CHAPTER 10 Probability - Prek 12

01/09/200611:10 PMPage 47CHAPTER10ProbabilityContent SummaryChapter 10 introduces basic probability concepts, including special kinds of events,expected value, and counting permutations and combinations.Relative Frequency Graphs and ProbabilityChapter 10 introduces relative frequency graphs, which display categorical data.Relative frequency bar and circle graphs display the percent or fraction of eachcategory relative to the total for all categories.Library CollectionOtherMedia6% 3%Children’s fiction18%Adultnonfiction14%35%Children’s nonfictionAir emissions (69%)24%Adult fictionLibrary Collection30PercentDA2GP 774 10.qxd20100Children’s Children’sfiction nonfictionAdultAdultfiction nonfictionCategoryMediaOtherThe chance of something occurring, or the probability of an outcome, can bedetermined from a relative frequency graph. For example, for a randomly chosenitem from the library collection, the probability that the item is adult fiction is 24%,or 0.24. An experimental probability, or observed probability, is based on data ornumber of occurrences of eventexperiments and is defined as . A theoretical probability, definedtotal number of trialsnumber of different ways the event can occuras ,usesknownquantities.For a fair coin, thetotal number of equally likely outcomes possibletheoretical probability of getting a head is 50%, because heads and tails are equallylikely. However, in flipping a coin, someone might get a run of heads or tails thatleads them to a different experimental probability for heads. After many coin tosses,the experimental probability for heads would approach 50%.(continued) 2007 Key Curriculum PressDiscovering Algebra: A Guide for Parents47

DA2GP 774 10.qxd01/09/200611:10 PMPage 48Chapter 10 Probability (continued)Independent EventsIf you flip a coin repeatedly and get 5 heads in a row, you might say that the chance,or probability, of getting a head on the next flip is very small. After all, the chance ofgetting 6 heads in a row is very small. Or, you might think that the chance of gettinga head on the next flip is large; that there’s a “run” on heads. In fact, however, thecoin has no memory; the chance of getting a head on the next flip is 0.5, as it hasbeen all along. You can say that the events are independent; the result of the sixth flipdoes not depend on the result of the fifth flip.In the case of independent events, the probability that both will occur is the productof the probabilities of the individual events. The probability of getting heads fivetimes in a row is 12 12 12 12 12 31 2 . This is also the probability of getting anyparticular string of heads or tails. In other words, the probability of gettingT T H T T is also 31 2 .Not all events are independent of previous events. Suppose you have a bag with sixbills, five 1-dollar bills and one 100-dollar bill. The chance of selecting the 100-dollarbill is 1 in 6, or about 0.17. However, if someone selects a one-dollar bill andremoves it from the bag, the next person has a 1 in 5, or 0.2 probability, of pickingthe 100-dollar bill. Of course, if the first person picks the 100-dollar bill, then thenext person has no chance, or 0 probability, of choosing the 100-dollar bill.Permutations and CombinationsDetermining the numbers to calculate theoretical probabilities can be challenging.Sometimes the outcomes to be counted are arrangements of things or people. Forexample, suppose ten people attend a meeting, and you randomly pick three of themto win different door prizes. Any of the ten could get door prize A, the most valuable.Any of the remaining nine could win door prize B, the next most valuable. And any ofthe remaining eight could win the third door prize, C. There are 10 9 8 720ways that three of the ten people could be arranged to get these door prizes. Thearrangements are called permutations; the number of permutations of ten people,three at a time, is abbreviated 10P3.If the door prizes were all the same, though, it wouldn’t matter who got which one.All that matters is the number of trios of people who win. These collections are calledcombinations. The six permutations ABC, ACB, BAC, BCA, CAB, and CBA wouldcount as one combination, because A, B, and C are the same prizes. The number ofcombinations of three people out of ten, written 10C3, is only 16 of 10P3, or 120.Multiple-Stage ExperimentsTree diagrams can be useful for determining probabilities of morecomplicated experiments. For two flips of a coin, the possibleoutcomes and their probabilities can be shown with a tree diagram.1st Flip2nd Flip1P (H) 21P(H and H) 41P (T) 21P(H and T) 41P (H) 21P(T and H) 41P (T) 21P(T and T) 41P (H) 21P (T) 2(continued)48Discovering Algebra: A Guide for Parents 2007 Key Curriculum Press

DA2GP 774 10.qxd01/09/200611:10 PMPage 49Chapter 10 Probability (continued)The expected value of an event is an average value found by multiplying thevalue of each possible event by its probability and adding the products. Forexample, the expected value of one spin on the spinner pictured would befound as shown: 2 511 4 (2) 4 (6) 2.5 0.5 1.5 0.5The expected value of the spinner is 0.50.1 ( 5)2 6Summary ProblemImagine you have a bag with colored blocks, three blue and four red. What kindsof questions can be asked and answered involving choosing blocks from the bag?Questions you can ask in your role as student to your student include: What is the probability of drawing a red block? A blue block?What is the probability of drawing two red blocks in a row? Do you need anymore information?If the red blocks are worth 2 and the blue blocks are worth 5, what is theexpected value of one draw?What values for each color block would give an expected value of 2 for onedraw? Try to find several possibilities.Sample AnswersThe probability of drawing a red block is 47 ; the probability of drawing a blue block is3 . To find the probability of drawing two red blocks in a row, you must know whether7the block will be replaced after is drawn. The probability of drawing two red blockswith replacement is 47 47 14 69 , whereas the probability of drawing two red blockswithout replacement is 47 36 14 22 , or 27 . If red blocks are worth 2 and blue blocks areworth 5, the expected value of one draw is 47 (2) 37 (5) 87 17 5 27 2 3.14, or 3.14. To have an expected value of 2 per draw, many combinations are possible.Some are 3.50 for red, 0 for blue; 2.75 for red, 1.00 for blue; 2.50 for red,and 8.00 for blue. 2007 Key Curriculum PressDiscovering Algebra: A Guide for Parents49

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DA2GP 774 10.qxd01/09/200611:10 PMPage 51Chapter 10 Review ExercisesNamePeriodDate1. (Lessons 10.1, 10.2) Sharon bought a bag of colored balloons for a party.The bag contained 9 white, 39 blue, 24 pink, 21 green, and 57 yellowballoons.a. Determine the percentage of each color of balloon, and use thatinformation to make a circle graph and a relative frequency bar graph.b. What percentage of the balloons are neither pink nor white?c. If Sharon reaches in the bag and pulls out one balloon at random,what is the probability that it will be green?2. (Lesson 10.3) Consider the figure at right.a. If you randomly plot a point in the large rectangle, what is thetheoretical probability that it will land in the shaded region?b. Suppose you randomly plot many points, and 40 of them land in theshaded region. Estimate the total number of points plotted.3. (Lesson 10.4) A high school is having a lottery, where three differentdigits are chosen from the digits 0–9 to create the winning number. Towin, you must correctly guess the winning number.a. Suppose the winning guess must have the same three digits, in thesame order, as the winning number. How many three-digit numberscan be made from the digits 0 to 9, where no digit is used twice? Whatis the probability of winning in this case?b. Now suppose the winning guess must have the same three digits asthe winning number, but the digits can be in any order. What is theprobability of winning in this case?4. (Lessons 10.5, 10.6) Brigham has a bag that contains seven numberedchips. There are five chips labeled with the number 7, and two chips withthe number 4. He reaches into the bag and pulls out a chip, sets that chipaside, and then pulls one more chip out of the bag.a. What is P 72 71 ? What is P 42 41 ?b. Create a tree diagram to calculate the probabilities of differentoutcomes of Brigham’s two-draw experiment.c. What is P 41 and 72 ?d. The numbers Brigham draws can add up to 8, 11, or 14. What is theprobability that the sum will be an even number?e. Find the expected value of the sum. 2007 Key Curriculum PressDiscovering Algebra: A Guide for Parents51

DA2GP 774 10.qxd01/09/200611:10 PMPage 52SOLUTIONS TO CHAPTER 10 REVIEW EXERCISES1. a. There are 150 balloons total. Find the percentage ofeach color by dividing the number of that color bynumber of green21 150. For example, total number 150 0.14, so14% of the balloons are green. Multiply the percentage by 360 to find the angle measure of eachsector. For example, 0.14 360 50.4, so the anglemeasure of the sector representing the relativenumber of green balloons is 50.4 . Sample graphsare shown rcentage30252015103. a. There are 10 choices for the first digit, 9 choices forthe second digit, and 8 choices for the third digit,so the total number of three-digit numbers formedfrom 0 to 9 without repetition is 10 9 8 720.You can also calculate 10P3. The probability of1 guessing the winning number is 720 0.001.b. For each three-digit number, there are 3 2 1 6ways to arrange the digits. Because the order of thedigits does not matter, divide the numberof permutations you found in 3a by 6, to get720 120. You can also calculate 10C3. The proba61 bility of guessing a winning number is 120 0.008.4. a. P 72 71 means “the probability that Brighamdraws a 7 on his second draw, given that he drewa 7 on his first draw.” If Brigham drew a 7 on thefirst draw, then there would be four 7’s and two4’s left in the bag, for a total of six chips. Therefore,P 72 71 46 , or 23 . If Brigham drew a 4 on the firstdraw, then there would be five 7’s and one 4 left inthe bag for his second draw, so P 42 41 16 .b. The first branch of the tree diagram indicatesthe probabilities of the possible outcomes forBrigham’s first draw, and the second branch showsthe probabilities of outcomes for his second draw.1st Draw50WhiteBluePink Green YellowColorb. The pink and white balloons together make up6% 16% 22% of the total, so the percentageof balloons that are neither pink nor white is100% 22% 78%.c. 14% of the balloons are green, so the probabilitythat she pulls out a green one is 14%, or 0.14.2. a. The area of the shaded region is 21 squares, andthe area of the whole rectangle is 8 14 112squares. Therefore, the probability of a randomlyplotted point landing in the shaded region is21 , or 0.1875.112b. Solve the proportion21 shaded squares40 pts in shaded .112 total squaresx total ptsx112 4021112x 40 402140x 213.3 Invert the proportion.Multiply both sides by 40.Multiply.Approximately 213 points were plotted.52Discovering Algebra: A Guide for Parents2nd Draw2P (72 71) 310P(71 and 72) 211P (42 71) 35P(71 and 42) 215P (72 41) 65P(41 and 72) 211P (42 41) 61P(41 and 42) 215P (71) 72P (41) 7c. Multiply the probabilities along the branchesleading to the outcome 41 and 72. P 41 and 72 2 5105 , or .7 64221d. Add the probabilities of outcomes that giveeven sums. P(sum is even) P 71 and 72 P 41 and 42 12 01 21 1 12 11 .e. For each outcome, multiply the sum of the numbers by the probability of the outcome, and thenfind the sum of the results. Expected value P 71 and 72 14 P 71 and 42 11 P 71 and 42 11 P 41 and 42 8 1055186 14 11 11 8 212121217The expected value of the sum of Brigham’s twodraws is 87 6 , or approximately 12.3. 2007 Key Curriculum Press

48 Discovering Algebra: A Guide for Parents 2007 Key Curriculum Press Chapter 10 Probability (continued) Independent Events If you flip a coin repeatedly and get 5 heads in a row, you might say that the chance, or probability, of getting a head on the next flip is very small. After all, the chance of getting 6 heads in a row is very small.