by generating 00 random integers from to and storing them in list L. Define list L as the product of lists L and L. The histogram shows the 00 products. The possible odd products are,, and 9. Adding the bin heights for these products gives , or 08. So, out of 00 simulated spins, 08 have an odd product. Therefore, P(odd product) The results of a random process are called outcomes. An event is a set of desired outcomes. The probability of an event is always between 0 and. The probability of an event that is certain to happen is. 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 59 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 with a pair of dice. Then, read the example below. [,,, 0, 0, 5] 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 grid points in the diagram at right. The four possible outcomes with an odd product are labeled A C B D A D. Point B, for example, represents an outcome of on the 4 first spinner and on the second, for a product of 9. 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. 4 So, P(odd product).,or %. This is a little less 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. 8 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

3 CONDENSED LESSON. Counting Outcomes and Tree Diagrams In this lesson, you Use tree diagrams to count outcomes and to find probabilities of compound events 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. In the tree diagram from Example A, each single branch represents a simple event. A sequence of simple events, represented by a path, is called a compound event. Investigation: The Multiplication Rule Work through the investigation in your book, and then compare your results with those below. Step Below is the tree diagram from Example A, part a, labeled with probabilities. The probability of each simple event (getting a particular toy in a box) is.5. Because there are four equally likely paths, the probability of any one path is,or 4.5. Notice that the probability of a path is the product of the probabilities of its branches. st Box.5 Toy nd Box.5 Toy Toy.5 a b P (Toy followed by Toy ).5 P (Toy followed by Toy ).5 Toy.5.5 Toy Toy.5 c d P (Toy followed by Toy ).5 P (Toy followed by Toy ).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. The sum of the probabilities of the six highlighted paths is, 7 or. 7 Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

4 Lesson. Counting Outcomes and Tree Diagrams Toy Toy Toy Toy Toy Toy Toy Toy Toy Toy Toy Toy P (Toy, Toy, Toy ) 7.07 P (Toy, Toy, Toy ) 7.07 P (Toy, Toy, Toy ) 7.07 P (Toy, Toy, Toy ) 7.07 P (Toy, Toy, Toy ) 7.07 P (Toy, Toy, Toy ) 7.07 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.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 5 different equally likely outcomes. The outcome (Toy, followed by Toy, followed by Toy 4, followed by Toy ) is one of these outcomes, so its probability is,or 5 about.004. You can also figure this out by realizing that a complete tree diagram would have 5 paths and the probability along each branch is.5. The specified outcome is one path with four branches, so its probability is (.5)(.5)(.5)(.5), or about.004. Step 4 To find the probability of a path, multiply the probabilities of its branches. 84 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

5 Lesson. 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 4 possibilities for the first toy, then there are possibilities for the second toy, then possibilities for the third toy, then possibility for the third. There are 4 such sequences.) So the 4 probability of getting a complete set is, 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 has 8 rock songs and 4 hip-hop songs. Disk has 5 rock songs and 5 hip-hop songs. Disk has rock songs and hip-hop songs. If Zak randomly selects song from Disk, then from Disk, and then from Disk, what is the probability he will play exactly 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 are those that include exactly two hip-hop songs. st CD nd CD rd CD R _ 5 R _ H 4 5 R _ 5 R H _ H 4 5 R _ 5 H _ R _ H 4 5 R _ 5 H _ 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 Key Curriculum Press

6 Lesson. 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. These events are called independent, meaning that the occurrence of one event has no influence on the occurrence 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 7 of your book. In part b of 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, 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, then P(rock song from Disk ).There is a probability he will play a hip-hop song from Disk, so the probability he plays a rock song from Disk is,or. 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 ) are independent, so P(rock on Disk hip-hop on Disk ) P(rock on Disk ) In Example B, the events (rock on Disk ) and (hip-hop on Disk ) are dependent, so P(rock on Disk hip-hop on Disk ) P(rock on Disk ) Page 7 of your book explains how you can use tree diagrams to break up dependent events into independent ones. Read that text carefully, and then try to make a similar diagram for the situation in Example B above. Then, read the multiplication rule. 8 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

7 CONDENSED LESSON. Mutually Exclusive Events and Venn Diagrams In this lesson, you 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., 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? Of the region labeled.5? 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 represents the probability that a student does not like football and does not like golf. The region labeled.5 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.40. 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 80. 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 Key Curriculum Press

8 Lesson. Mutually Exclusive Events and Venn Diagrams Investigation: Addition Rule Work through Steps 5 of the investigation in your book. Then, compare your results with those below. Step The events are not mutually exclusive because a student can take both math and science. Step Math 40 0 Science 0 Step Neither 0 Math.4. Science. Neither. 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 Suppose two dice are rolled. Let A sum is 7 and B both dice. In the diagram below, the circled points represent rolls in event A and the points in the shaded region represent rolls in event B. 5 Second die First die 88 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

9 Lesson. Mutually Exclusive Events and Venn Diagrams This Venn diagram shows how the possible rolls are distributed. A B 4 4 Use this information to find the probabilities in parts a f of Step in your book. Then, compare your results with those below. a. P(A).7 b. P(B).444 c. P(A and B).05 d. P(A or B) 0.55 e. P(not A and not B).444 f. P(A or B) P(A) P(B) P(A and B) 0.55 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 8 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). 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 Key Curriculum Press

10 CONDENSED LESSON.4 Random Variables and Expected Value In this lesson, you 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.78. From the diagram, you see that the probability she will miss her first shot and make her second shot is (.)(.78), or.7. 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. st Shot Make.78 Miss. nd Shot Make.78 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, do 0 trials on your own and find the mean. Step The results will vary, but the mean number of rolls should be about. Step Based on the experiment in Step, you would expect a 4 to come up on the th roll. Step In this perfect sequence, a 4 comes up every rolls. Step 4 The probability of success on any one roll is, and the probability of failure is 5.So, Miss. P(rolling first 4 on st roll).7 P(rolling first 4 on nd roll) 5 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 Step The sum should be about. This is the average of the number of rolls it will take to roll a 4. Step 7 The sum you found in Step should be close to your estimates in Steps and. n Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

11 Lesson.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 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, which means it is acount of the number of trials before something (a success or a failure) happens. 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 89 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 5 n. n The value you found in Step of n the investigation was an estimate of the expected value. (It was the sum of the first 00 terms.) Example B in your book demonstrates that even if the random variable is discrete, 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) 4 Outcome x Probability P(x) The expected value is CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

12 CONDENSED LESSON.5 Permutations and Probability In this lesson, you 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: Order and Arrange 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 If n and r, then there is one object, 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, then there are two objects, 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 objects, A and B, and two slots. In this case there are two things you can do put A in slot and B in slot or put B in slot 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 : A, B, C ( possibilities) n and r : AB, AC, BA, BC, CA, CB ( possibilities) n and r : ABC, ACB, BAC, BCA, CAB, CBA ( possibilities) Here is the completed table at right: Number of items, n Step Possible patterns: The numbers in n n n n 4 n 5 the first row are consecutive integers. In the second row, the values increase by r 4 5 consecutive even numbers (4 then then r 0 8). In the third row, the numbers increase by multiples of 8. In every column, the r 4 0 results for r n and r n are the r same. This is because there is only one r 5 0 way to place the last object. Number of spaces, r There is a shortcut for counting the possibilities in the investigation. For example, suppose n 4 and r. Then there are four objects and three slots to fill. There are 4 choices of objects to go in the first slot. After the first slot is filled, there are choices left for the second slot. After the second slot is filled, there are 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 example about Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

13 Lesson.5 Permutations and Probability choosing an outfit on page 95 of your book.) This shortcut, known as the counting principle, is stated on page 95. 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 )(n ) (n r ). 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 choices for the first dog, 5 choices for the second dog, 4 choices for the third dog, and so on. The total number of possibilities is P b. First, count how many orderings have Abby first and Fargus last. Draw six slots to represent the dogs. There is possibility for the first slot (Abby) and possibility for the last slot (Fargus). There are then 4 choices for the second slot, then choices for the third slot, then choices for the fourth slot, then choice for the fifth slot. 4 Using the counting principle, there are 4, or 4 orderings in which Abby is first and Fargus is last. Thus, the probability Theo will 4 walk Abby first and Fargus last is 7, or. 0 0 In part a of the example above, you can see that P is the product of all the integers from down to. The product of integers from n down to 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. 94 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

14 CONDENSED LESSON. Combinations and Probability In this lesson, you 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 8 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 8choices for the first player, and then there are 7 choices left for her opponent, for a total of 8 7, or 5 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 objects without regard to order, you are counting combinations. In the tennis-tournament example, you found the number of combinations of 8 people taken at a time. This is written 8 C.Although there are 8 P 5 permutations of 8 people taken at a time, there are only half that many combinations: 8 C 8P 5 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 4people taken 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 books taken 4 at a time. However, first think about the number of permutations of books (call them A F) taken 4 at a time: P 4 0. This number counts each combination of 4 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, C 4 P 4! There are 5 different combinations of 4 books Jason can take on his trip. Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

15 Lesson. Combinations and Probability Read the text in the Combinations box on page 705 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 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 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 47!!(47 )! The probability that any one combination will win is , 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 4 students in your class, your group generated a total of 00 combinations of six numbers, and there are 000 students in your school. The results below are based on these assumptions. Steps Your group invested $00, and your class invested $00. Assuming the 00 combinations are all different, the probability that someone in your class wins 00 is 07, 757 or about Assuming each person in your school generated 5 combinations, for a total of ,000 combinations, the probability that someone in your school wins is 0 7, 757 or about.00. Step 7 If each of the 0,77,57 combinations were written on a -inch chip and the chips were laid end to end, then the line of chips would be about 9 miles long (07757 in ft 9 mi). 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 0,77,57 different names (including yours, of course)! 9 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

16 CONDENSED LESSON.7 The Binomial Theorem and Pascal s Triangle In this lesson, you 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 few rows of Pascal s triangle. The triangle contains many different patterns that have been studied for centuries. For example, notice 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 can find. In Lesson., you studied numbers of combinations. These numbers can be found in Pascal s triangle. For example, the numbers, 5, 0, 0, 5, in the sixth row are the values of 5 C r : C 0 5 C 5 5 C 0 5 C 0 5 C C 5 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 5 C 0 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. 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. 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 C r n C r Step In Pascal s triangle, each entry (except a ) is the sum of the two entries above it. The rth entry in row n is n C r, and the two entries above it are n C r and n C r.therefore, n C r n C r n C r, which is precisely the pattern you found in Step 5. Discovering Advanced Algebra Condensed Lessons CHAPTER Key Curriculum Press

17 Lesson.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 x x y xy y.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 7. 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 A coin is bent so that the probability of getting a head on any one flip is.4. Suppose the coin is flipped five times. Find the probabilities of getting 0,,,, 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 5x in the expansion of (p q) 5. Specifically, P(0) 5 C 0 p 0 q P() 5 C p 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 Example C in your book shows how the ideas from this lesson are used in sampling. Read the text on page 7, and then read Example C. Here is another example. EXAMPLE B A company produces widgets. Yesterday, the quality-control manager selected arandom sample of 50 widgets and found that were defective. What is the probability that a widget selected randomly from those produced yesterday is defective? Solution A ratio of 5, 0 or.0 would be highly likely if the probability that a randomly selected widget is defective is also.0. But this ratio could also happen for other probabilities. Let p be the unknown probability that a widget is defective. The probability that n widgets in a sample of 50 are defective is 50 C n p n q 50n.Ifn, then the most likely value of p is.0. To find other values of p that give a higher probability of defective widgets than of or 4, you need to solve these inequalities: 50 C p q C p q 48 and 50 C p q C 4 p 4 q 4 Solve these inequalities yourself. (Use the solution to Example C in your book as 4 a guide.) You should get 7 p 5, or.0588 p So, the probability that a randomly selected widget is defective is between about 5.9% and 7.8%. 98 CHAPTER Discovering Advanced Algebra Condensed Lessons 004 Key Curriculum Press

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