AB BC CD DE EF FG AC BD CE DF EG AD BE CF DG AE BF CG AF BG AG

Transcription

1 VIII. Make a systematic list A systematic list provides a way to organize the information of a problem in a methodical way. The system employed used should be understandable and clear so that the person making the list can verify its accuracy quickly. Additionally, others should be able to understand the system and verify the solution without too much effort. Many systematic lists are in the form of tables whose columns are labeled with the information in the problem; the rows are used to indicate possible combinations. Example: Andrew and his friends have formed a fantasy basketball league in which each team will play the others three times each. There are seven teams: the (San Diego State) Aztecs, the (St. Bonaventure) Bonnies, the (George Washington) Colonials, the (Duquesne) Dukes, the (LaSalle) Explorers, the (Dayton) Flyers, and the (St. Mary s College) Gaels. How many games will be played in all? Solution: A systematic listing of the matchups will be valuable in this problem. It will help us first to ensure that every possible matchup appears in our list, and secondly, it will prevent us from listing certain matchups repeatedly, which would throw off our count. As should be apparent, the names of the teams are irrelevant for the counting, so we can easily abbreviate the teams with initial letters A, B, C, D, E, F, and G. Then, an alphabetical method for listing matchups presents itself: first list all possible matchups in which team A plays, then continue with all the remaining matchups in which B plays, then C, etc. This leads to the following list, prepared sequentially column by column: AB BC CD DE EF FG AC BD CE DF EG AD BE CF DG AE BF CG AF BG AG We count = 2 different pairs of teams. Since each pair plays three games, there will be a total of 6 games in the league. 0

2 Example: Penny has 2 dimes in her change purse. She arranges them into three piles, putting an odd number of dimes in each pile. How many ways can she do this? Solution: Make a systematic list of the possible combinations. A natural method would be to construct a table in which the columns correspond to the three piles, marking in each row the number of coins in each pile. Begin by putting the fewest possible number of dimes in the first pile, the fewest remaining possible number in the second pile, so that all the other dimes must go in the third pile. (Note that the size of the third pile is no longer a choice, since the sum of the sizes of the piles must equal 2.) In subsequent rows, we repeat this process, making sure not to repeat the choices marked in an earlier row. This is controlled by upping the number in the second pile at each row, remembering that only odd numbers of coins are allowed in each pile. Pile Pile 2 Pile This procedure work fine, until we come to the row,,. At this point, we recognize that since the third number is now smaller than the second, we have repeated the choice made in the previous row: having coins in pile 2 and in pile is exactly the same as having in pile 2 and in pile the piles are indistinguishable, so their numbering is irrelevant to the solution of the problem. For this reason, we strike that row from the list. The same reasoning applies to the next row,,, 2. This time we notice a repeat with the second row of the table, as now the second number in the row is less than the first. So we strike that row from the table as well. In fact, we observe that if any row contains a number in a higher numbered pile which is smaller than a number in a lower numbered pile, then it must repeat some earlier row of the table. So as we continue adding to the table, we make sure to never enter numbers in a row that decrease as we proceed to higher numbered columns. The process is complete when we reach the row,,. A quick count now reveals that there are 6 rows in the table, hence there are 6 ways for Penny to arrange her piles. (If Penny chooses to label her piles, then the duplications noticed above are in fact distinguishable from each other. It follows that there will be many more possible combinations. A similar method to the one used above will find 8 possibilities!)

3 Example: For her Shakespeare course, Kristen must read all of the following plays and choose three to write a paper about: Richard III, A Midsummer Night s Dream, The Tempest, Macbeth, and Othello. How many different sets of books can Kristen do her papers on? Solution: Make a table with five columns, one for each of the five plays. Mark three s in each row to identify some collection of the five columns. Do so systematically by selecting the leftmost columns first, then later columns in order from left to right Richard III Midsummer Tempest Macbeth Othello We find that there are 0 ways for Kristen to choose the subjects of her papers. Another solution method makes use of a tree diagram to organize the listing spatially. This method emphasizes order of selection, and is called a tree diagram since it looks like branches of a tree. The first set of branches identifies the first option selected, the second tier of branches the second option, and the third and final set of branches, the leaves of the tree, mark the final selection. The number of possibilities overall corresponds to the number of leaves showing. Example: In the previous example, once Kristen has chosen her three plays say she s chosen Richard III, The Tempest, and Macbeth. In how many different orders can she write her three papers? 2

4 Solution: We use a tree diagram to organize this information: Richard III The Tempest Macbeth The Tempest Macbeth Richard III Macbeth The Tempest Richard III Macbeth The Tempest Macbeth Richard III Richard III The Tempest The diagram clearly shows all six possibilities, where the order in which Kristen writes her papers corresponds to one of the six paths from the root of the tree (at the top) down to one of the leaves. Tree diagrams are also very useful for solving problems about computing probabilities, as the following example shows. Example: A bag contains one white ball and two red balls. A ball is drawn at random from the bag. If the ball is white, then it is returned to the bag with a second white ball, but if it is red, then it is returned with two additional red balls. Then another ball is drawn at random. What is the probability that this second ball drawn from the bag is red? Solution: A simple tree diagram can neatly organize the possible choices at each stage: /2 W The numbers on the edges correspond to the probabilities attached to the choice represented by that edge. For instance, since there is one white ball and two red balls initially in the bag, there is a / chance that the white ball is selected and a 2/ chance that a red / 2/ W R /2 / 4/ R W R

5 ball is chosen. On the second stage edges, we notice that after a white ball was chosen first, there will then be two white and two red balls in the bag, giving an equal /2 probability that either a white or red ball is chosen second. However, if a red ball is chosen first, there will then be one white and four red balls in the bag, so there is a / chance that the white ball is selected and a 4/ chance that a red ball will be chosen. It follows that the probability of choosing two white balls is P (W,W ) = 2 = ; the probability 6 of choosing a white ball first and a red ball second is P (W, R) = 2 = 6 ; that of choosing a red ball first then a white ball is P ( R,W ) = 2 = 2 ; and finally, the probability of choosing two red balls is P ( R, R) = 2 4 = 8. (It is straightforward to verify that these four numbers do, as expected, add up to.) Example: A company that conducts individual drug tests to identify the existence of certain illegal substances in a person s system claims that its drug test is 0% accurate. That is, given a group of people that use these illegal substances, the test will correctly identify 0% of the group as users. Furthermore, given a group of people who don t use these drugs, the test will correctly identify 0% of them as non-users. These numbers sound pretty good, but suppose someone has just taken this drug test and receives a positive result (suggesting that they are a user). What is the probability that this person actually is a user? To answer this question, you will need to know that only % of the population is made up of illegal drug users. Solution: Consider a hypothetical group of 0,000 people. If % of them use drugs, then there are 00 drug users in the group. If all 0,000 individuals take the drug test and it is correct 0% of the time, then of the 00 drug users, 40 will be correctly identified as users, meaning that 0 will be misidentified as being non-users. They represent the false negatives for the test. The tree diagram illustrates this situation. % 00 drug users 0% 0% 0 negative testers 40 positive testers % 00 non-users 0% 0 positive testers 0% 80 negative testers 4

6 Now let s consider the 00 people who do not use drugs. The test is 0% accurate, so it will correctly identify 80 people as non-users. However 0% of the people, 0 in the group, will register as false positives, indicating that they are using drugs when in fact they are not. Notice that = 400 people in the group who tested positive, but only 40 of them are actually drug users; the other 0 are false positives. This means that only 40/400 2% of those who test positive are truly drug users! (What are the implications of this information for the people who use such drug tests to make decisions and for the people who are affected by these decisions? Indeed, many drug tests now take information about false positives like this into account.)