Calculate the Probability

12-3 Probability with Permutations and Combinations

1. GEOMETRY Five students are asked to randomly 2. PLAYS A high school performs a production of A select and name a polygon from the group shown Raisin in the Sun with each freshman English class below. What is the probability that the first two of 18 students. If the three members of the crew are students choose the triangle and quadrilateral, in that decided at random, what is the probability that Chase order? is selected for lighting, Jaden is selected for props, and Emelina for spotlighting?

SOLUTION: Since choosing students for the particular roles is a SOLUTION: way of ranking the members, order in this situation is Find the total number of outcomes. There are 5 important. polygons. Therefore, 5 polygons can be chosen in 5! or 120 ways. The number of possible outcomes in the sample space is the number of permutations of 18 people Find the number of favorable outcomes. This is the taken 3 at a time, 15P3. number of permutations of the other polygons if the triangle is chosen first and the quadrilateral is chosen second: (5 – 2)! or 3!.

Calculate the probability. Among these, there is only one particular arrangement in which Chase is selected for lighting, Jaden is selected for props, and Emelina for spotlighting. Therefore, the probability is

ANSWER:

The probability that the first two students choose the triangle and then the quadrilateral is .

ANSWER:

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3. DRIVING What is the probability that a license 4. CHEMISTRY In chemistry lab, you need to test six plate using the letters C, F, and F and numbers 3, 3, 3, samples that are randomly arranged on a circular and 1 will be CFF3133? tray.

SOLUTION: There are 3 letters and 4 digits This is a permutation with repetition.

There are a total of 7 objects in which F appears twice and 3 appears three times.

The number of distinguishable permutations is a. What is the probability that the arrangement shown is produced? b. What is the probability that test tube 2 will be in The total number of possible outcomes is 420 and the top middle position? there is only one favorable outcome which is CFF3133. Therefore, the probability is SOLUTION: a. The number of distinguishable permutations of n ANSWER: objects arranged in a circle with no fixed reference point is The 6 samples can be arranged in 5! = 120 different ways. There is only one favorable outcome as shown, so the probability is

b. Because the samples are arranged with a fixed reference point (the top middle position), this is a linear permutation. So, there are 6! or 720 ways in which the samples can be arranged. The number of favorable outcomes is the number of permutations of the other 5 samples given that test tube 2 will be in the top middle position, 5! or 120. Therefore, the probability is

ANSWER:

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5. Five hundred boys, including Josh and Sokka, entered 6. CONCERTS Nia and Chad are going to a concert a drawing for two football game tickets. What is the with their high school’s key club. If they choose a probability that the tickets were won by Josh and seat on the row below at random, what is the Sokka? probability that Chad will be in seat C11 and Nia will SOLUTION: be in C12? Two students are chosen out of 500. The order in which the two students are chosen does not matter, so the total number of outcomes is: SOLUTION: Since choosing seats is a way of arranging the students, order in this situation is important.

There is only 1 favorable outcome, so the probability There are 12 seats. The number of possible outcomes is . in the sample space is the number of permutations of 12 seats taken 2 at a time, P . ANSWER: 12 2

Among these, there is only one particular arrangement in which Chad will be in seat C11 and Nia will be in C12. Therefore, the probability is

ANSWER:

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7. FAIRS Alfonso and Colin each bought one raffle second, and Kimi is third? ticket at the state fair. If 50 tickets were randomly SOLUTION: sold, what is the probability that Alfonso got ticket 14 a. The number of possible outcomes is the number of and Colin got ticket 23? arrangements of 7 performers taken 7 at a time. So, SOLUTION: the number of possible outcomes is 7! = 5040. Since buying tickets is a way of arranging the people, order in this situation is important. The first 3 finalists can be arranged in 3! = 6 ways. The rest of the finalists can be arranged 4! ways. There are 50 tickets. The number of possible outcomes in the sample space is the number of Therefore, the number of favorable outcomes is (4!) permutations of 50 tickets taken 2 at a time, 50P2. (3!) = (24)(6) = 144. So, the probability is

b. The number of possible outcomes is 5040. The

number of favorable outcomes with Cecilia first, Among these, there is only one particular Annie second, and Kimi third is 1. After the first arrangement in which Alfonso gets ticket 14 and three places are set, the rest of the finalists can be Colin gets ticket 23. Therefore, the probability is arranged 4! ways. Therefore, the probability is

ANSWER: ANSWER:

8. MODELING The table shows the finalists for a b. floor exercises competition. The order in which they will perform will be chosen randomly.

a. What is the probability that Cecilia, Annie, and Kimi are the first three gymnasts to perform, in any order? b. What is the probability that Cecilia is first, Annie is eSolutions Manual - Powered by Cognero Page 4 12-3 Probability with Permutations and Combinations

9. JOBS A store randomly assigns their employees 11. MAGNETS Santiago bought some letter magnets work identification numbers to track productivity. that he can arrange to form words on his fridge. If he Each number consists of 5 digits ranging from 1–9. If randomly selected a permutation of the letters shown the digits cannot repeat, find the probability that a below, what is the probability that they would form randomly generated number is 25,938. the word BASKETBALL?

SOLUTION: The number of possible outcomes is the number of arrangements of 9 digits taken 5 at a time, 9P5.

SOLUTION: The only favorable outcome is the arrangement There are 10 letters in which A, B, and L each 25,398, so the probability is appear twice.

ANSWER: The number of distinguishable permutations is

The total number of possible outcomes is 453,600 and 10. GROUPS Two people are chosen randomly from a there is only one favorable outcome which is group of ten. What is the probability that Jimmy was BASKETBALL. Therefore, the probability is selected first and George second?

SOLUTION: Because choosing people is a way of ranking the ANSWER: members, order in this situation is important.

The number of possible outcomes in the sample space is the number of permutations of 10 people taken 2 at a time, 10P2.

Among these, there is only one particular arrangement in which Jimmy is selected first and George second. Therefore, the probability is

ANSWER:

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12. ZIP CODES What is the probability that a zip code 14. AMUSEMENT PARKS Sylvie is at an amusement randomly generated from among the digits 3, 7, 3, 9, park with her friends. They go on a ride that has 5, 7, 2, and 3 is the number 39372? bucket seats in a circle. If there are 8 seats, what is the probability that Sylvie will be in the seat farthest SOLUTION: from the entrance to the ride? There are 8 digits in which 3 appears three times and 7 appears twice. SOLUTION: Because the people are seated with a fixed reference The number of distinguishable permutations is point, this is a linear permutation.

So, there are 8! or 40,320 ways in which the people

can be seated. The number of favorable outcomes is The total number of possible outcomes is 3360 and the number of permutations of the other 7 people there is only one favorable outcome which is 39372. given that Sylvie will be in the seat farthest from the Therefore, the probability is entrance to the ride, 7! or 5040. Therefore, the probability is ANSWER: ANSWER:

13. GROUPS Keith is randomly arranging desks into circles for group activities. If there are 7 desks in his circle, what is the probability that Keith will be in the desk closest to the door?

SOLUTION: Because the people are seated with a fixed reference point, this is a linear permutation.

So, there are 7! or 5040 ways in which the people can be seated. The number of favorable outcomes is the number of permutations of the other 6 people given that Keith will be in the desk closest to the door, 6! or 720. Therefore, the probability is

ANSWER:

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15. PHOTOGRAPHY If you are randomly placing 24 16. ROAD TRIPS Rita is going on a road trip across photos in a photo album and you can place four the United States. She needs to choose from 15 cities photos on the first page, what is the probability that where she will stay for one night. If she randomly you choose the photos shown? pulls 3 city brochures from a pile of 15, what is the probability that she chooses Austin, Cheyenne, and Savannah?

SOLUTION: You are choosing 3 cities from a set of 15 cities, and order is not important. So, the number of possible outcomes is 15C3. SOLUTION: You are choosing 4 photos from a set of 24 photos, and the 4 chosen photos can be placed in any order.

So, the number of possible outcomes is 24C4. The number of favorable outcomes is only one, which is choosing the cities Austin, Cheyenne, and Savannah. Therefore, the probability is

There is only one favorable outcome, so the ANSWER: probability is

ANSWER: 17. STRUCTURE Use the figure below. Assume that the balls are aligned at random.

a. What is the probability that in a row of 8 pool balls, the solid 2 and striped 11 would be first and second from the left? b. What is the probability that if the 8 pool balls were mixed up at random, they would end up in the order shown? c. What is the probability that in a row of seven balls, with three 8-balls, three 9-balls, and one 6-ball, the three 8-balls would be to the left of the 6-ball and the three 9-balls would be on the right? d. If the eight original balls were randomly rearranged and formed a circle, what is the probability that the 6-ball is next to the 7-ball?

SOLUTION: eSolutions Manual - Powered by Cognero Page 7 12-3 Probability with Permutations and Combinations

a. The number of possible outcomes in the sample b. space is the number of permutations of 8 balls taken 8 at a time, 8! = 40,320. c.

The number of favorable outcomes is the number of d. permutations of the remaining 6 balls after fixing the solid 2 and striped 11 at the first and second from the 18. How many lines are determined by 10 randomly left, 6! = 720. selected points, no 3 of which are collinear? Explain your calculation. Therefore, the probability is SOLUTION: Two points determine a line. If no 3 points are collinear, then each line will consist of exactly 2 b. The number of possible outcomes is 40,320 and the points. Because there are 10 total points, the number number of favorable outcomes is the only one as of lines is the combination of 10 points taken 2 at a shown. Therefore, the probability is time, which is or 45.

c. The required probability is the probability of getting ANSWER: an arrangement of 8886999 from three 8-balls, three 45; Sample answer: The number of lines is the 9-balls and one 6-ball. So, the number of combination of 10 objects taken 2 at a time, which is distinguishable permutations is or 45. There is only one favorable outcome, which is 8886999, so the probability is

d. Eight balls are arranged in a circle with one fixed ball (7 ball). So, the circular permutation of arranging 8 balls with 1 fixed ball is 7!.

The 6-ball can be arranged either to the left of ball 7 or right of ball 7. So, the circular permutation of arranging 8 balls with 2 fixed balls is .

ANSWER:

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19. Suppose 7 points on a circle are chosen at random, as b. If the carousel is filled randomly, what is the shown. probability that you and your friend will end up in the bench seat? c. If 6 of the 9 people randomly filling the carousel are under the age of 8, what is the probability that a person under the age of 8 will end up on the one horse that does not move up or down?

SOLUTION: a. The number of ways in which the 9 seats can be a. Using the letters A through E, how many ways can filled is the permutation of 9 people taken 9 at a time. the points on the circle be named? So, the number of ways is 9! = 362,880. b. If one point on the circle is fixed, how many arrangements are possible? b. The total number of possible outcomes is 362,880. Because the carousel is filled with a fixed reference SOLUTION: point, this is a linear permutation. a. The number of distinguishable permutations of n objects arranged in a circle with no fixed reference So, there are 7! or 5040 ways to arrange the point is remaining 7 people after you and your friend are placed in the bench seat. Also, for each arrangement there is an alternate arrangement by interchanging The 7 points can be named in a circular way in 6! = the position of you and your friend. Therefore, the 720 different ways. probability is

b. Because the points are named with a fixed reference point, this is a linear permutation. There are c. The total number of arrangements of the 9 people 7! or 5040 ways in which the points can be named. is or 504. The number of favorable outcomes is the number of distinguishable permutations of the ANSWER: other eight places if a child under 8 is on the horse a. 720 b. 5040 that does not move up or down: or 336.

20. RIDES A carousel has 7 horses and one bench seat Calculate the probability. that will hold two people. One of the horses does not move up or down.

ANSWER: a. 362,880 b. c.

a. How many ways can the seats on the carousel be randomly filled by 9 people? eSolutions Manual - Powered by Cognero Page 9 12-3 Probability with Permutations and Combinations

21. LICENSES A camera positioned above a traffic Straight: She has only one chance of winning: light photographs cars that fail to stop at a red light. P(straight) = 1 ÷ 1000 or 0.001. In one unclear photograph, an officer could see that the first letter of the license plate was a Q, the 3-Way Boxed: There are three winning numbers: P(3-Way) = 3 ÷ 1000 or 0.003. second letter was an M or an N, and the third letter was a B, P, or D. The first number was a 0, but the 6-Way Boxed: There are six winning numbers: P(6- last two numbers were blurry. How many possible Way) = 6 ÷ 1000 or 0.006. license plates fit this description?

SOLUTION: Straight-Box: There are still 6 winning combinations. P(Straight-Box) = 6 ÷ 1000 or 0.006. By the Fundamental Counting Principle, the number of possible outcomes in a sample space can be found Because the odds of winning are the same for a 6- by multiplying the number of possible outcomes from Way Boxed and a Straight-Box, Corie should choose each stage or event. either of these options to play.

The first letter of the license plate is Q. ANSWER: There are 2 choices for the second letter. First, find the probability of winning with each option. There are 10³ or 1000 three-digit numbers, so there There are 3 choices for the third letter. are 1000 possible choices. The first number is 0. There are 10 choices for the second digit. Straight: She has only one chance of winning: There are 10 choices for the third digit. P(straight) = 1 ÷ 1000 or 0.001.

Therefore, the total number of choices is 1 × 2 × 3 × 3-Way Boxed: There are three winning numbers: 1 × 10 × 10 = 600. P(3-Way) = 3 ÷ 1000 or 0.003.

ANSWER: 6-Way Boxed: There are six winning numbers: P(6- 600 Way) = 6 ÷ 1000 or 0.006.

22. MULTI-STEP Corie is interested in playing a state Straight-Box: There are still 6 winning lottery game in which she needs to choose a certain combinations. P(Straight-Box) = 6 ÷ 1000 or 0.006. three-digit number to win. The different game options are shown in the table. Which option should Because the odds of winning are the same for a 6- she choose? Explain your solution process. Way Boxed and a Straight-Box, Corie should choose either of these options to play.

SOLUTION: First, find the probability of winning with each option. There are 10³ or 1000 three-digit numbers, so there are 1000 possible choices.

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23. MODELING Fifteen boys and fifteen girls entered 24. REASONING A student claimed that permutations a drawing for four free movie tickets. What is the and combinations were related by . Use probability that all four tickets were won by girls? algebra to show that this is true. Then explain why and differ by the factor r!. SOLUTION:

Four people can be chosen from 30 people in 30C4 SOLUTION: ways and 4 girls can be chosen from 15 girls in 15C4 Sample answer: ways.

The number of possible outcomes is 27,405 and the number of favorable outcomes is 1365. Therefore, nCr and nPr differ by the factor r! because there are the probability is always r! ways to order the groups that are selected. ANSWER: Therefore, there are r! permutations of each combination.

ANSWER: Sample answer:

nCr and nPr differ by the factor r! because there are always r! ways to order the groups that are selected. Therefore, there are r! permutations of each combination.

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25. OPEN-ENDED Describe a situation in which the 26. CONSTRUCT ARGUMENTS Is the following probability is given by . statement sometimes, always, or never true? Explain. SOLUTION: In this situation, you have 7 objects taken 3 at a time nPr = nCr and order is not important because it is a combination.

SOLUTION: Sample answer: A bag contains seven marbles that are red, orange, yellow, green, blue, purple, and black. The probability that the orange, blue, and black marbles will be chosen if three marbles are drawn at random can be calculated using a combination.

ANSWER: Sample answer: A bag contains seven marbles that are red, orange, yellow, green, blue, purple, and black. The probability that the orange, blue, and black marbles will be chosen if three marbles are drawn at random can be calculated using a combination. r! = 1 when r = 0 or 1. However, objects cannot be taken 0 at a time.

The statement is true when r is 1. Therefore, the statement is sometimes true.

ANSWER: Sometimes; sample answer: The statement is true when r is 1.

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27. PROOF Prove that . 29. Steven, Yasmin, Tyler, Vanessa, and Zack have been nominated to be club officers (president and vice SOLUTION: president). Assuming the officers are chosen at Set the two sides equal to each other. Develop each random from among the five nominees, what is the side using the formulas in this lesson and simplify. probability that Steven is chosen to be the president Working with the left side will eventually reduce it to and Yasmin is chosen to be the vice president?

the right side. A

SOLUTION: Step 1: Because choosing officers is a way of ANSWER: ranking team members, order in this situation is important.

The number of possible outcomes in the sample space is the number of permutations of 5 people taken 2 at a time, which is .

Step 2: The number of favorable outcomes is the number of permutations of the 2 students in their 28. WRITING IN MATH Compare and contrast specific positions. This is 1!, or 1. permutations and combinations. Step 3: The probability of Steven being chosen as SOLUTION: Sample answer: Both permutations and combinations the president and Yasmin as the vice president is .

are used to find the number of possible arrangements

of a group of objects. The order of the objects is So, the correct answer is choice D. important in permutations but not in combinations. ANSWER: ANSWER: D Sample answer: Both permutations and combinations are used to find the number of possible arrangements of a group of objects. The order of the objects is important in permutations but not in combinations.

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30. Manuel must choose two books from a list of seven 31. Every student at Shellie’s school is assigned a 4-digit to read for book reports in his literature class. How PIN to access the school’s computers. The first digit many different pairs of books could he choose? of the code is 0 or 1. The remaining digits are chosen from the numbers 2 through 9, and no number may be A 10 used more than once in a code. What is the B 21 probability that Shellie is assigned the PIN 1234? C 42 D 5040 A

SOLUTION: B Because the order in which the books are chosen does not matter, the number of possible outcomes in the sample space is the number of combinations of 7 C books taken 2 at a time.

So, the correct answer is choice B. SOLUTION: Step 1: First, consider how many choices Shellie has ANSWER: for each digit in her PIN: B For the first digit, there are 2 numbers to pick from, 0 or 1. So, this is .

For the last three digits, there are 8 numbers to pick from, 2 through 9. So, this is .

So, there are 672 possible PINs.

Step 2: There is only 1 favorable arrangement, 1234.

Step 3: The probability that Shellie is assigned PIN 1234 = .

So, the correct answer is choice D.

ANSWER: D

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32. The coach is going to select a team of 5 from among 33. Marsala inserts three $5-bills into 3 of 10 envelopes. 10 players. Find the probability that John, a particular The other 7 she leaves empty. She has each of her 10 player, is on the team. soccer teammates choose one of the envelopes. How many different ways can the money be won by her SOLUTION: teammates? Because order does not matter, the number of possible outcomes in the sample space is the number A 120 of combinations of 10 people taken 5 at a time, . B 35 C 720 D 3,628,800

SOLUTION: Because John must occupy one of the places on the Because the order of the envelopes with the $5 in team, you then need to determine . them does not matter, the number of possible outcomes in the sample space is the number of combinations of 10 envelopes taken 3 at a time, . So, the probability that John will be on the team is .

So, the correct answer is choice A. ANSWER: ANSWER: A

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34. Midori has tiles with the letters J, K, L, M, N, P, and 35. MULTI-STEP Assuming that any arrangement of Q. She chooses three of the tiles at random. What is letters forms a ‘word’, what is the probability that a the probability that the tiles name three collinear ‘word’ formed from the letters of the word points on the coordinate plane shown here? SQUARE:

a. starts with E? b. starts with E and ends with A?

SOLUTION: a. Order is important in this situation, so the number of possible outcomes in the sample space is the number of permutations of 6 letters taken 1 at a time, .

The number of favorable outcomes is 1, which is the B letter E, so the probability is .

C b. Given that the probability of the word starting with E is , we must now consider the remainder of the D word. Order is important, so the number of possible outcomes in the new sample space is the number of SOLUTION: permutations of 5 letters taken 1 at a time, . Because the order does not matter, the number of possible outcomes in the sample space is the number of combinations of 7 letters taken 3 at a time, .

Again, the number of favorable outcomes is 1, which is the letter A, so the probability is .

There are only 2 favorable outcomes, collinear points To find the probability of both events, multiply. J, K, L or collinear points L, M, N, so the probability

is .

The correct answer is choice C. ANSWER: ANSWER: a. C b.

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36. 15 student council members select a committee of 3 for a project, and then select one of the three to be the liaison for the project.

a. In how many ways is this possible? b. What is the probability that Carmen is in the committee? c. What is the probability that Carmen is the liaison for the project?

SOLUTION: a. Because the order does not matter, the number of possible outcomes in the sample space is the number of combinations of 15 members taken 3 at a time, .

Because there are 3 possible favorable outcomes, multiply 455 × 3, or 1365.

b. The probability that Carmen is in the committee is .

c. To determine the probability that Carmen is the liaison, multiply .

ANSWER: a. 1365 b. c.