Chapter 4. Exercise 4.1, page 209

Transcription

1 Chapter Exercise., page 09.(a) Answers may vary. or example, toss a coin 0 times and record the number of times 7 or more heads occur. Answers may vary. or example, 0.7. Roll a die and record the number of times occurs. Answers may vary. or example, 0.7..(a) Answers may vary. or example, (i) Drawing a card. (ii) Answers may vary. or example, 50. (iii) Drawing a queen from the deck..(a) Answers may vary. An example is given below. (i) Number of Repetitions otal Number of Heads Observed Probability (ii) ossing a coin. (iii) he coin turns up heads. (iv) he probability gets closer to 0.5 since the probability of getting a head and a tail is the same..(a) Answers may vary. or example, Answer depends on the length of the shaft of the tack, roundness of the top, and diameter of the shaft. he answer is (Answer of (a)). If the answer in (a) is 0.50, then answer in is tacks. 5.(a) oss 0 coins and record the number of times at least 5 heads occurs and the total number of trials. Instruction. Answers may vary. or example, 0.0..(a) Answers may vary. or example, 0.5. Answers may vary. or example, (a) Answers may vary. or example, Answers may vary. or example, Answers may vary. or example, make a spinner with 5 equal sectors of angle 7, labelled A, B, C, D, and E. Spin 5 times, recording the outcome and compare to the correct given answers. 9.(a) Let 0 be a head and let be a tail. he probability of getting three heads 8 9. Increase the number of trials in the experiment can make the result more accurate. (d) Let 0 represent a male and let represent a female. (e) You can do more trials if needed. 0. Answers may vary. or example, make a spinner with sector angles 7 and 88. he larger sector represents a field goal. Answers may vary. or example, Answers may vary. or example, make a spinner with sector represents a defective keyboard. Answers may vary. or example, (a) Make a spinner with sector angles 08 and 5. he larger sector represents a seat belt wearer. Answers may vary. or example, No, because the result is only a experimental probability and is not accurate enough in the long run.. Answers may vary. or example, roll a die and record how many rolls it takes to get all numbers. Create a spreadsheet of random integers to. How many rows until you have all numbers to? Answers may vary. or example,.. Answers may vary. or example, create a spreadsheet of random integers to. Let represent a green light. How many rows have number? Answers may vary. or example, Answers may vary. A sample solution is shown. Program Program:Light :0 G :0 R :or(i,,00) :If randint(,) :hen :G G :Else :R R :End :End :{0,} L :{G,R} L :DispGraph : Required Settings Part : Making Sense of the Data 9

2 . Answers may vary. or example, let heads represent a male and tails represent a female. oss 5 coins and record the number of times 5 heads occurs and the total number of trials. 7. Probability of a box lands on its end if all faces are equal. Since the end is a bit thinner, a good estimate would be less than. 8. Buffon s needle problem determines the probability of a needle landing on one of a family of parallel lines when dropped at random on a plane; when the length of the needle is l, and the lines are a units apart, the probability appropriately formalized is l π a. his provides a poor method of computing π. 9. Advantages: when the number of trials is large enough it can efficiently show the probability of some realworld events actually happening; generally, the simulation is easy to run than the real-world event. Disadvantages: the result can be biased; the game can be unfair in some circumstances; the result is not representative to the given situation.. Exercises, page 8.(a) 7 of diamonds. ace of spades, ace of hearts, ace of clubs, ace of diamonds (an ace).,,, 5,, 7, 8, 9, 0 (numbered) of clubs. (d),,, 8, 0 (even-numbered) of clubs, diamonds, hearts, or spades (card of any suit)..(a) Only (a) is a simple event since a 7 of diamonds is one outcome only. (i) In the deck there is only one 7 of diamonds and there are 5 possible outcomes. herefore, the answer is 5. (ii) In the deck there are desired outcomes of getting an ace and there are 5 possible outcomes. herefore, the answer is 5 or. (iii) In the deck there are 9 desired outcomes (,,, 5,, 7, 8, 9, 0 of clubs) of getting a numbered club and there are 5 possible outcomes. herefore, the answer is 9 5. (iv) In the deck there are 5 (even-numbered:,,, 8, 0) (all suits) 0 desired outcomes and 5 possible outcomes. herefore, the answer is 0 5 or 5..(a) here are three black marbles and two red marbles in the box. herefore, there are 5 possible outcomes. here are desired outcomes (black marbles) and 5 possible outcomes. herefore, the probability is 5. here are desired outcomes (red marbles) and 5 possible outcomes. herefore, the probability is 5. (d) here are 0 desired outcomes since all marbles are either red or black. herefore, the probability is 0..(a) here are desired outcomes (two jokers) and 5 possible outcomes. herefore, the probability is 5 or 7. here is outcome (red joker) and 5 possible outcomes. herefore, the probability is 5. here are four queens (queen in all suits) and 5 possible outcomes. herefore, the probability is 5 or 7. (d) here are 7 desired outcomes (all black cards plus the black joker) and 5 possible outcomes. herefore, the probability is 7 5 or. (e) here are 9 desired outcomes (,,,, 5,, 7, 8, 9 of all suits) and 5 possible outcomes. herefore, the probability is 5 or. (f) here are desired outcomes ( red aces plus the red joker) and 5 possible outcomes. herefore, the probability is 5 or 8. 5.(a) here is desired outcome (tails) and possible outcomes. herefore, the probability is. here is outcome (rolling ) and possible outcomes. herefore, the probability is. here are desired outcomes (all red cards) and 5 possible outcomes. herefore, the probability is. (d) here are desired outcomes (seven of club and seven of spade) and 5 possible outcomes. herefore, the probability is 5 or. (e) here are 0 desired outcomes (,,,, 5,, 7, 8, 9, and 0 of all suits) and 5 possible outcomes. herefore, the probability is 0 or 0. 5.(a) here are desired outcomes (odd numbers:,, 5, 7) and 8 possible outcomes. herefore, the probability is 8 or. here are desired outcomes ( and 8, which are divisible by ) and 8 possible outcomes. herefore, the probability is 8 or. 50 Nelson Data Management Solutions Manual

3 . Answers may vary. or example, let heads represent a male and tails represent a female. oss 5 coins and record the number of times 5 heads occurs and the total number of trials. 7. Probability of a box lands on its end if all faces are equal. Since the end is a bit thinner, a good estimate would be less than. 8. Buffon s needle problem determines the probability of a needle landing on one of a family of parallel lines when dropped at random on a plane; when the length of the needle is l, and the lines are a units apart, the probability appropriately formalized is l π a. his provides a poor method of computing π. 9. Advantages: when the number of trials is large enough it can efficiently show the probability of some realworld events actually happening; generally, the simulation is easy to run than the real-world event. Disadvantages: the result can be biased; the game can be unfair in some circumstances; the result is not representative to the given situation.. Exercises, page 8.(a) 7 of diamonds. ace of spades, ace of hearts, ace of clubs, ace of diamonds (an ace).,,, 5,, 7, 8, 9, 0 (numbered) of clubs. (d),,, 8, 0 (even-numbered) of clubs, diamonds, hearts, or spades (card of any suit)..(a) Only (a) is a simple event since a 7 of diamonds is one outcome only. (i) In the deck there is only one 7 of diamonds and there are 5 possible outcomes. herefore, the answer is 5. (ii) In the deck there are desired outcomes of getting an ace and there are 5 possible outcomes. herefore, the answer is 5 or. (iii) In the deck there are 9 desired outcomes (,,, 5,, 7, 8, 9, 0 of clubs) of getting a numbered club and there are 5 possible outcomes. herefore, the answer is 9 5. (iv) In the deck there are 5 (even-numbered:,,, 8, 0) (all suits) 0 desired outcomes and 5 possible outcomes. herefore, the answer is 0 5 or 5..(a) here are three black marbles and two red marbles in the box. herefore, there are 5 possible outcomes. here are desired outcomes (black marbles) and 5 possible outcomes. herefore, the probability is 5. here are desired outcomes (red marbles) and 5 possible outcomes. herefore, the probability is 5. (d) here are 0 desired outcomes since all marbles are either red or black. herefore, the probability is 0..(a) here are desired outcomes (two jokers) and 5 possible outcomes. herefore, the probability is 5 or 7. here is outcome (red joker) and 5 possible outcomes. herefore, the probability is 5. here are four queens (queen in all suits) and 5 possible outcomes. herefore, the probability is 5 or 7. (d) here are 7 desired outcomes (all black cards plus the black joker) and 5 possible outcomes. herefore, the probability is 7 5 or. (e) here are 9 desired outcomes (,,,, 5,, 7, 8, 9 of all suits) and 5 possible outcomes. herefore, the probability is 5 or. (f) here are desired outcomes ( red aces plus the red joker) and 5 possible outcomes. herefore, the probability is 5 or 8. 5.(a) here is desired outcome (tails) and possible outcomes. herefore, the probability is. here is outcome (rolling ) and possible outcomes. herefore, the probability is. here are desired outcomes (all red cards) and 5 possible outcomes. herefore, the probability is. (d) here are desired outcomes (seven of club and seven of spade) and 5 possible outcomes. herefore, the probability is 5 or. (e) here are 0 desired outcomes (,,,, 5,, 7, 8, 9, and 0 of all suits) and 5 possible outcomes. herefore, the probability is 0 or 0. 5.(a) here are desired outcomes (odd numbers:,, 5, 7) and 8 possible outcomes. herefore, the probability is 8 or. here are desired outcomes ( and 8, which are divisible by ) and 8 possible outcomes. herefore, the probability is 8 or. 50 Nelson Data Management Solutions Manual

4 here are desired outcomes ( and, which are less than ) and 8 possible outcomes. herefore, the probability is 8 or. 7.(a) here are desired outcomes (red blocks) and possible outcomes. herefore, the probability is or. here are 9 desired outcomes (not red blocks, i.e. green blocks) and possible outcomes. herefore, the probability is 9 or. 8.(a) here is outcome (one S ) and possible outcomes ( letters). herefore, the probability is. here are desired outcomes (two M s) and possible outcomes ( letters). herefore, the probability is. here are desired outcomes (four letters: two A s, one E and one I ) and possible outcomes ( letters). herefore, the probability is. 9.(a) here are 5 desired outcomes (fifteen O s) and 75 possible outcomes. herefore, the probability is 5 75 or 5. here are 5 desired outcomes (5, 0, 5, 0, 5, 0, 5, 0, 5, 50, 55, 0, 5, 70, 75) and 75 possible outcomes. herefore, the probability is 5 75 or 5. Odd number is more likely to occur since there are 8 odd numbers and 7 even numbers. 0.(a) he probability that one coin comes up heads is, so the probability that all three coins come up heads is 8. he probability of all three coins coming up heads is, so the probability that at least one of 8 three coins comes up tails is he probability that exactly two come up heads is 8. Note: Multiply by at the end since the order can be HH, HH, and HH.. here are 59 desired outcomes (0 80 exclusive) and 0 possible outcomes. herefore, the probability 59 is 0..(a) here are possible outcomes ( dice). here are desired outcomes (, 5,,, 5, ) and possible outcomes. herefore, the probability is or. he probability of getting a sum of seven is as calculated in. herefore, the probability of getting a sum of anything but seven is 5.. here is outcome (only one number is correct) and 0 possible outcomes. herefore, the probability is 0.. Let x be the number of caffeine-free diet colas. x x 0. x 9. 0.x 0.5x. x herefore, there are caffeine-free diet colas. 5. People do not choose randomly. he outside numbers are chosen less often and the higher number is selected more often from the two numbers left.. here are desired outcomes ( rap CDs) and 8 possible outcomes (8 CDs). herefore, the probability is 8 or Answers may vary. In the example of tossing a coin, each toss of the coin represents an event. he result, which is heads or tails of the coin, is an outcome. 8. here are desired outcomes ( orders in which boys start and end the presentation: MRB, MRB, BRM, and BRM) and possible outcomes. herefore, the probability is or. 9. here are 57 desired outcomes (,,, 7, 8,, 90, 9, 9, 9, 9, 98, 00) and 00 possible outcomes. 57 herefore, the probability is 00. Exercise., page 8.(a) {, 9} {, 5,, 9, 0, } {9, 0} (d) {,,, 9, 0, } (e) (f) {9}.(a) P(A) P(B) P(A B), therefore A and B are mutually exclusive and P(A B) 0. P(A B) 0, therefore A and B are not mutually exclusive. P(A B) P(A) P(B) P(A B) Part : Making Sense of the Data 5

5 here are desired outcomes ( and, which are less than ) and 8 possible outcomes. herefore, the probability is 8 or. 7.(a) here are desired outcomes (red blocks) and possible outcomes. herefore, the probability is or. here are 9 desired outcomes (not red blocks, i.e. green blocks) and possible outcomes. herefore, the probability is 9 or. 8.(a) here is outcome (one S ) and possible outcomes ( letters). herefore, the probability is. here are desired outcomes (two M s) and possible outcomes ( letters). herefore, the probability is. here are desired outcomes (four letters: two A s, one E and one I ) and possible outcomes ( letters). herefore, the probability is. 9.(a) here are 5 desired outcomes (fifteen O s) and 75 possible outcomes. herefore, the probability is 5 75 or 5. here are 5 desired outcomes (5, 0, 5, 0, 5, 0, 5, 0, 5, 50, 55, 0, 5, 70, 75) and 75 possible outcomes. herefore, the probability is 5 75 or 5. Odd number is more likely to occur since there are 8 odd numbers and 7 even numbers. 0.(a) he probability that one coin comes up heads is, so the probability that all three coins come up heads is 8. he probability of all three coins coming up heads is, so the probability that at least one of 8 three coins comes up tails is he probability that exactly two come up heads is 8. Note: Multiply by at the end since the order can be HH, HH, and HH.. here are 59 desired outcomes (0 80 exclusive) and 0 possible outcomes. herefore, the probability 59 is 0..(a) here are possible outcomes ( dice). here are desired outcomes (, 5,,, 5, ) and possible outcomes. herefore, the probability is or. he probability of getting a sum of seven is as calculated in. herefore, the probability of getting a sum of anything but seven is 5.. here is outcome (only one number is correct) and 0 possible outcomes. herefore, the probability is 0.. Let x be the number of caffeine-free diet colas. x x 0. x 9. 0.x 0.5x. x herefore, there are caffeine-free diet colas. 5. People do not choose randomly. he outside numbers are chosen less often and the higher number is selected more often from the two numbers left.. here are desired outcomes ( rap CDs) and 8 possible outcomes (8 CDs). herefore, the probability is 8 or Answers may vary. In the example of tossing a coin, each toss of the coin represents an event. he result, which is heads or tails of the coin, is an outcome. 8. here are desired outcomes ( orders in which boys start and end the presentation: MRB, MRB, BRM, and BRM) and possible outcomes. herefore, the probability is or. 9. here are 57 desired outcomes (,,, 7, 8,, 90, 9, 9, 9, 9, 98, 00) and 00 possible outcomes. 57 herefore, the probability is 00. Exercise., page 8.(a) {, 9} {, 5,, 9, 0, } {9, 0} (d) {,,, 9, 0, } (e) (f) {9}.(a) P(A) P(B) P(A B), therefore A and B are mutually exclusive and P(A B) 0. P(A B) 0, therefore A and B are not mutually exclusive. P(A B) P(A) P(B) P(A B) Part : Making Sense of the Data 5

6 P(A B) 0, therefore A and B are mutually exclusive. P(B) P(A B) P(A) (a) If A and B are mutually exclusive, P(A and B) 0 If A and B are mutually exclusive, P(A or B) P(A) P(B) If A and B are mutually exclusive, then P(A and B) 0 and P(C) P(A) P(B) he probability of making at least one team P(volleyball OR field hockey) P(volleyball) P(field hockey) P(volleyball AND field hockey) he probability that it is a female or a tetra P(female OR tetra) P(female) P(tetra) P(female AND tetra) (a) C 0 7 (i) here are 0 desired outcomes (0 chocolate bars) and 0 possible outcomes. herefore, the probability is. (ii) here are desired outcomes ( packages of toffee) and 0 possible outcomes. herefore, the probability is 0. (iii) here are desired outcomes (0 chocolate bars and packages of toffee) and 0 possible outcomes. herefore, the probability is he probability of encountering any type of defect P(mechanical defect OR other defect) P(mechanical defect) P(other defect) P(mechanical defect AND other defect) S 8.(a) H 8 C (i) here are desired outcomes (headache) and 50 possible outcomes (50 patients). herefore, the probability is 50 or 7 5. (ii) here are 8 desired outcomes (headache OR cold) and 50 possible outcomes (50 patients). herefore, the probability is 8 9 or (iii) here are desired outcomes ( only headache and no symptoms) and 50 possible outcomes (50 patients). herefore, the probability is S or here are 8 desired outcomes (9, 0 of all suits) and 5 possible outcomes (a deck of playing cards). herefore, the probability of choosing 9 and 0 8 is 5 or. 0. he probability of choosing an ace or an club P(ace OR club) P(ace) P(club) P(ace AND club) Probability of an outcome in which the sum is greater than or 5 on one of the dices P(sum greater than OR 5 on one die) P(sum greater than ) P(5 on one die) P(sum greater than AND 5 on one die) Probability it will rain today or tomorrow P(rain today OR rain tomorrow) P(rain today) P(rain tomorrow) P(rain today AND rain tomorrow) P(A B) is the sum of the probabilities of each event since P(A B) 0. An example is the probability rolling a die getting a number greater than and the probability of tossing a coin getting a head..(a) Number of students who are taller than 85 cm and have dark hair n(taller than 85 cm AND dark hair) n(taller than 85 cm) n(dark hair) n(taller than 85 cm OR dark hair) 8 9 (5 5) students P(taller than 85 cm AND has dark hair) Nelson Data Management Solutions Manual

7 As in (a), it can be observed that n(taller than 85 cm OR dark hair) herefore, P(taller than 85 cm OR dark hair) Number of students not taller than 85 cm 5 9 students herefore, P(not taller than 85 cm) he probability of selecting a week in which you paid overtime wages AND had hired temporary help P(overtime wages AND temporary help) P(overtime wages) P(temporary help) P(overtime wages OR temporary help) Answers may vary. An example is getting a spade and getting a face card from a deck of cards. 7. Number of people who go to either dance n(first dance OR second dance) n(first dance) n(second dance) n(first dance AND second dance) students Number of people did not go to either dance n(not go to either dance) n(all students) n(first dance OR second dance) students 05 students did not go to either dance. 8.(a) Answers may vary. An example is the probability rolling a die getting a number greater than and the probability of tossing a coin getting a head. Answers may vary. An example is getting a spade and getting an ace from a deck of cards. 9. he probability of seeing at least one of these media P(oronto Star OR CityV) P(oronto Star) P(CityV) P(oronto Star AND CityV) 70% 0% 0% 90% 90% of people see at least one of these media. 0. Given three sets, A, B and C, the number of elements in A B C can be found by totaling the number of elements in three sets and then subtracting the number that have been counted two or three times. he double counted elements and the triple counted elements will be found in the intersection of the three sets. n(a B C) n(a) n(b) n(c) n(a B) n(a C) n(b C) n(a B C) Exercise., page.(a) We are given that there are 9 male students and of them like school. herefore, the probability is 9. We are given that 8 students dislike school and of them are female. herefore, the probability is 8.. he probability of blonde hair given wear glasses P(blonde hair wears P ( blonde hair wears glasses) P( wears glasses) he probability ia wins P(both tail at least one tails) P(both tails) P(at least one tail). he probability being dealt with clubs in a row P(first club second club) P(second club first club) P(first club) (a) Probability of dying from lung cancer given smoker P(dying from lung cancer smoker) P (dying from lung cancer smoker) P(smoker) Probability of dying from lung cancer given nonsmoker P(dying from lung cancer non-smoker) P (dying from lung cancer non - smoker) P(non - smoker) Part : Making Sense of the Data 5

8 As in (a), it can be observed that n(taller than 85 cm OR dark hair) herefore, P(taller than 85 cm OR dark hair) Number of students not taller than 85 cm 5 9 students herefore, P(not taller than 85 cm) he probability of selecting a week in which you paid overtime wages AND had hired temporary help P(overtime wages AND temporary help) P(overtime wages) P(temporary help) P(overtime wages OR temporary help) Answers may vary. An example is getting a spade and getting a face card from a deck of cards. 7. Number of people who go to either dance n(first dance OR second dance) n(first dance) n(second dance) n(first dance AND second dance) students Number of people did not go to either dance n(not go to either dance) n(all students) n(first dance OR second dance) students 05 students did not go to either dance. 8.(a) Answers may vary. An example is the probability rolling a die getting a number greater than and the probability of tossing a coin getting a head. Answers may vary. An example is getting a spade and getting an ace from a deck of cards. 9. he probability of seeing at least one of these media P(oronto Star OR CityV) P(oronto Star) P(CityV) P(oronto Star AND CityV) 70% 0% 0% 90% 90% of people see at least one of these media. 0. Given three sets, A, B and C, the number of elements in A B C can be found by totaling the number of elements in three sets and then subtracting the number that have been counted two or three times. he double counted elements and the triple counted elements will be found in the intersection of the three sets. n(a B C) n(a) n(b) n(c) n(a B) n(a C) n(b C) n(a B C) Exercise., page.(a) We are given that there are 9 male students and of them like school. herefore, the probability is 9. We are given that 8 students dislike school and of them are female. herefore, the probability is 8.. he probability of blonde hair given wear glasses P(blonde hair wears P ( blonde hair wears glasses) P( wears glasses) he probability ia wins P(both tail at least one tails) P(both tails) P(at least one tail). he probability being dealt with clubs in a row P(first club second club) P(second club first club) P(first club) (a) Probability of dying from lung cancer given smoker P(dying from lung cancer smoker) P (dying from lung cancer smoker) P(smoker) Probability of dying from lung cancer given nonsmoker P(dying from lung cancer non-smoker) P (dying from lung cancer non - smoker) P(non - smoker) Part : Making Sense of the Data 5

9 . Probability of drawing two red marbles P(first marble second marble) P(second marble first marble) P(first marble) Probability of getting a green light at both intersection P(first green second green) P(second green first green) P(first green) (a) he probability he or she should wear eyeglasses while driving P(should wear glasses) n (need to wear glasses) n(number of people) (d) he probability he or she wears eyeglasses while driving P(wears glasses) n(wears glasses) n(number of people) he probability he or she wears glasses while driving even though he or she does not need to P(wears glasses but does not need to) n (wears glasses but does not need to) n(number of people) he probability he or she does not wear glasses while driving even though he or she needs to P(does not wear glasses but needs to) n (does not wear glasses but needs to) n(number of people) (a) (d) (e) (f) Probability of getting joker on first draw and ace on second draw P(joker first ace second) P(ace second joker first) P(joker first) Probability of getting numbered card on first draw and the red joker on second draw P(numbered card first red joker second) P(red joker second numbered card first) P(numbered card first) Probability of getting a queen on both draws P(queen first queen second) P(queen second queen first) P(queen first) Probability of getting any black card on both draws P(black first black second) P(black second black first) P(black first) Probability of getting any numbered card below 0 on first draw and the same number on a card on second draw P(numbered card below 0 first same number second) P(same number second any numbered card below 0 first) P(any numbered card below 0 first) Probability of getting the red joker or a red ace on either draw P(red joker red ace) P(NO getting red joker or red ace on either draw) Nelson Data Management Solutions Manual

10 0.(a) he possible pairings are: (, ); (, ); (, ); (, ); (, ); (, ). (i) Probability of including at least one of the top two candidates P(at least one of the top two candidates) P(NO top two candidates) 5 (ii) Probability of including both of the top candidates P(both top candidates) P(select top candidate in first draw) P(select top candidate in second draw) (iii) Probability of including neither of the top candidates P(NO top candidates for two draws) P(select non top candidate in first draw) P(select non top candidate in second draw) (iv) Probability of including Emil, if you know Sara has been selected P(Emil Sara) P (Emil Sara) P(Sara) (v) Probability of including Emil or ran, if you know Sara has been selected P(Emil OR ran Sara) P (Emil OR ran Sara) P(Sara).(a) he probability a contract is settled in two weeks given that the strike fund is large enough to support a strike P(settle in two weeks large strike fund) P (settle in two weeks large strike fund) P(large strike fund) he probability the union strike is large enough to support a strike given that a contract will be negotiated within two weeks P(large strike fund settle in two weeks) P (large strike fund settle in two weeks) P(settle in two weeks) (a) Region A B satisfies both A and B conditions, so events are not mutually exclusive 0. A You get the same answers for the probabilities, so the condition has no influence..(a) he probability Gwen s well will not be dry 0.8 (given in the question: One in five wells that were dug recently in the vicinity were dry ) Probability that the well will be uncontaminated, given that it is not dry P(uncontaminated not dry) P (uncontaminated not dry) P(not dry) Probability Gwen will have safe-drinking water from her well (d) Probability Gwen will not have safe-drinking water from her well P(safe-drinking water) 0..(a) Probability of a reader of he News also read Info 0. (given in the question) Probability of a reader of he News reads Info but not he Chronicle Probability of the back of the card is white P(white color face down black color face up) P (white colour face down black colour face up) P(black colour face up).(a) he probability that a good shipment is rejected P(rejected) n (reject shipment that graded good) n(good shipment) B 0. S 5 Part : Making Sense of the Data 55

11 he probability that an inferior shipment is accepted P(accepted inferior shipment) n (accepted AND inferior) n(accepted) Probability that the sum is greater than given that the first roll is P(sum is greater than first roll is ) P (sum is greater than first roll is ) P(first roll is ) 8. Answers may vary. or example, what is the probability of a retired person in favour of building a new community centre in town? Note: he condition is the person is retired. 9.(a) Probability of second marble is green given that the first is red P(green second red first) (d) P (green second red first) P(red first) Probability of getting both marbles red P(red first red second) P(red second red first) P(red first) Probability of getting both marbles green P(green first green second) P(green second green first) P(green first) Probability of second marble is red P(second marble is red) P(red second red first) P(red first) P(red second green first) P(green first) R.H.S. PA ( ) P(B A) PB ( ) PA ( ) PB ( ) PB ( A ) PA ( ) PA ( B ) PB ( ) P(A B) L.H.S. Exercise.5, page 5. Number of routes routes..(a).(a) (d).(a) Probability of getting all four questions correctly by guessing. Probability of getting three questions correctly by guessing. (he incorrect answer could be on any of the four positions. herefore, there is a multiplication of at the end.) he outcome is: 7D7D (D diamond). he outcome is: AS7D (A ace, S spade). he outcome is: CC, CC, CC,, 0C9C, 0C0C (C club). he outcome is: ASC, ASC, ASC,, AD8C, AD0C (A ace, S spade, D diamond, C club). here are branches. It is because the only possible answer is red is and white is. It is because the only two possible answers are () red is and white is ; and () red is and white is. 5 Nelson Data Management Solutions Manual

12 he probability that an inferior shipment is accepted P(accepted inferior shipment) n (accepted AND inferior) n(accepted) Probability that the sum is greater than given that the first roll is P(sum is greater than first roll is ) P (sum is greater than first roll is ) P(first roll is ) 8. Answers may vary. or example, what is the probability of a retired person in favour of building a new community centre in town? Note: he condition is the person is retired. 9.(a) Probability of second marble is green given that the first is red P(green second red first) (d) P (green second red first) P(red first) Probability of getting both marbles red P(red first red second) P(red second red first) P(red first) Probability of getting both marbles green P(green first green second) P(green second green first) P(green first) Probability of second marble is red P(second marble is red) P(red second red first) P(red first) P(red second green first) P(green first) R.H.S. PA ( ) P(B A) PB ( ) PA ( ) PB ( ) PB ( A ) PA ( ) PA ( B ) PB ( ) P(A B) L.H.S. Exercise.5, page 5. Number of routes routes..(a).(a) (d).(a) Probability of getting all four questions correctly by guessing. Probability of getting three questions correctly by guessing. (he incorrect answer could be on any of the four positions. herefore, there is a multiplication of at the end.) he outcome is: 7D7D (D diamond). he outcome is: AS7D (A ace, S spade). he outcome is: CC, CC, CC,, 0C9C, 0C0C (C club). he outcome is: ASC, ASC, ASC,, AD8C, AD0C (A ace, S spade, D diamond, C club). here are branches. It is because the only possible answer is red is and white is. It is because the only two possible answers are () red is and white is ; and () red is and white is. 5 Nelson Data Management Solutions Manual

13 (d) Sum Required Rolls (Red, White) (, ) (, ); (, ) (, ); (, ); (, ) 5 (, ); (, ); (, ); (, ) (, 5); (, ); (, ); (, ); (5, ) 7 (, ); (, 5); (, ); (, ); (5, ); (, ) 8 (, ); (, 5); (, ); (5, ); (, ) 9 (, ); (, 5); (5, ); (, ) 0 (, ); (5, 5); (, ) (5, ); (, 5) (. ) (e) Sum Probability (a) otal outcomes 0 (playing cards) (die) 0. he elements of sample space is (card number and suit, die number) (i) Probability of an even card and an even roll of the die P(an even card an even roll) P(an even card) P(an even roll) (ii) Probability of an even card and an odd roll of the die P(an even card an odd roll) P(an even card) P(an odd roll) (iii) Probability of a card of and a roll of the die of or less P(card of die of or less) P(card of ) P(die of or less) (iv) Probability that the sum of the card and the die is 7 P(sum 7) P(number of card) P(number of die) number of arrangements [ possible arrangements: (, ); (, 5); (, ); (, ); (5, ); (, )] (v) Probability of the sum of the card and the die is P(sum ) P(number of card) P(number of die) number of arrangements [ possible arrangements: (0, ); (9, ); (8, ); (7, ); (, 5); (5, )].(a) Probability of one of the jokers on the first draw and an ace on the second P(jokers first ace second) P(jokers first) P(ace second) Probability of the red joker on the second draw and a numbered card of any suit on the first P(red joker second numbered card first) P(red joker second) P(numbered card first) Probability of a queen on both draws P(queen on both draws) P(queen first) P(queen second) (d) Probability of any black cards on both draws P(black cards on both draws) P(black card first) P(black card second) (e) Probability of any numbered card less than 0 on the first draw and a card with the same number on the second P(numbered card less than 0 first same number second) P(numbered card less than 0 first) P(same number second) (f) Probability of the red joker or a red ace on either draw P(red joker or red ace on either draw) P(NO red joker and red ace on both draws) Probability of a safe -engine plane P(safe -engine plane) P(unsafe -engine plane) Part : Making Sense of the Data 57

14 8.(a) (d) 9.(a) Probability of a safe -engine plane P(safe -engine plane) P(unsafe -engine plane) ( ) (: number of combination when engines fail to work; : number of combination when engines fail to work) herefore, the -engine plane is safer since >. Probability that the first person is a union member P(first person is a union member) 5 0 Probability that the first two people selected from the list are union members P(union member first union member second) P(union member first) P(union member second) Probability that all committee members are union members P(union member first union member second union member third union member fourth) P(union member first) P(union member second) P(union member third) P(union member fourth) Let U be a union member. Let N be a non union member. Probability that three of the four committee members are union members P(NUUU UNUU UUNU UUUN) P(NUUU) P(UNUU) P(UUNU) P(UUUN) Probability you will be able to keep a chocolate bar P(chocolate bar first chocolate bar second) P(chocolate bar first) P(chocolate bar second) Probability you will be able to keep any candy P(keep any candy) P(chocolate bar first) P(chocolate bar second) P(fruit bar first) P(fruit bar second) P(toffee first) P(toffee second) Probability you won t be able to keep any candy P(not keep any candy) P(keep any candy) (a) Probability of spending $000 P(spending $000) P(first month) P(same as first month) P(same as second month) 8 Probability of spending more than $000 P(spending more than $000) P(spending $5000) P(spending $000) 8 Probability of spending more than $000 P(spending more than $000) P(spending $000) (a) It can create numbers. Probability of ending with 5 P(ending with 5 ) Probability of starting with or P(starting with OR ) P(starting with ) P(starting with ) (a) he number of possible postal codes he first letter determines the postal codes of oronto. herefore the probability is. here are possible postal codes as found in (a), the probability is therefore (a)(i) (ii) (iii) Probability of making three out of five attempts P(making three out of five attempts) (he number of combination of three successes and two failures is 0) Probability of missing all five attempts P(missing all five attempts) Probability of making the first three attempts and missing the next two Nelson Data Management Solutions Manual

15 (iv). (a) Probability of making at least three out of five attempts P(making at least three out of five attempts) P(making three attempts) P(making four attempts) P(making five attempts) Answers may vary. Students may use I 8 to simulate the events. 5 red blue green yellow red blue green yellow red blue green yellow red blue green yellow red blue green yellow red blue green yellow Probability a is rolled and green is spun P( is rolled green is spun) P( is rolled) P(green is spun) Probability an even number is rolled or red is spun P(even number OR red is spun) P(NO even number AND NO red is spun) 5 5. Probability Joshua wins P(Joshua wins) P(Joshua wins three times) P(Joshua wins four times) P(Joshua wins five times) Prove A and B are independent P(A B) P(A B ) ( ) If A and B are independent, then P(A B) P(A) and P(A B ) P(A), therefore P(A B) P(A B ). ( ) P(A B) P(A B ) PA ( B) PA ( B ) PB ( ) PB ( ) P(A B) P(B ) P(A B ) P(B) P(A B) ( P(B)) P(A B ) P(B) P(A B) P(A B) P(B) P(A B ) P(B) P(A B) P(B)[P(A B) P(A B )] P(A B) P(B) P(A) herefore, according to the multiplicative principle for probabilities of independent events, A and B are independent events. 7. wo events are independent of each other if the probability of one happening is not affected by the other. wo events are dependent of each other if the probability of one happening is affected by the other. Exercise., page 55.(a) 5 0 5P 5 0 ( 5 )! 8!! (8 7 5 ) ( 5 ) (d)! 5 (e) 0! P(0, ) ( 0 )! (f)! 58!!! ! ( 5 ) ( 5 ) 8! 00.(a) n! n ( n )! n ( n )! ( n )! Part : Making Sense of the Data 59

16 (iv). (a) Probability of making at least three out of five attempts P(making at least three out of five attempts) P(making three attempts) P(making four attempts) P(making five attempts) Answers may vary. Students may use I 8 to simulate the events. 5 red blue green yellow red blue green yellow red blue green yellow red blue green yellow red blue green yellow red blue green yellow Probability a is rolled and green is spun P( is rolled green is spun) P( is rolled) P(green is spun) Probability an even number is rolled or red is spun P(even number OR red is spun) P(NO even number AND NO red is spun) 5 5. Probability Joshua wins P(Joshua wins) P(Joshua wins three times) P(Joshua wins four times) P(Joshua wins five times) Prove A and B are independent P(A B) P(A B ) ( ) If A and B are independent, then P(A B) P(A) and P(A B ) P(A), therefore P(A B) P(A B ). ( ) P(A B) P(A B ) PA ( B) PA ( B ) PB ( ) PB ( ) P(A B) P(B ) P(A B ) P(B) P(A B) ( P(B)) P(A B ) P(B) P(A B) P(A B) P(B) P(A B ) P(B) P(A B) P(B)[P(A B) P(A B )] P(A B) P(B) P(A) herefore, according to the multiplicative principle for probabilities of independent events, A and B are independent events. 7. wo events are independent of each other if the probability of one happening is not affected by the other. wo events are dependent of each other if the probability of one happening is affected by the other. Exercise., page 55.(a) 5 0 5P 5 0 ( 5 )! 8!! (8 7 5 ) ( 5 ) (d)! 5 (e) 0! P(0, ) ( 0 )! (f)! 58!!! ! ( 5 ) ( 5 ) 8! 00.(a) n! n ( n )! n ( n )! ( n )! Part : Making Sense of the Data 59

17 n! n ( n )! n ( n )! ( n )! ( n r)! ( n r ) ( n r )! (n r) ( n r )! ( n r )!.(a) ! ! 0! 7!! 7!! 0! (d)! 0! 0!.(a) 0! (0 )! 9! (0 )! 8! Number of way of choosing people from a group of 5 people where order matters P(5, ) 5! 5 70 ( 5 )!!.(a)! 5 70! P(, )! ( )!! he word MISSISSAUGA has letters, including I s, S s, and A s. Number of arrangements! 5 800!!! 8.(a) L.H.S. P! 5! ( )!! ( 5 P ) R.H.S. ( 5 )! (i) n! ( n )! 5 n ( n )! ( n )! 5 n 5 (ii) n! ( n )! 90 n ( n ) ( n )! ( n )! 90 n(n ) 90 n n 90 0 (n 0)(n 9) 0 n 0 or n 9 (reject 9, since n 0) herefore, n 0. (iii) np 5 ( n P ) n! n! ( n 5)! ( n )! n! ( n 5)! n n 8 n! ( n )( n 5)! (iv) np 7( n P ) n! n! 7 ( n )! ( n )! n! n! 7 ( n )! ( n )( n )! n 7 n 9 9.(a) Number of possible outcomes 0 [first playing card] 9 [second playing card] [ sides of a die] 90 (i) Probability one even card drawn and even number rolled P(one even card even number rolled) (ii) Probability two even cards drawn and odd number rolled P(two even cards odd number rolled) (iii) Probability a card of drawn and or less rolled P(a card of or less rolled) (iv) Let the set be (card, card, die) Probability sum of cards and die is 7 P(sum is 7) 0 9 [for (, 5, ); (,, ); (,, ); (5,, )] 0 9 [for (,, )] 0 9 [for (,, ); (,, ); (,, ); (,, )] [for (,, ); (,, )] 0 9 [for (,, )] [for (,, ); (,, )] 0 9 [for (,, 5)] (v) Let the set be (card, card, die) Probability sum of the cards and die is less than 5 P(sum is less than 5) P(sum is ) P(sum is ) 0 9 [for (,, )] 0 9 [for (,, )] 0 9 [for (,, ); (,, )] (a) Number of possible outcomes 0 [first playing card] 0 [second playing card] [ sides of a die] Nelson Data Management Solutions Manual

18 (i) (ii) (iii) (iv) Probability that one even card is drawn and an even number is rolled P(one even card even number rolled) Probability that two even cards are drawn and an odd number is rolled P(two even cards odd number rolled) Probability that one card of is drawn and or less is rolled P(one card of or less rolled) Let the set be (card, card, die) Probability that the sum of the cards and the die is 7 P(sum is 7) 0 0 [for (, 5, ); (,, ); (,, ); (5,, )] 0 0 [for (,, )] 0 0 [for (,, ); (,, ); (,, ); (,, )] [for (,, ); (,, )] 0 0 [for (,, )] [for (,, ); (,, )] 0 0 [for (,, 5)] (v) Let the set be (card, card, die) Probability that the sum of the cards and the die is less than 5 P(sum is less than 5) P(sum is ) P(sum is ) 0 0 [for (,, )] 0 0 [for (,, )] 0 0 [for (,, ); (,, )] (a) Number of arrangements Probability of getting the correct arrangements P(correct arrangements) Number of arrangements Probability of getting the correct arrangements P(correct arrangements) (a) Probability of getting yellow, red, and green, in order Probability of getting yellow, green, and green, in order Probability of getting yellow, yellow, and red, in order Number of possible arrangements of 0 elements 0! Number of correct arrangements herefore, the probability is P(correct answer) (a) Original number of arrangements New number of arrangements 7 78 he difference is Number of arrangements using four letters and three numbers Number of arrangements () ( n )! 5. 0 ( n )! ( n )( n )( n )! 0 ( n )! (n )(n ) 0 n n 0 n n 8 0 (n )(n ) 0 n or n (reject, since n 0) herefore n..(a) Number of ways! P(, )! ( )! 9! 0 9! 0 9! Number of ways 0 0 P(rancesca as publisher and Marc as editor) 0 0 Part : Making Sense of the Data

19 7. Let Dr. Eisen and Dr. Bugada be an object which is to be arranged. here are five objects to be arranged (). Also, there are! combination for the arrangement of the object of Dr. Eisen and Dr. Bugada. herefore, the number of arrangements is! 0 8. here are letters to be selected (C, A, N, D), and A can be selected up to three times. Number of arrangements using A at most once! P(, ) ( )!! Number of arrangements using A twice 9 Number of arrangements using A three times herefore, there are 9 three-letter arrangements possible. 9. Answers may vary. or example, what is the probability that a random arrangement of the letters E, I, L, L,, and will spell little?.7 Exercises, page 8! 8!.(a) C(8, )!( 8 )! 5!! 8 7 5! 5 7! 7! 7C!( 7 )!!! 7 5! 5!!!!!( )!!!! 5 (d)!( 5 )!!! 5! 0! 0! (e) C(0, ) 0!!( 0 )! 7!! ! 0 7! 5 (f) 5 ( 5 5)! 0!. Number of ways!!!( )!!! 9. Number of ways 0! 0!!( )!!( 0 )!!!! 0!!! 5! !!!! 50.(a) Number of ways 8 8! 8!!( 8 )!!! 5 Number of ways 0 0! 0!!( 0 )! 8!! 5 Probability that jelly beans selected at random are red P( red jelly beans) Number of different eight card hands !( 5 8)! 8!! 9!.(a) R.H.S. C(9, ) C(9, 5) 5!! 9! 5!! 9! 5!! 9! 5 0 9! 0!! 5 55!! 55!! C(0, 5) L.H.S. R.H.S. n n ( n )! n!( n 5)! nn ( )! 5 5!( n 5)! 5 n! 5!( n 5 )! 5 n 5 L.H.S. R.H.S. n n r n ( n )! ( r )![ n ( r )]! ( n )! nn ( )! n r ( r )!( n r)! r ( r )!( n r)! n! r r!( n r)! r n r L.H.S. (d) L.H.S. n ( n )! n ( n )![ n ( n )]! 7.(a) (d) (e) ( n )! n ( n )! ( n )! n! n ( n )! n! n ( n )! ( n)! n R.H.S. nn!! n!( n n)! n Number of ways to form a committee 9 5 9! 9! ( 9 5)! 5!! Number of ways to form a committee 5! 0! ( 5 )!!( )! Number of ways to form a committee 5! 0!( 5 )!!( )! Number of ways to form a committee 5 5 0! ( 5 5)! 0!( 0)! Number of ways to form a committee n(all combination) n(combination with one men) 5!!( 5 )!!( )! 5 Nelson Data Management Solutions Manual

20 7. Let Dr. Eisen and Dr. Bugada be an object which is to be arranged. here are five objects to be arranged (). Also, there are! combination for the arrangement of the object of Dr. Eisen and Dr. Bugada. herefore, the number of arrangements is! 0 8. here are letters to be selected (C, A, N, D), and A can be selected up to three times. Number of arrangements using A at most once! P(, ) ( )!! Number of arrangements using A twice 9 Number of arrangements using A three times herefore, there are 9 three-letter arrangements possible. 9. Answers may vary. or example, what is the probability that a random arrangement of the letters E, I, L, L,, and will spell little?.7 Exercises, page 8! 8!.(a) C(8, )!( 8 )! 5!! 8 7 5! 5 7! 7! 7C!( 7 )!!! 7 5! 5!!!!!( )!!!! 5 (d)!( 5 )!!! 5! 0! 0! (e) C(0, ) 0!!( 0 )! 7!! ! 0 7! 5 (f) 5 ( 5 5)! 0!. Number of ways!!!( )!!! 9. Number of ways 0! 0!!( )!!( 0 )!!!! 0!!! 5! !!!! 50.(a) Number of ways 8 8! 8!!( 8 )!!! 5 Number of ways 0 0! 0!!( 0 )! 8!! 5 Probability that jelly beans selected at random are red P( red jelly beans) Number of different eight card hands !( 5 8)! 8!! 9!.(a) R.H.S. C(9, ) C(9, 5) 5!! 9! 5!! 9! 5!! 9! 5 0 9! 0!! 5 55!! 55!! C(0, 5) L.H.S. R.H.S. n n ( n )! n!( n 5)! nn ( )! 5 5!( n 5)! 5 n! 5!( n 5 )! 5 n 5 L.H.S. R.H.S. n n r n ( n )! ( r )![ n ( r )]! ( n )! nn ( )! n r ( r )!( n r)! r ( r )!( n r)! n! r r!( n r)! r n r L.H.S. (d) L.H.S. n ( n )! n ( n )![ n ( n )]! 7.(a) (d) (e) ( n )! n ( n )! ( n )! n! n ( n )! n! n ( n )! ( n)! n R.H.S. nn!! n!( n n)! n Number of ways to form a committee 9 5 9! 9! ( 9 5)! 5!! Number of ways to form a committee 5! 0! ( 5 )!!( )! Number of ways to form a committee 5! 0!( 5 )!!( )! Number of ways to form a committee 5 5 0! ( 5 5)! 0!( 0)! Number of ways to form a committee n(all combination) n(combination with one men) 5!!( 5 )!!( )! 5 Nelson Data Management Solutions Manual

21 (f) (g) Number of ways to form a committee n(combination with three women) n(combination with four women) 5 5!!( 5 )!!( )!!( 5 )! he probability of each parts from to (f): 0 Probability 0 Probability 0 0 (d) Probability (e) Probability!!( )! 5 (f) Probability 5 8.(a) Number of ways to assign bedrooms 9 5 9!!!( 9 )!!( 5 )!!( )! 0 Number of ways of assigning bedrooms such that Renaldo will be assigned to the room with two beds 8 8!!!!( 8 )!!( )!!( )! 80 he probability that Renaldo will be assigned to the room with two beds P(Renaldo will be assigned to the room with two beds) Number of ways to draw three cards from a standard deck of 5 playing cards (a) 5!( 5 )! 9!! 00 Number of ways to draw three hearts!!!( )! 0!! 8 Probability of drawing three hearts P(drawing three hearts) Number of ways to draw three black cards!!!( )!!! 00 Probability of drawing three black cards P(drawing three black cards) Number of ways to draw three aces!!!( )!!! Probability of drawing three aces P(drawing three aces) (d) Number of ways to draw three face cards!!!( )! 9!! 0 Probability of drawing three face cards P(drawing three face cards) Number of ways to draw two candies from 0 candies 0 0! 0!!( 0 )! 8!! 90 (a) Number of ways to draw two gum ball 0 0! 0!!( 0 )! 8!! 5 Probability of keeping a gum ball 5 P(keeping a gum ball) Number of ways to draw two candy bars 7 7! 7!!( 7 )! 5!! Number of ways to draw two packages of toffee!!!( )!!! Probability of keeping a candy 5 9 P(keeping a candy) Probability of not keeping a candy P(not keeping an candy) P(keeping an candy) Number of ways to draw balls from 9 balls 9 9! 9! 98 8!( 9 )!!! P(winning) P(losing) Odds: : : Part : Making Sense of the Data

22 .(a) Probability of tossing a head Probability of tossing a tail Odds of tossing a head : : Odds of tossing a tail : : Answers may vary. or example, the probability of x event A is and the probability of event B is x. herefore, the odds of event A compared to 00 event B is x : 00 x..(a) P(a quarter) 5 P(a dime) Odds of the coin being a quarter 5 : 5 : Odds of the coin being a dime : 5 : 5. P(raining) 0.8 P(not raining) Odds of rain tomorrow 0.8 : 0. : 5. P(winning) 0. P(not winning) Odds the team will win 0. : 0. :.(a) he odds are in the format of (probability of unfavourable outcome : probability of favourable outcome). he probability the horse will win (a) Both represent the number of arrangement of different objects. In permutations, order is important and in combinations, order is not important. An example of permutation is the number of ways to arrange out of 5 cars. An example of combination is the number of possible winning numbers for lotto /9. nc r is r objects selected from n objects; order is not important. n P r is r objects selected from n different objects; order is important. n C r n P r r!. 8.(a) otal number of songs 5 7 Number of ways songs can be selected 7 7! 7! ( 7 5)! 5!! Number of ways songs can be selected from CDs and 5 9 9! 9! ( 9 5)! 5!! Probability of selecting songs from CDs and 5 P(selected from CDs and 5) Number of ways songs can be selected when one song from each of the CDs is played 5!!!( )!!( 5 )!!( )!!!!( )!!( )! 5 80 Probability one song from each of the CDs is played P(one song from each CDs played) Number of ways to be played Probability one song from each of the CDs is played P(one favourite song from each CDs played) Number of games 0 0!!( 0 )! 0! 5 5 games 8!! No. Need to play 5 games in 0 cities. 0. Proof: n L.H.S. n n n! n! n!!( n )!! ( n )!!( n )! n![ ( n )( n ) ( n ) ] ( n )! n n![ n n ] ( n )! n n n ( )!( ) n ( n )! R.H.S. n n must be larger than because of, n is only defined when n. Nelson Data Management Solutions Manual