1. Copyright © Cengage Learning. All rights reserved. CHAPTER 7 Uncertainty: Data and Chance

2. Copyright © Cengage Learning. All rights reserved. SECTION 7.4 Counting and Chance

3. 3 What Do You Think? What does “counting” mean in the context of probability? What is the difference between a combination and a permutation?

4. 4 Investigation A – How Many Ways to Take the Picture? Let’s say your college has a chapter of Kappa Delta Pi, the national education society. Let’s say this was a new society in your college, and in the first year there were 9 members. Let’s also say that the members decided to have a group picture taken and wanted only one row—that is, they wanted everyone in the first row. In how many different ways can they line up for the picture? Discussion: This is one of the many problems in mathematics where many people find that if they see a solution, they see it right away, and if they don’t, they’re stuck.

5. 5 Investigation A – Discussion Even if you were systematic, you could become stuck trying to do this one: 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 8 etc. Stuck, that is, unless you have problem-solving tools. If you are stuck, look at the steps for problem-solving on the inside front cover of this book. What strategies might help? cont’d

6. 6 Investigation A – Discussion Strategy 1: Make a Simpler Problem There are several different strategies that can be successfully applied to this problem. The first one we will discuss is “make a similar, simpler problem and then work up.” If you didn’t make much progress on your own and you like this method, use it to finish the problem. cont’d

7. 7 Investigation A – Discussion Let’s see how it works. Let’s start with a club with only 3 members: 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 This gives us a total of 6 different arrangements. cont’d

8. 8 Investigation A – Discussion Where did the 6 come from? Most people see 3 groups of 2, and hence 3  2. Some people can jump to the solution of the original problem from here; others need more data from which to generalize. cont’d

9. 9 Investigation A – Discussion If you don’t yet feel confident that you can connect this to the original problem, try a club with 4 members. 1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 2 1 4 2 3 1 4 3 2 This gives us 6 possibilities in which person number 1 is on the left. cont’d

10. 10 Investigation A – Discussion Either by reasoning or by working out the other possibilities yourself, you can see that there will be 6 outcomes in which each of the other 3 members is on the left. This gives us a total of 24 outcomes. Where did the 24 come from? As before, we have 4 groups, each of which consists of 2 groups of 3. If we put the numbers in order, we have either 4  3  2 or 2  3  4. cont’d

11. 11 Investigation A – Discussion Strategy 2: Make a Tree Diagram In this case, a tree diagram (Figure 7.60) is not terribly helpful for most students. I mention this here because many students who struggle in math often think that the better students and the teacher always know what to do and always pick the “right” strategy, and this is not true. cont’d Figure 7.60

12. 12 Investigation A – Discussion I know many students who have done well in the course who admit that they frequently found that their first attempt at a challenging problem was not productive and that they needed to try another strategy. Strategy 3: Connect This Problem to Something Familiar What about the multiplication principle? Many people understand this strategy better if we simultaneously use the “act it out” strategy. Let’s begin with the first person. We have nine possibilities. Once we have that person set, how many choices do we have for the second position? There are eight possibilities. cont’d

13. 13 Investigation A – Discussion Do you see how to finish the problem using this line of reasoning? Using this line of reasoning, the number of possibilities will be 9 times 8 times 7 times 6 and so on, so that the total number of possibilities is 9  8  7  6  5  4  3  2 New terminology will make our discussion of this problem and other probability problems easier. How would you describe the computation above to someone—let’s say a friend in class who didn’t have this book? cont’d

14. 14 Investigation A – Discussion One way: “Start with 9, then 8, then 7, and keep reducing the number by one till you can’t anymore. Now compute the product.” This kind of computation is common in probability and has a name: factorial. That is, 9  8  7  6  5  4  3  2 can be written as 9! Mathematicians insert a 1 at the end of this product, and define any number n factorial, n!, as n! = n  (n – 1)  (n – 2)  …  3  2  1 cont’d

15. 15 Investigation B – How Many Different Election Outcomes? Let’s say Kappa Delta Pi has its first election: for president and treasurer. How many possible ways can a president and treasurer be elected from a pool of 9 people? Discussion: Strategy 1: Be Systematic, Make a Table, and Look for Patterns Table 7.20

16. 16 Investigation B – Discussion Note that the numbers in Table 7.20 represent possible outcomes. For example, 1, 2 represents the outcome “person 1 as president and person 2 as treasurer.” Why is {2, 1} not simply a duplicate of {1, 2}? Why did we skip {2, 2}? When you complete this table, you find 9 columns that each has 8 outcomes, giving a total of 72 ways. cont’d

17. 17 Investigation B – Discussion Strategy 2: Make a Tree Diagram Some students find a tree diagram (Figure 7.61) very helpful. cont’d Figure 7.61

18. 18 Investigation B – Discussion Strategy 3: Connect To Previous Problems We can use the multiplication principle again. There are 9 different possibilities for president (each member), but only 8 possibilities for treasurer (assuming that no one will be both president and treasurer). Without too much difficulty, most students can see this as 8 groups of 9, or 9  8, which gives us 72 ways. This is like the beginning of factorial. cont’d

19. 19 Investigation B – Discussion Electing 3 officers: Now let’s see how confident you are about this idea. What if the group decided to have a president, a treasurer, and a secretary? How many different outcomes are there for this scenario? cont’d

20. 20 Investigation B – Discussion Each of the three strategies from before still applies (Figure 7.62). Figure 7.62 cont’d

21. 21 Investigation B – Discussion The total number is 9  8  7 This question (about president, treasurer, secretary) is more complex than the question in Investigation 7.4A about the picture. Instead of looking at all the possible permutations of a set of specified size, which we did in the picture investigation, we are now asking how many different permutations of specified size we can make from a set of a given size. In this situation, the outcome “1 2 3” is considered to be different from the outcome “1 3 2.” cont’d

22. 22 Investigation B – Discussion Using symbols, we say that our election question seeks the number of different subsets of size r that we can make from a set of size n. We can also refer to this amount as the number of permutations of n things taken r at a time. The shorthand for this situation is n P r. In the case of the different possibilities for president and treasurer, we were looking for 9 P 2. In the case of the different possibilities for president, treasurer, and secretary, we were looking for 9 P 3. cont’d

23. 23 Investigation B – Discussion Can you now generalize a formula for the number of permutations of a subset of size r from a set of size n? The generalization is n P r = n(n – 1)... (n – r + 1) This observation leads to an alternative formula: cont’d

24. 24 Investigation C – How Many Outcomes This Time? Suppose Kappa Delta Pi decided to elect two members to send to the national convention. How many possible outcomes are there now? Discussion: Do you understand why this is not the same as the problem with the president and treasurer? Combinations: Using the notation of the investigation with the president and treasurer, the outcomes {1, 2} and {2, 1} represent different outcomes.

25. 25 Investigation C – Discussion In the first outcome, student 1 is the president and student 2 is the treasurer, whereas in the second outcome, student 2 is the president and student 1 the treasurer. However, delegates {1, 2} and {2, 1} represent the same outcome. Mathematically, we distinguish between permutations and combinations. Permutations: In probability situations in which order matters, we speak of permutations. When order doesn’t matter, we speak of combinations. cont’d

26. 26 Investigation C – Discussion For example, if outcomes “abc” and “bac” are considered to be different, we are in a permutation situation; if they are considered to be the same, we are in a combination situation. Using symbols, we can write the number of combinations of n things taken r at a time as n C r. Now let us discuss the solution to how many outcomes there are for this question. First, we will be systematic and look for patterns (Table 7.21). cont’d Table 7.21

27. 27 Investigation C – Discussion Can you finish the problem? Does this problem remind you of a problem or a pattern that we have encountered before? If you thought about the handshakes problem, you are right. We find that there are 36 different outcomes—that is, 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1. cont’d

28. 28 Investigation C – Discussion We would like to be able to generalize this procedure so that we can use it in a variety of situations; for example, what if we had a set of 20 elements (instead of 9) and we wanted all outcomes consisting of 3 elements at a time (instead of 2)? One way to determine the general procedure is to analyze the results of the simpler problems. Thus we will explore where the 36 comes from. cont’d

29. 29 Investigation C – Discussion From one perspective, we can see that 36 comes from 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1. Thus we could generalize that if there were 20 members and we wanted all outcomes of two at a time, there would be 19 + 18 + 17 + … + 3 + 2 + 1. What if we wanted all possible outcomes of three at a time? In order to develop a procedure that will work in more situations, we have to look more deeply. What other ways do you see to determine the 36? cont’d

30. 30 Investigation C – Discussion Using the algorithm developed in Exploration 1.1, we can say that Now this should look similar to the permutation algorithm: 9 P 2 = 9(8) = 72 Why do we divide by 2? Because each combination is connected to two permutations, we can see that in this case, the number of combinations will be half the number of permutations. cont’d

31. 31 Investigation D – Pick a Card, Any Card! Consider a standard deck of playing cards. Pick two cards. What is the probability of getting two queens? Can we use any of the algorithms? Can you explain why one of the algorithms applies or why none of them applies? Discussion: This is a combination (not a permutation) problem. Does this mean that the probability is ?

32. 32 Investigation D – Discussion The answer to this question is not. The number 1326 means that there are 1326 different possible combinations. Because we want to know the probability of getting two queens, we also have to determine how many of those 1326 combinations involve two queens. One strategy is to list them systematically—for example, SD, SH, SC, HD, HC, DC. Another strategy is to use the appropriate combination algorithm appropriately. cont’d

33. 33 Investigation D – Discussion In either case, we find that there are 6 different combinations that involve 2 queens. Thus we find that the probability of two queens is cont’d

34. 34 Investigation E – So You Think You’re Going to Win the Lottery? Since 1980, the number of states with lotteries has increased considerably. What factor(s) do you think might have led to the increase in states having lotteries? During the 1980s, federal aid to states and cities shrank considerably. Many states “solved” this problem by establishing or increasing lotteries and designating the profits from the lottery for expenses such as education. The state lotteries have been called voluntary taxes.

35. 35 Investigation E – So You Think You’re Going to Win the Lottery? Let’s begin with a rather straightforward lottery in which players select four digits. The winning number is commonly selected in the following fashion: 10 Ping-Pong balls (each with a digit 0 through 9 written on it) are placed in a hopper. One Ping-Pong ball is selected. That digit now goes in the first place. That Ping-Pong ball is put back in the hopper, and the process is repeated three times. What is the probability of picking a winning number? cont’d

36. 36 Investigation E – Discussion Strategy 1: Connect to the Multiplication Principle Each time, there are 10 possibilities, so the sample space is 10  10  10  10. Because all outcomes are equally likely, the probability of any outcome is Strategy 2: Act it Out What are some possible winning numbers? Possibilities include 1234, 4612—in fact, any number from 0000 to 9999.