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2 Answers Chapter 10 Probability prepared by subject experts adhering to the latest Common Core Curriculum. Solve the BIM Book Algebra 2 Chapter 10 Probability Questions available here and improve your conceptual knowledge. To make it simple for you we have sequenced the Big Ideas Math Algebra 2 Chapter 10 Probability Answer Key as per the BIM Algebra 2 Textbooks. Have a sneak peek into the Topicwise Big Ideas Math Algebra 2 Answers Ch 10 Probability through the quick links
available. Refer to the Big Ideas Math Book Algebra 2 Ch 10 Probability Solutions prepared keeping in mind student’s level of understanding. All of them are given by subject expertise after ample research. So you don’t have to worry about their accuracy. Assess your preparation standard using the Chapter 10 Probability Big Ideas Math Answer Key and work on the areas you are lagging by allotting time accordingly. Prepare whichever topic you face difficulty and practice them on a regular basis so that you
no longer feel them difficult. Write
and solve a proportion to answer the question. Question 3. 34.4:86*100 = (34.4*100):86 = 3440:86 =
40 Display the data in a histogram. Answer: Question 5. Probability Mathematical PracticesMathematically proficient students apply the mathematics they know to solve real-life problems. Monitoring Progress In Exercises 1 and 2, describe the event as unlikely, equally likely to happen or not happen, or likely. Explain your reasoning. Question 2. Question 3. Lesson 10.1 Sample Spaces and ProbabilityEssential Question How can you list the possible outcomes in the sample space of an experiment? EXPLORATION 1 Finding the Sample Space of an Experiment The sample space of a fair coin flip is {H, T}. The sample space of a sequence of three fair coin flips is all 23 possible sequences of outcomes: {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}. The sample space of a sequence of five fair coin flips in which at least four flips are heads is {HHHHH,HHHHT,HHHTH,HHTHH,HTHHH,THHHH}. EXPLORATION 2 Finding the Sample Space of an Experiment a)1,2,3,4,5,6 b) Total 36 EXPLORATION 3 Finding the Sample Space of an Experiment a. How many ways can you spin a 1? 2? 3? 4? 5? Answer: 1, 2, 3, 2, 4 b. List the sample space. Answer: 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5 c. What is the total number of outcomes? Answer: 12 EXPLORATION 4 Finding the Sample Space of an Experiment a. How many ways can you choose two blue? a red then blue? a blue then red? two red? Answer: BB – 2, RB – 10, BR – 10, RR – 20 b. List the sample space. c. What is the total number of outcomes? Answer: 42 Communicate Your Answer Question 5. Question 6. Answer: Number 1 could be spun in 1 way, out of a total of 12 ways r1 = 1/12 Number 2 could be spun in 2 way, out of a total of 12 ways r2 = 2/12 Number 3 could be spun in 3 way, out of a total of 12 ways r3 = 3/12 Number 4 could be spun in 2 way, out of a total of 12 ways r4 = 2/12 Number 5 could be spun in 4 way, out of a total of 12 ways r5 = 4/12 The sum of the ratios r1, r2, r3, r4 and r5 is 1/12 + 2/12 + 3/12 + 2/12 + 4/12 = 1 Two red balls can be drawn in 20 ways, out of a total of 42 ways rBB = 20/42 Red then Blue can be drawn in 10 ways, out of a total of 42 ways rRB = 10/42 Blue then Red can be drawn in 10 ways, out of a total of 42 ways rRB = 10/42 Two Red balls can be drawn in 2 ways, out of a total of 42 ways rRR = 2/42 20/42 + 10/42 + 10/42 + 2/42 = 1 Thus the sum of both explorations are same. Monitoring Progress Find the number of possible outcomes in the sample space. Then list the possible outcomes. -All of the outcomes of this experiment are shown below as a list . if “H” represent getting a head , and “T” represents getting a tail ; the following represent all of outcomes: (HH,HT,TH,TT) Question 2. Question 3. Find P(\(\bar{A}\)). Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Sample Spaces and Probability 10.1 ExercisesVocabulary and Core Concept Check Question 2. Theoretical probability is when the outcomes are equally likely and can be found using a formula. Monitoring Progress and Modeling with Mathematics In Exercises 3–6, find the number of possible outcomes in the sample space. Then list the possible outcomes. Question 4. Question 5. Question 6. Question 7. Answer: Question 8. Question 9. Question 10. a. A person chosen at random is at least 15 years old. b. A person chosen at random is from 25 to 44 years old. Answer: a)For this we will calculate the complementary probability. The probability of someone being under 5 is 0.07. The probability of someone being between 5 and 14 is 0.13. To get the probability of a random person being over 15 we will add these probabilities up and substract them from 1. 1-(0.07+0.13)=0.8 The probability of a random person being at least 15 is 0.8. b) To calculate this probability we add up the probability of someone being between 25 and 34, and 35 and 44 . Both possibilities are 0.13. 0.13+0.13=0.26 The probability of someone being between 25 and 44 is 0.26. Question 11. Answer: Question 12. Answer: The probability is 13/15. Explanation: To calculate the complementary probability we are looking for the probability of a number being less or equal to 4. The probability of this is 4/30. Now we can substract this value from 1 1-4/30= 13/15. Question 13. Answer: Question 14. a. Texas b. Alabama c. Florida d. Louisiana Answer: First , from the given image. let’s define the length of shoreline for each state: Texas=367mi Louisiana= 397mi Alabama=53mi Florida=770mi Mississippi=44mi Explanation: Total length of shoreline =367+397+56+770+44=1631mi By using formula P(The ship land in three state)= shoreline length of the state/Total length of shoreline We can determine the probability that a ship coming ashore at a random point in the gulf of Mexico lands in the given state. 2) a)P(The ship lands in Texas )=shoreline length of the Texas/Total length of shoreline=367/1631 =0.23 =23% b)P(The ship lands in Alabama)=shoreline length of the Alabama/Total length of shoreline =53/1631=0.03=3% c)P(The ship lands in Florida ) = shore length of the Florida /Total length of shoreline =770/1631=0.47=47% d)P(The ship lands in Louisiana)= shore length of the Louisiana/Total length of shoreline=397/1631=0.24=24% Question 15. Answer: Question 16. Answer: Black Explanation: Question 17. a. What is the theoretical probability that the spinner stops on a multiple of 3? b. You spin the spinner 30 times. It stops on a multiple of 3 twenty times. What is the experimental probability of stopping on a multiple of 3? c. Explain why the probability you found in part (b) is different than the probability you found in part (a). Answer: Question 18. Question 19. Answer: Question 20. Answer: Italian = 526 Explanation: To get the probability of this happening, we need to divide the number of events that satisfy this event by the number of total events 526/2392=0.22 Question 21. Question 22. A. It rains on Sunday. B. It does not rain on Saturday. C. It rains on Monday. D. It does not rain on Friday. Answer: P(B)<P(A)<P(C)<P(D) Explanation: A. The probability it rains on Sunday is 80%= 0.8 ;P(a)=0.8 B. The probability it rains on Saturday is 30%=0.3,so the probability that it doesn’t rain on Saturday is 1- 0.3=0.7 ;P(B)=0.7. C. The probability it rains on Monday is 90 %=0.9; P (c)=0.9 D. The probability it rains on Friday is 5% =0.05 So, the probability that it doesn’t rain on Saturday is 1-0.05= 0.95 ; P(D)=0.95 Question 23. USING TOOLS Use the figure in Example 3 to answer each question. a. List the possible sums that result from rolling two six-sided dice. b. Find the theoretical probability of rolling each sum. c. The table below shows a simulation of rolling two six-sided dice three times. Use a random number generator to simulate rolling two six-sided dice 50 times. Compare the experimental probabilities of rolling each sum with the theoretical probabilities. Answer: Question 24. Question 25. Answer: Question 26. Answer: Given that y=f(x)+c intersect x- axis Question 27. Question 28. Answer: Maintaining Mathematical Proficiency Find the
product or quotient. Question 30. Question 31. Question 32. Question 33. Question 34. Lesson 10.2 Independent and Dependent EventsEssential Question How can you determine whether two events are independent or dependent? EXPLORATION 1 Identifying
Independent and Dependent Events a. Two six-sided dice are rolled. Answer: P = Number of outcomes that satisfy the requirements/Total number of possible outcomes Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die. When two dice are thrown simultaneously, thus the number of event can be 62 = 36 because each die has 1 to 6 number on its faces. This is independent event b. Six pieces of paper, numbered 1 through 6, are in a bag. Two pieces of paper are selected one at a time without replacement. Answer: EXPLORATION 2 Finding Experimental Probabilities The experimental probability that the sum of two number rolled is 7 P(sum = 7) = 3/30 = 1/10 = 10% b. In Exploration 1(b), experimentally estimate the probability that the sum of the two numbers selected is 7. Describe your experiment. Answer:
The experimental probability that the sum of the two numbers selected is 7, EXPLORATION 3 Finding Theoretical Probabilities Communicate Your Answer Question 4. Question 5. Answer: Independent b. Your teacher chooses a student to lead a group. chooses another student to lead a second group. and chooses a third student to lead a third group. Monitoring Progress Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Independent and Dependent Events 10.2 ExercisesVocabulary and Core Concept Check Question 2. Monitoring Progress and Modeling with Mathematics In Exercises 3–6, tell
whether the events are independent or dependent. Explain your reasoning. Question 4. Answer: The two event that considered in this experiment are on example of independent events. The occurrence of the four in the throwing of the die does not influence the out come of Tails in tossing of the coin. The Two events that considered in this experiment are Independent Events. Question 5. Answer: Question 6. In Exercises 7–10, determine whether the events are independent. Answer: Question 8. Question 9. Question 10. Question 11. Answer: Question 12. Question 13. Question 14. Answer: Consonants (A) = 56 Vowels (B) = 42 Blank = 2 Total letters = 100 Two letters are drawn with replacement. P(A) = 56/100 = 0.56 P(B) = 42/(100-1) = 0.424 P(A).P(B) = 0.56*0.424 = 0.2376 Question
15. Answer: Question 16. Answer: Question 17. Question 18. Question 19. Question 20. Question 21. Answer: Question 22. Answer: Let A be the event that a future tropical cyclone in the Northern Hemisphere is a hurricane. There are 821 total cyclones in Northern hemisphere Total outcomes = 821 Total number of hurricane cyclone in northern hemisphere is a hurricane P(A) = 379/821 = 0.462 Question 23. Question 24. Question 25. Question 26. Answer: It is given that a bag contains one red marble and one blue marble. The diagrams show the possible outcomes of randomly choosing two marbles using different methods. As per the diagram, we observe that If firstly red marble is drawn randomly, then we get blue marble on the 2nd draw. Blue marble is drawn randomly, then we get red marble on the 2nd draw. That means after choosing red marble we then get the blue marble only. After choosing blue marble we then get the red marble only. It means that red marble was not put back in the bag after getting selected. Blue marble was not put back in the bag after getting selected. b. We have to determine whether the marbles were selected with or without replacement, for each method. “Without replacement ” means that you don’t put the ball or balls back in the box so that the number of balls in the box gets less as each ball is removed. “With replacement ” means that you put the ball or balls back in the box. According to the diagram, we can see that If firstly, red marble is drawn randomly, then we get either red marble or blue marble on the 2nd draw. If firstly, blue marble is drawn randomly, then we get also get either red marble or blue marble on the 2nd draw. That means, after choosing red marble, we then get any of the marbles, either red or blue. And, after choosing blue marble, we then also get any of the marbles, either red or blue. That means red marble was put back in the bag after getting selected. Similarly, blue marble was put back in the bag after getting selected. This shows that marbles were selected “With Replacement” in each method. Question 27. MAKING AN ARGUMENT A meteorologist claims that there is a 70% chance of rain. When it rains, there is a 75% chance that your softball game will be rescheduled. Your friend believes the game is more likely to be rescheduled than played. Is your friend correct? Explain your reasoning. Answer: Question 28. Answer: Question 29. a. If the team goes for 1 point after each touchdown, what is the probability that the team wins? loses? ties? b. If the team goes for 2 points after each touchdown, what is the probability that the team wins? loses? ties? c. Can you develop a strategy so that the coach’s team has a probability of winning the game that is greater than the probability of losing? If so, explain your strategy and calculate the probabilities of winning and losing the game. Answer: Question 30. Maintaining Mathematical Proficiency Solve the equation. Check your solution. Question 32. Question 33. Lesson 10.3 Two-Way Tables and ProbabilityEssential Question How can you construct and interpret a two-way table? EXPLORATION 1 Completing and Using a Two-Way Table a. How many students play an instrument? Answer: 41 b. How many students speak a foreign language? c. How many students play an instrument and speak a foreign language? d. How many students do not play an instrument and do not speak a foreign language? e. How many students play an instrument and do not speak a foreign language? EXPLORATION 2 Two-Way Tables and Probability Answer: 41/80 b. speaks a foreign language. c. plays an instrument and speaks a foreign language. d. does not play an instrument and does not speak a foreign language. e. plays an
instrument and does not speak a foreign language. EXPLORATION 3 Conducting a Survey Answer: Now the Venn diagram and two-way table interpret that out of 50 students surveyed, 28 students like maths 22 does not like maths, 26 students like history and 24 do not like history. Communicate Your Answer Question 4. Identify the variables. There are two variables of interest here: the commercial viewed and opinion. Question 5.
Monitoring Progress Question 1. Question 2. Question 3. Question 4. Question 5. Question
6. Question 7. Answer: Two-Way Tables and Probability 10.3 ExercisesVocabulary and Core Concept Check Question 2. Monitoring Progress and Modeling with Mathematics In
Exercises 3 and 4, complete the two-way table. Answer: Question 4. Answer: Number of students who said no is 49 – 7 = 42 Total number of students is 56 + 42 = 98 Total number of people is 98 + 10 = 108 Out of the total number of people, 49 of them said no, Total number of people who said yes is 108 – 49 = 59 Number of teachers who said yes is 59 – 56 = 3 Question 5. Answer: Question 6. USING
STRUCTURE In Exercises 7 and 8, use the two-way table to create a two-way table that shows the joint and marginal relative frequencies. Answer: Question 8. Answer: Question 9. Question 10. Answer: Question 11. Question 12. Question 13. a. What is the probability that a randomly selected patient located in Saratoga was satisfied with the communication of the physician? b. What is the probability that a randomly selected patient who was not satisfied with the physician’s communication is located in Glens Falls? c. Determine whether being satisfied with the communication of the physician and living in Saratoga are independent events. Answer: Question 14. a. What is the probability that a randomly selected student who lives in Nebraska plans to stay in his or her home state after graduation? b. What is the probability that a randomly selected student who does not plan to stay in his or her home state after graduation lives in North Carolina? c. Determine whether planning to stay in their home state and living in Nebraska are independent events. Answer: ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in finding the given conditional probability. Question 15. Answer: Question 16. Answer: Question 17. Answer: Question 18. Answer: Question 19. Question 20. a. What does 120 represent? b. What does 1336 represent? c. What does 1501 represent? Answer: Question 21. Answer: Question 22. Question 23. a.What is the probability that a randomly selected person does not own either pet? b. What is the probability that a randomly selected person who owns a dog also owns a cat? Answer: Question 24. Question 25. The company is deciding whether it should try to improve the snack before marketing it, and to whom the snack should be marketed. Use probability to explain the decisions the company should make when the total size of the snack’s market is expected to (a) change very little, and (b) expand very rapidly. Answer: Question 26. P(Cat owner) = 61/210 = 0.29 P(Dog owner) = 93/210 = 0.442 P(Cat Owner/Dog owner) = P(Dog owner and cat owner)/P(Dog owner) = 0.387 P(Dog owner/Cat owner) = P(Cat owner/Dog owner)P(Dog owner)/P(Cat owner) = 0.387 × 0.442/0.29 = 0.5898 Maintaining Mathematical Proficiency Draw a Venn diagram of the sets described. Question 28. Question
29. Probability Study Skills: Making a Mental Cheat Sheet10.1–10.3 What Did You Learn? Core Vocabulary Core Concepts Section 10.2 Section 10.3 Mathematical Practices Question 2. Study Skills: Making a Mental Cheat Sheet
Probability 10.1–10.3 QuizQuestion 1. Find P(\(\bar{A}\)). Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. a. the center circle Answer: b. outside the square c.
inside the square but outside the center circle Question 9. a. Find the probability that a female student, chosen at random from the students surveyed, eats fruit every day. Answer: b. Find the probability that a 15 – year – old student. chosen at random from the students surveyed, eats fruit every day. Question 10. Question 11. Lesson 10.4 Probability of Disjoint and Overlapping EventsEssential Question How can you find probabilities of disjoint and overlapping events? EXPLORATION 1 Disjoint Events and Overlapping Events a. Event A: The result is an even number. Event B: The result is a prime number. Answer: b. Event A: The result is 2 or 4. EXPLORATION 2 Finding the Probability that Two Events Occur Answer: b. Event A: The result is 2 or 4. EXPLORATION 3 Discovering Probability Formulas Answer: b. In general, if event A and event B are overlapping, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion. c. Conduct an
experiment using a six-sided die. Roll the die 50 times and record the results. Then use the results to find the probabilities described in Exploration 2. How closely do your experimental probabilities compare to the theoretical probabilities you found in Exploration 2? a. P(A) = \(\frac{1}{2}\) = 50% P(A) = \(\frac{21}{50}\) = 42% P(B) = \(\frac{1}{2}\) = 50% P(B) = \(\frac{32}{50}\) = 64% P(A and B) = \(\frac{1}{6}\) ≈ 16.7% P(A and B) = \(\frac{9}{50}\) ≈ 18% P(A or B) = \(\frac{5}{6}\) ≈ 83.3% P(A or B) = \(\frac{44}{50}\) ≈ 88% P(A) = \(\frac{1}{3}\) ≈ 33.3% P(A) = \(\frac{17}{50}\) = 34% P(B) = \(\frac{1}{2}\) = 50% P(B) = \(\frac{29}{50}\) = 58% P(A and B) = 0 = 0% P(A and B) = \(\frac{0}{50}\) = 0% P(A or B) = \(\frac{5}{6}\) ≈ 83.3% P(A or B) = \(\frac{46}{50}\) = 92% Communicate Your Answer Question 4. Question 5. a. Event A: The result is an even number. b. Event A: The result is 2 or 4. Monitoring Progress A card is randomly selected from a standard deck of 52 playing cards. Find the probability of the event. Question 2. Question 3. Question 4. Question 5. Probability of Disjoint and Overlapping Events 10.4 ExercisesVocabulary and Core Concept Check Question 2. Answer: The union of events A and B , denoted by AUB, consists of all outcomes that are in A or in B or in both A and B . so, there are 4+2+3=9 outcomes in the union of A and B. Outcomes in intersection of A and B =Outcomes shared by both A and B = Outcomes in B also in A =2 Outcomes in union of A nd B =9. Monitoring Progress and Modeling with Mathematics In Exercises 3–6, events A and B
are disjoint. Find P(A or B). Question 4. Question 5. Question 6. Question 7. PROBLEM SOLVING Your dart is equally likely to hit any point inside the board shown. You throw a dart and pop a balloon. What is the probability that the balloon is red or blue? Answer: Question 8. Question 9. PROBLEM SOLVING You are performing an experiment to determine how well plants grow under different light sources. Of the 30 plants in the experiment, 12 receive visible light, 15 receive ultraviolet light, and 6 receive both visible and ultraviolet light. What is the probability that a plant in the experiment receives visible or ultraviolet light? Answer: Question 10. ERROR ANALYSIS In Exercises 11 and 12, describe and correct the error in finding the probability of randomly drawing the given card from a standard deck of 52 playing cards. Answer: Question 12. In Exercises 13 and 14, you roll a six-sided die. Find P(A or B). Question
14. Question 15. Answer: Question 16. Question 17. Question 18. a. Find the probability that the Redbirds score and the Bluebirds do not score when the coach leaves the goalie in. b. Find the probability that the Redbirds score and the Bluebirds do not score when the coach takes the goalie out. c. Based on parts (a) and (b), what should the coach do? Answer: Question 19. Question 20. Answer: No; The intersection of A and B is not empty. Question 21. Answer: Question 22. Question 23. Maintaining Mathematical Proficiency Write the first six terms of the sequence. Question 25. Question 26. Lesson 10.5 Permutations and CombinationsEssential Question How can a tree diagram help you visualize the number of ways in which two or more events can occur? EXPLORATION 1 Reading a Tree
Diagram a. How many outcomes are possible? Answer: There are 4 possible outcomes when two coins are flipped. There are three possible outcomes for the spinner. Therefore the possible number of outcomes are, 4 × 3 = 12 b. List the possible outcomes. Answer: The possible outcomes when two coins are flipped are HH, HT, TH, TT. Since the numbers on the spinner are 1, 2 and 3. Therefore the possible outcomes when the spinner is spun are 1, 2, 3. The possible outcomes are (H, H, 1), (H, H, 2), (H, H, 3) (H, T, 1), (H, T, 2), (H, T, 3) (T, H, 1), (T, H, 2), (T, H, 3) (T, T, 1), (T, T, 2), (T, T, 3) EXPLORATION 2 Reading a Tree Diagram a. How many events are shown? Answer: In the given tree diagram, there is an initial event having two branches and its each branch is followed by the second event having three branches. Further, each of the three branches of the second event are followed by the third event having two branches. Lastly, these two branches are followed by a fourth event having two branches. Thus there are total of four branches in a given tree diagram. b. What outcomes are possible for each event? Answer: We know that the number of nodes at the end of the branch(s) representing an event in a tree diagram gives the number of its possible outcomes. Number of possible outcomes of a first event =2 Number of possible outcomes of a second event =3 Number of possible outcomes of a third event =2 Number of possible outcomes of a fourth event =2 Thus the number of possible outcomes in first, second, third and fourth event are 2, 3, 2 and 2. c. How many outcomes are possible? Answer: The number of nodes at the end of the tree diagram gives the number of its possible outcomes Number of final nodes = Number of possible outcomes Number of possible outcomes =24 d. List the possible outcomes. Answer: There are four events. The first event has 2 outcomes which are A and B. The second event has 3 outcomes which are1,2 and3. The third event has 2 outcomes which are X and Y. The fourth event has 2 outcomes which are A and B.The number of possible outcomes =24 All the possible outcomes are given by {A1, XA, A1XB, A1YA, A1YB, A2XA, A2XB, A2YA, A2YB, A3XA, A3XB, A3YA, A3YB, B1XA, B1XB, B1YB, B1YB, B2XA, B2XB, B2YA, B2YB, B3XA, B3XB, B3YA, B3YB} EXPLORATION 3 Writing a Conjecture a. Consider the following general problem: Event 1 can occur in m ways and event 2 can occur in n ways. Write a conjecture about the number of ways the two events can occur. Explain your reasoning. b. Use the conjecture you wrote in part (a) to write a conjecture about the number of ways more than two events can occur. Explain your reasoning. c. Use the results of Explorations 1(a) and 2(c) to verify your conjectures. Communicate Your Answer Question 4. Question 5. Monitoring Progress Question 1. Question 2. In how many ways can you arrange 3 of the letters in the word MARCH? Answer: Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Permutations and Combinations 10.5 ExercisesVocabulary and Core Concept Check Question 2. Answer: 7!/2!.5 This option is the only one that is not a form of a combination Monitoring Progress and Modeling with Mathematics In Exercises 3–8, find the number of ways you can arrange (a) all of the letters and (b) 2 of the letters in the given word. Question 4. Question 5. Question 6. Question 7. Question 8. In Exercises 9–16, evaluate the expression. Question 10. Question 11. 9P1 Answer: Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Answer: Question 19. Question 20. Answer: In Exercises 21–24, count the possible combinations of r letters chosen from the given list. Question 22. Question 23. Question 24. In Exercises 25–32, evaluate the expression. Question 26. Question 27. Question 28. Question 29. Question 30. Question 31. Question 32. Question 33. PROBLEM SOLVING Each year, 64 golfers participate in a golf tournament. The golfers play in groups of 4.How many groups of 4 golfers are possible? Answer: Question 34. ERROR ANALYSIS In Exercises 35 and 36, describe and correct the error in evaluating the expression. Answer: Question 36. Answer: The permutation formula was used instead of the combination formula. REASONING In Exercises 37–40, tell whether the question can be answered using permutations or combinations. Explain your reasoning. Then answer the question. Question 38. Question 39. Question 40. Question 41. Question 42. Question 43. Question 44. Question 45. Answer: Question 46. Question 47. Answer: Question 48. REASONING In Exercises 49 and 50, find the probability of winning a lottery using the given rules. Assume that lottery numbers are selected at random. Question 50. In Exercises 51–58, use the Binomial Theorem to write the binomial expansion. Question 52. Question
53. Question 54. Question 55. Question 56. Question
57. Question 58. In Exercises 59–66, use the given value of n to find the coefficient of xn in the expansion of the binomial. Question 60. Question 61. Question 62. Question 63. Question 64. Question 65. Question 66. Question 67. Question 68. b. Use your result from part (a) to write an explicit rule for the nth triangular number Tn. Answer: Question
69. a. Use the combinations formula to write an expression for the number of diagonals in an n-sided polygon. b. Use your result from part (a) to write a formula for the number of diagonals of an n-sided convex polygon. Answer: Question 70. Answer: You are ordering a burrito with 2 main ingredients and 3 toppings. Total number of main ingredients = 6 As the order in which the ingredients are chosen is not important, the total number of ways to select 2 main ingredients out of 6 can be found by using the combinations formula. nCa = n!/a!(n – a)! 6C2 = \(\frac{6!}{(6-2) !}\) = \(\frac{6!}{4!}\) . \(\frac{1}{2!}\) =6 . 5 . 4 . 3 . 2 . 1/(4 . 3 . 2 . 1)(2 . 1) = 3 . 5 6C2 = 15 Total number of toppings = 8 Number of toppings to be chosen = 3 nCa = n!/a!(n – a)! 8C3 = \(\frac{8!}{(8-3) !}\) = \(\frac{8!}{5!}\) . \(\frac{1}{3!}\) =8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1)(3 . 2 . 1) = 8 . 7 8C3 = 56 Total possible selections = ways to select main ingredients × Ways to select toppings = 15 × 56 = 840 Question 71. Answer: Question 72. Question 73. Question 74. a. How many combinations of three marbles can be drawn from the bag? Explain. b. How many permutations of three marbles can be drawn from the bag? Explain. Answer: Question 75. Question 76. Question 77. Answer: Question 78. Question 79. Maintaining Mathematical Proficiency Question 80. Question 81. Answer: Lesson 10.6 Binomial DistributionsEssential Question How can you determine the frequency of each outcome of an event? EXPLORATION 1 Analyzing Histograms a. In how many ways can 3 heads occur when 5 coins are flipped? Answer: b. Draw a histogram that shows the numbers of heads that can occur when 6 coins are flipped. c. In how many ways can 3 heads occur when 6 coins are flipped?
6C3 = (6×5×4)/(1×2×3) EXPLORATION 2 Determining the Number of Occurrences Answer: b. Determine the pattern shown in the table. Use your result to find the number of ways in which 2 heads can occur when 8 coins are flipped. Communicate Your Answer Question 3. Question 4. Monitoring Progress An octahedral die has eight sides numbered 1 through 8. Let x be a random variable that represents the sum when two such dice are rolled. Question 1. Make a table and draw a histogram showing the probability distribution for x. Answer: Question 2. Question 3. What is the probability that the sum of the two dice is at most 3? Answer: According to a survey, about 85% of people ages 18 and older in the U.S. use the Internet or e-mail. You ask 4 randomly chosen people (ages 18 and older) whether they use the Internet or e-mail. Question 5. Question
6. Binomial Distributions 10.6 ExercisesVocabulary and Core Concept Check Question 2. Monitoring Progress and Modeling with Mathematics In Exercises 3–6, make a table and draw a histogram showing
the probability distribution for the random variable. Question 4. w = 1 when a randomly chosen letter from the English alphabet is a vowel and w = 2 otherwise. Question 6. In
Exercises 7 and 8, use the probability distribution to determine (a) the number that is most likely to be spun on a spinner, and (b) the probability of spinning an even number. Answer: Question 8. Answer: USING EQUATIONS In Exercises 9–12, calculate the probability of flipping a coin 20 times and getting the given number of
heads. Question 10. Question 11. Question 12. Question 13. a. Draw a histogram of the binomial distribution for your survey. b. What is the most likely outcome of your survey? c. What is the probability that at most 2 people have a class ring? Answer: Question 14. ERROR
ANALYSIS In Exercises 15 and 16, describe and correct the error in calculating the probability of rolling a 1 exactly 3 times in 5 rolls of a six-sided die. Answer: Question 16. Answer: Question 17. a. Find P(x) for x= 0, 1, 2, . . . , 7. b. Make a table showing the probability distribution for x. c. Make a histogram showing the probability distribution for x. Answer: Question 18. Answer: Question 19. Answer: Question 20. Question 21. Question 22. Maintaining Mathematical Proficiency List the possible outcomes for the situation. Question 24. Probability Performance Task: A New Dartboard10.4–10.6 What Did You Learn? Core Vocabulary Core Concepts Section 10.6 Mathematical
Practices Question 2. Question 3. Performance Task A New Dartboard You are a graphic artist working for a company on a new design for the board in the game of darts. You are eager to begin the project, but the team cannot decide on the terms of the game. Everyone agrees that the board should have four colors. But some want the probabilities of hitting each color to be equal, while others want them to be different. You offer to design two boards, one for each group. How do you get started? How creative can you be with your designs? Probability Chapter Review10.1 Sample Spaces and Probability (pp. 537–544) Question 1. Question 2. Answer: 10.2 Independent and Dependent Events (pp. 545–552) Find the probability of randomly selecting the given marbles from a bag of 5 red, 8 green, and 3 blue marbles when (a) you replace the first marble before drawing the second, and (b) you do not replace the first
marble. Compare the probabilities. Question 4. Question 5. 10.3 Two-Way Tables and Probability (pp. 553–560) Question 6. Question 7. 10.4 Probability of Disjoint and Overlapping Events (pp. 563–568) Question 8. Question 9. 92/100 = 14/100 + 80/100 – P(A and B) 10.5 Permutations and Combinations (pp. 569–578) Evaluate the expression. Question 11. Question 12. 6C2 Answer: Question 13. Question 14. Question 15. 10.6 Binomial Distributions (pp. 579–584) Question 16. Question 17. Probability Chapter TestYou roll a six-sided die. Find
the probability of the event described. Explain your reasoning. Question 2. You roll a multiple of 3. Answer: Evaluate the expression. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Answer: Question 12. Question 13. a. You choose one piece at random. Find the probability that you choose a black piece or a queen. b. You choose one piece at random, do not replace it, then choose a second piece at random. Find the probability that you choose a king, then a pawn. Answer: Question 14. Probability Cumulative AssessmentQuestion 1. Question 2. Answer: Question 3. Answer: Question 4. Question 5. Answer: Question 6. Question 7. a. Complete the two-way table. b. What is the probability that a randomly selected student is female and prefers choir? c. What is the probability that a randomly selected male student prefers gym class? Answer: Question 8. Answer: b. Find the probability that at least one of the mowers is unusable on a given day. c. Suppose the least-reliable mower stops working completely. How does this
affect the probability that the lawn-moving business can be productive on a given day? Question 9. Answer: |