hapter 4 Probability Distributions for Discrete Variables hapter 4 Prerequisite Skills hapter 4 Prerequisite Skills Question Page 42 a) Four of the 52 cards in a standard deck are kings. The probability of selecting a king is or 3. b) Twenty-six of the 52 cards in a standard deck are red cards. The probability of selecting a red card is 26 52, or 2. c) Thirteen of the 52 cards in a standard deck are spades. The probability of selecting a spade is 3 52, or 4. hapter 4 Prerequisite Skills Question 2 Page 42 a) If you were to list all 26 possible outcomes, the possible sums are any natural from 3 to 8. When three different numbers form a grouping, there are 3! different ways of arranging these numbers. When two different numbers form the grouping, there are three different ways of arranging these numbers. Number of Sum Possible Groupings Outcomes Probability 3 (,,) 4 (,2,) 3 5 (,3,), (,2,2) 6 6 (,4,), (,3,2), (2,2,2) 0 7 (,4,2), (,3,3), (5,,), (3,2,2) 5 8 (,4,3), (,2,5), (,,6), (4,2,2), (3,3,2) 2 9 (6,2,), (5,3,), (5,2,2), (4,4,), (4,3,2), (3,3,3) 25 0 (6,3,), (6,2,2), (5,3,2), (5,4,), (4,4,2), (4,3,3) 27 (6,4,), (6,3,2), (5,5,), (5,4,2), (5,3,3), (4,4,3) 27 2 (6,5,), (6,4,2), (6,3,3), (5,5,2), (5,4,3), (4,4,4) 25 3 (6,6,), (6,5,2), (6,4,3), (5,5,3), (5,4,4) 2 4 (6,4,4), (6,5,3), (5,5,4), (6,6,2) 5 5 (6,6,3), (6,4,5), (5,5,5) 0 6 (6,6,4), (6,5,5) 6 7 (6,6,5) 3 8 (6,6,6) 26 3 26 6 26 0 26 5 26 2 26 25 26 27 26 27 26 25 26 2 26 5 26 0 26 6 26 3 26 26 4 52, MHR Data Management 2 Solutions
b) The sum of the probabilities is. hapter 4 Prerequisite Skills Question 3 Page 42 a) Since three of the eight doors are red, the probability that the food was placed behind a red door is 3 8. b) Since the food is behind only one of the are eight doors, the probability that the rat will select the correct door is 8. hapter 4 Prerequisite Skills Question 4 Page 42 a) There are 4 marbles with 6 red and green. For n(s) = 4 and n(a) = 7, na ( ) PA ( ) ns ( ) 7 4 2 The probability that the marble is red or green is 2. b) There are 4 marbles with green and 3 yellow (neither red nor blue). For n(s) = 4 and n(a) = 4, na ( ) PA ( ) ns ( ) 4 4 2 7 The probability that the marble is neither red nor blue is 2 7. hapter 4 Prerequisite Skills Question 5 Page 42 a) Of the 2 face cards, 4 are queens and 4 are kings. The probability of selecting a queen or a king is 8 2, or 2 3. b) Of the 2 face cards, 6 are red cards and queen of spades. The probability of selecting a red card or the queen of spades is 7 2. c) There are no cards that are red and a spade. The probability of selecting a red card and a spade is 0. 2 MHR Data Management 2 Solutions
hapter 4 Prerequisite Skills Question 6 Page 42 a) Since the outcome of rolling a die does not affect the outcome of rolling another die, these are independent events. b) Since the outcome of dealing a first card affects the second card dealt, these are dependent events. c) Since the outcome of flipping a coin does not affect the outcome of flipping another coin, these are independent events. d) Since the outcome of selecting a first male affects the selecting of a second male, these are dependent events. hapter 4 Prerequisite Skills Question 7 Page 42 a) b) Yes. The events (head, tail, tail) are independent. The outcome of flipping a coin does not affect the outcome of flipping another coin. hapter 4 Prerequisite Skills Question 8 Page 42 a) There are 3 hearts. The probability that a player will select a queen followed by a king is, or 3 2 56. b) Selecting a queen and a king can be done in two ways: queen then king or king then queen. The probability that a player will select a queen and a king is 2 56, or 78. c) There are 3 hearts with 0 non-face cards. The probability that a player will not select a face card on either draw is 0 9, or 5 3 2 26. MHR Data Management 2 Solutions 3
hapter 4 Prerequisite Skills Question 9 Page 42 a) There are 2 men and 0 women. Three men and three women are selected. 2! 0! 23 03 9!3! 7!3! 26 400 Three men and three women can be selected in 26 400 ways. b) For more men than women out of six names, select 4, 5, or 6 men. 2! 0! 2! 0! 2! 0! 24 02 25 0 26 00 8!4! 8!2! 7!5! 9!! 6!6! 0!0! 22 275 7920 924 More men than women can be selected in 22 275 + 7920 + 924, or 3 9 ways. hapter 4 Prerequisite Skills Question 0 Page 42 a) There are 3 chocolate chip, 4 peanut butter, and 6 butterscotch cookies. Select 2 chocolate chip, 2 peanut butter, and butterscotch cookie. 3! 4! 6! 32 42 6!2! 2!2! 5!! 08 Two chocolate chips, two peanut butter, and one butterscotch cookie can be selected in 08 ways. b) There are 3 chocolate chip, 4 peanut butter, and 6 butterscotch cookies. Select 0 chocolate chips. There are 0 non-chocolate chip. No chocolate chips cookies can be selected in 0 5, or 252 ways. c) There are 3 chocolate chip, 4 peanut butter, and 6 butterscotch cookies. Select at least of each type of cookie. ase : chocolate chip, peanut butter, 3 butterscotch: 3 4 6 3 ase 2: chocolate chip, 2 peanut butter, 2 butterscotch: 3 4 2 6 2 ase 3: chocolate chip, 3 peanut butter, butterscotch: 3 4 3 6 ase 4: 2 chocolate chip, peanut butter, 2 butterscotch: 3 2 4 6 2 ase 5: 3 chocolate chip, peanut butter, butterscotch: 3 3 4 6 ase 6: 2 chocolate chip, 2 peanut butter, butterscotch: 3 2 4 2 6 3 4 6 3 + 3 4 2 6 2 + 3 4 3 6 + 3 2 4 6 2 + 3 3 4 6 + 3 2 4 2 6 = 894 At least one of each type of cookie can be selected in 894 ways. hapter 4 Prerequisite Skills Question Page 42 a) There are 5! ways to assign the books. Order of assignment does not matter. For example, AB is the same as BA, AB, BA, BA, and AB. So, there are 3! ways for the books to be assigned to each person. The number of ways that 5 books can be assigned to five people is This is the same as 5 3 2 3 9 3 6 3 3 3. 5!, or 68 68 000. 3!3!3!3!3! 4 MHR Data Management 2 Solutions
b) The number of ways that 5 books can be assigned to three people is 5 5 0 5 5 5, or 756 756. hapter 4 Prerequisite Skills Question 2 Page 43 a) 25 83 2! 8! 7!5! 5!3! 20 98 8 7 6 5! 3! 44 352 b) 8 5 8! 3!5! 2! 5!7! 8! 5!7! 3!5! 2! 8 7 65 4 20 98 0.0707... 2 7 c) 7 3 9 5 6 8 7! 9! 4!3! 4!5! 6! 8!8! 7! 9! 8!8! 4!3! 4!5! 6! 0.3426... d) 2 3 5! 2 3 (0.6) (0.4) (0.6) (0.4) 3!2! 5 4 (0.6) 2 (0.4) 3 2 0.2304 5 2 e) 6 4 4 2 4 2 2 6! 2 3 3 2!4! 3 3 2 65 2 6 2 3 0.0823... hapter 4 Prerequisite Skills Question 3 Page 43 a) For n = 4, 4 4 0 3 2 2 3 0 4 ( x y) x y x y x y x y x y 4 0 4 4 2 4 3 4 4 x y 4x y 6x y 4x y x y 4 0 3 2 2 3 0 4 x 4x y 6x y 4xy y b) Use 4x and 3y. 4 3 2 2 3 4 (4x 3 y) (4 x) (3 y) (4 x) (3 y) (4 x) (3 y) (4 x) (3 y) (4 x) (3 y) (4 x) (3 y) 5 5 0 4 3 2 2 3 4 0 5 5 0 5 5 2 5 3 5 4 5 5 (4 x) (3 y) 5(4 x) (3 y) 0(4 x) (3 y) 0(4 x) (3 y) 5(4 x) (3 y) (4 x) (3 y) 5 0 4 3 2 2 3 4 0 5 5 4 3 2 2 3 024x 3840x y 5760x y 4320x y 620xy 243y 4 5 MHR Data Management 2 Solutions 5
c) Use 0.3 and 0.7. (0.3 0.7) (0.3) (0.7) (0.3) (0.7) (0.3) (0.7) (0.3) (0.7) (0.3) (0.7) 6 6 0 5 4 2 3 3 2 4 6 0 6 6 2 6 3 6 4 (0.3) (0.7) (0.3) (0.7) 5 0 6 6 5 6 6 6 0 5 4 2 3 3 2 4 (0.3) (0.7) 6(0.3) (0.7) 5(0.3) (0.7) 20(0.3) (0.7) 5(0.3) (0.7) 6(0. 0.000 729.00 206 0.059 535 0.85 22 0.32435 0.302 526 0.7 649 3) (0.7) (0.3) (0.7) 5 0 6 d) Use 4 and 3 4. 5 5 0 4 3 2 2 3 4 0 5 3 3 3 3 3 3 3 5 0 5 5 2 5 3 5 4 5 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 0 4 3 2 2 3 4 0 5 3 3 3 3 3 3 5 0 0 5 4 4 4 4 4 4 4 4 4 4 4 4 5 90 270 405 243 024 024 024 024 024 024 hapter 4 Prerequisite Skills Question 4 Page 43 hapter 4 Section Probability Distributions hapter 4 Section Example Your Turn Page 47 a) The random variable, X, is the number of rooms in apartments in a particular complex. b) c) The area of each bar represents its probability, as a percent. The width of each bar is, so the probability, as a percent is shown on the vertical axis. 6 MHR Data Management 2 Solutions
d) Four-room apartments occur most frequently, and the probability decreases as the room value increases or decreases from four. e) Sum = 00%. So the sum of the probabilities is. This confirms the results in the example. hapter 4 Section Example 2 Your Turn Page 49 a) MHR Data Management 2 Solutions 7
Number of Blue, x Distribution of Blue Frequency Probability, P(x) b) 0 RRRR BRRR RBRR RRBR RRRB 4 2 RRBB RBRB RBBR BRRB BRBR BBRR 3 RBBB BRBB BBRB BBBR 4 4 BBBB 6 6 4 3 8 4 6 c) 4 x0 3 x P( x) 0 2 3 4 6 4 8 4 6 6 3 4 0 4 8 4 6 4 2 2 4 6 2 The expected number of times the spinner lands on blue is 2. d) On average, the spinner would land on blue 2 out of the 4 spins. hapter 4 Section Example 3 Your Turn Page 50 a) total value of all prizes EX ( ) price per ticket number of tickets sold 0 000 000 500 000 050 000 5 4 000 000 2.75 5 2.25 The expected value per ticket is $2.25. b) Answers may vary. The results of both Example 3 and this question show that you lose money if you buy a ticket. They are not a good investment. 8 MHR Data Management 2 Solutions
c) Answers may vary. To make buying a ticket more attractive, the price could be reduced or more prizes could be given away. hapter 4 Section R Page 5 While it is impossible to have.8 children, expected values are predicted average values and should not be rounded. hapter 4 Section R2 Page 5 Answers may vary. Two examples of a discrete probability distribution are the number of email messages you receive each day of the week and the number of students in each mathematics class at your school. These examples are discrete because each must be a whole number. hapter 4 Section R3 Page 5 Answers may vary. reate a table showing all possible sums of the two 2-sided dice. Determine the frequency of each sum and its probability. Then, construct a histogram to illustrate the probability distribution. hapter 4 Section Question Page 5 a) Discrete: The number of points scored in a basketball game must be a whole number. b) ontinuous: The time spent playing a basketball game can be measured in fractions of a second. c) Discrete: While mass in general is continuous, the mass of weights in a weight room are typically whole numbers of kilograms, not fractions. d) Discrete: The number of windows in a classroom will be a whole number. e) ontinuous: The area of a window in a classroom can be measured in fractions of square centimetres. hapter 4 Section Question 2 Page 5 Expectation can be found using n E( x) x P( x ). i0 i It is the predicted average of all possible outcomes. It does equal the mean of the outcomes weighted according to their respective frequencies. The sum in the expected value calculations does not equal. Answer A. hapter 4 Section Question 3 Page 52 In Example 2, the discrete random variable is the number of girls in a family of three children. Answer. i MHR Data Management 2 Solutions 9
hapter 4 Section Question 4 Page 52 a) b) For ease of graphing, convert probabilities to decimal values. hapter 4 Section Question 5 Page 52 a) 4 x x P( x ) (0.3) 2(0.2) 3(0.) 4(0.4) 2.6 The expected value of x is 2.6. 3 b) Ex ( ) 0 2 4 6 8 0 5 0 5 0 0 0 6 4 6 8 0 0 5 0 0 0 6 8 6 8 0 0 3.8 The expected value of x is 3.8. hapter 4 Section Question 6 Page 52 a) The random variable, X, is the diameter of the marbles in a bag. 0 MHR Data Management 2 Solutions
b) Discrete: The number of marbles in a bag is a whole number. c) First calculate the probabilities. Diameter (mm) Frequency Probability, P(x) 2.0 5 3.0 4.0 24 20.0 5 25.0 5 2 60 2 5 4 2 d) The area of each bar represents its probability. The width of each bar is, so the probability is shown on the vertical axis. 2.0(5) 3.0() 4.0(24) 20.0(5) 25.0(5) e) weighted mean 60 6.0666... The weighted mean of the outcomes equals the expectation. hapter 4 Section Question 7 Page 52 a) reate a table showing all possible sums of the two 8-sided dice. Determine the frequency of each sum. First Die 2 3 4 5 6 7 8 Second Die 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 0 4 5 6 7 8 9 0 2 5 6 7 8 9 0 2 3 6 7 8 9 0 2 3 4 7 8 9 0 2 3 4 5 8 9 0 2 3 4 5 6 MHR Data Management 2 Solutions
Sum, x Frequency 2 3 2 4 3 5 4 6 5 7 6 8 7 9 8 0 7 6 2 5 3 4 4 3 5 2 6 b) Use a graphing calculator. c) 3 5 3 7 7 3 5 Ex ( ) 2 3 4 5 6 7 8 9 0 2 3 64 32 64 6 64 32 64 8 64 32 64 6 3 4 5 6 64 32 64 2 3 2 5 30 2 56 9 70 33 60 3 42 5 6 64 32 64 6 64 32 64 8 64 32 64 6 64 32 64 2 6 2 20 30 42 56 72 70 66 60 52 42 30 6 64 9 On average, the expected sum of two dice is 9. 2 MHR Data Management 2 Solutions
hapter 4 Section Question 8 Page 52 Answers may vary. The rectangle has a perimeter of 24 cm. Rectangle Width (cm), x Distribution of Dimensions Frequency Probability, P(x) by 2 2 by 0 3 3 by 9 4 4 by 8 5 5 by 7 6 6 by 6 6 6 6 6 6 6 hapter 4 Section Question 9 Page 52 a) A total of 2000 tickets are old, n(s) = 2000. There are nine prizes, n(a) = 9. The probability 9 of winning a prize is 2000. total value of all prizes b) Expected payout number of tickets sold 500 3(00) 5(50) 2000 0.525 The expected payout per ticket is $0.525. c) Determine the difference between the ticket price and payout per ticket. $5 $0.525 = $4.475 The expected profit per ticket is $4.475. d) A 90% profit per ticket would mean $4.50 per ticket. Solving x 0.525 = 4.5, gives x = 5.025. This means the price of a ticket would have to be $5.03. MHR Data Management 2 Solutions 3
hapter 4 Section Question 0 Page 53 a) Let F represent a face card and NF represent not a face card. Number of Face ards, x Distribution of Face ards Frequency 0 NF,NF,NF F,NF,NF NF,F,NF NF,NF,F 3 2 F,F,NF F,NF,F NF,F,F 3 3 F,F,F Probability, P(x) 8 3 8 3 8 8 b) c) Discrete: The number of face cards must be a whole number. 3 3 d) Ex ( ) 0 2 3 8 8 8 8 3 6 3 8 2 8.5 The expected value is.5. On average, the expected number of face cards in three trials is.5. hapter 4 Section Question Page 53 a) Let D represent rolling doubles and ND represent not rolling doubles. The probability of rolling doubles on two standard dice is. The probability of not rolling doubles on two standard 6 dice is 5 6. 4 MHR Data Management 2 Solutions
b) To calculate the probability of each outcome, multiply the probabilities along each branch. Number of Doubles Distribution of Doubles Probability 0 ND,ND,ND D,ND,ND ND,D,ND ND,ND,D 2 D,D,ND D,ND,D ND,D,D 3 D,D,D 25 75 5 c) Ex ( ) 0 2 3 26 26 26 26 75 30 3 26 08 26 0.5 The expected number of doubles in three rolls is 0.5. hapter 4 Section Question 2 Page 53 Refer to question 2 of Prerequisite Skills. 25 26 75 26 5 26 26 MHR Data Management 2 Solutions 5
hapter 4 Section Question 3 Page 53 Answers may vary. hapter 4 Section Question 4 Page 53 a) Answers may vary. There are many more values in these data sets that start with one compared to other digits. (30.) 2(7.6) 3(2.5) 4(9.7) 5(7.9) 6(6.7) 7(5.8) 8(5.) 9(4.6) b) Ex ( ) 00 344. 00 3.44 The expected value is 3.44. On average, the expected first digit is 3.44. hapter 4 Section Question 5 Page 53 a) The probability of cutting the first ace on the first try is 4 52, or 3. b) If the first ace appears on the second try, the card on the first try was anything other than an ace. 48 4 92 52 52 2704 2 69 The probability of cutting the first ace on the second try is 2 69. c) If the first ace appears on the third try, the card on the first and second try was anything other than an ace. 48 48 4 926 52 52 52 40 608 44 297 The probability of cutting the first ace on the third try is 44 297. d) If the first ace appears on the n try, the card on the first, second, third, (n )th try was anything other than an ace. 48 4 The probability of cutting the first ace on the nth try is. 52 52 n 6 MHR Data Management 2 Solutions
hapter 4 Section Question 6 Page 53 Divide the circle into a total of 5 equal sectors of 24. Assign the number to a sector of 24. Assign the number 2 to a sector size of 48, 3 to sector of 72, 4 to a sector of 96, and 5 to sector of 20. Then, the probability of each sector is 2 3 4 5,,,, and 5 5 5 5 5, respectively. hapter 4 Section Question 7 Page 53 Let the probability of rolling a, 2, 3, 4, and 6 be 7. Then, the probability of rolling a 5 is 2 7. To calculate the probability of each outcome, multiply the probabilities of rolling each value on the dice. Then, add results for each sum. For example, probability of a sum of 6 is 2 3 2 7 7 7 7. First Die 2 3 4 5 6 Second Die 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 0 5 6 7 8 9 0 6 7 8 9 0 2 MHR Data Management 2 Solutions 7
Sum Frequency Probability 2 3 2 4 3 5 4 6 5 7 6 8 5 9 4 0 3 2 2 49 2 49 3 49 4 49 7 49 8 49 7 49 6 49 4 49 4 49 49 2 3 4 7 8 7 6 4 4 Ex ( ) 2 3 4 5 6 7 8 9 0 2 49 49 49 49 49 49 49 49 49 49 49 2 6 2 20 42 56 56 54 40 44 2 49 7.020... The expected sum of the two weighted dice is approximately 7.02. hapter 4 Section 2 Uniform Distributions hapter 4 Section 2 Example Your Turn Page 56 a) Yes, the probability distribution of areas is uniform. Each randomly generated radius is equally likely and there is a single trial. 8 MHR Data Management 2 Solutions
b) Random Number, x P(x) x P(x) c) 2 3 4 5 6 7 8 8 8 2 8 8 3 8 8 4 8 8 5 8 8 6 8 8 7 8 8 8 8 8 d) Use the sum of the values of x P(x). 2 3 4 5 6 7 8 Ex ( ) 8 8 8 8 8 8 8 8 36 8 4.5 The expectation is 4.5. On average, the expected radius length will be 4.5. MHR Data Management 2 Solutions 9
hapter 4 Section 2 Example 2 Your Turn Page 56 alculate the expectation. Points received indicate a positive value for the random variable, x. Points lost indicate a negative value for x. Spin Point Value, x P(x) x P(x) 2 4 3 3 4 8 5 5 6 6 7 7 8 6 8 8 4 8 8 8 3 8 8 8 8 8 5 8 8 6 8 8 7 8 6 8 8 Use the sum of the values of x P(x). 4 3 8 5 6 7 6 Ex ( ) 8 8 8 8 8 8 8 8 6 8 0.75 A fair game will have an expectation equal to 0. This is not a fair game because the player will win 0.75 point on each turn, on average. hapter 4 Section 2 R Page 57 Yes, randomly selecting students by their student number is uniform. Each randomly generated student number is equally likely in a single trial. hapter 4 Section 2 R2 Page 57 a) A spinner with a uniform distribution will have equal sectors. b) A spinner with a non-uniform distribution will have unequal sectors. 20 MHR Data Management 2 Solutions
hapter 4 Section 2 R3 Page 57 total value of all prizes Expected payout number of tickets sold 500 2000 0.75 The expected payout per ticket is $0.75. Determine the difference between the ticket price and payout per ticket. $2 $0.75 = $.25 The expected profit per ticket is $.25 (or loss to the customer of $.25). This gives an advantage to the school. hapter 4 Section 2 Question Page 57 a) Since there are different probabilities for different sums in a trial, this is not a uniform distribution. b) Since there is an equal probability of selecting each card in a trial, this is a uniform distribution. c) Since there is an equal probability of each song being randomly selected in a trial, this is a uniform distribution. d) Since there are different probabilities for different numbers of boys in a family of five, this is not a uniform distribution. e) Since there is an equal probability of each person being randomly selected in a trial, this is a uniform distribution. hapter 4 Section 2 Question 2 Page 57 Uniform distribution: In the case of political parties using robo-callers to telephone all constituents in a riding, the probability is for each person. Uniform distribution: In the case of three people being selected at random from a group of four girls and five boys, there is equal probability of each person being randomly selected. Not a uniform distribution: In the case of dealing one card, face down, to each of five players, the probability changes with each card dealt. Uniform distribution: In the case of a school randomly selecting a student to attend a conference, there is equal probability of each student being randomly selected. Answer. MHR Data Management 2 Solutions 2
hapter 4 Section 2 Question 3 Page 57 The probability of each number is. Use this common fraction to determine the expected value. 0 Ex ( ) ( 2 3 4 5 6 7 8 9 0) 0 0(0 ) 0 2 5.5 The expected outcome is 5.5. Answer D. hapter 4 Section 2 Question 4 Page 58 alculate the expectation. Points received indicate a positive value for the random variable, x. Points lost indicate a negative value for x. Let n represent the number of green balls, n > 4. Ball olour Number of Balls Point Value, x P(x) x P(x) red 4 3 green n 2 4 2 4 + n 4 + n n 2n 4 + n 4 + n Use the sum of the values of x P(x). Set E(X) = 0 for a fair game. E( x) x P( x ) x P( x ) 2 2 2 2n 0 4n 4n 2 2n 0 4 n 0 2 2n 2n 2 n 6 For a fair game, there must be 6 green balls. hapter 4 Section 2 Question 5 Page 58 a) The probability of each value is. Use this common fraction to determine the expected 5 value. Ex ( ) (5 0 5 20 25) 5 (75) 5 5 The expected value is 5. b) The probability of each value is 2.5%, or 0.25. Use this common value to determine the expected value. 22 MHR Data Management 2 Solutions
Ex ( ) 0.25(0 2 3 4 5 6 7) 7(7 ) 0.25 2 3.5 The expected value is 3.5. hapter 4 Section 2 Question 6 Page 58 a) Random Number, x P(x) x P(x) 2 3 4 5 6 7 8 9 0 2 2 2 2 2 2 3 2 2 4 2 2 5 2 2 6 2 2 7 2 2 8 2 2 2 2 2 2 Use this common fraction to determine the expected value. Ex ( ) ( 2 3 4 5 6 7 8 9 0 2) 2 2(2 ) 2 2 6.5 The expected outcome is 6.5. 9 2 0 2 2 2 2 b) This does not mean that the time represented by the expectation is most likely. The expectation is simply the predicted average of all outcomes. Each number between and 2 is equally likely. hapter 4 Section 2 Question 7 Page 58 a) If you assume that any specific card means a card, the probability of any specific card in the deck is 52. MHR Data Management 2 Solutions 23
b) Since there is an equal probability of selecting any specific card in a trial, this is a uniform distribution. c) The probability of any specific card on the second selection is 5. d) Since the probability of any specific card changes after a card is removed, this is not a uniform distribution. hapter 4 Section 2 Question 8 Page 58 Let A =, B = 2, = 3, D = 4, and E = 5. Random Answer, x P(x) 2 3 4 5 5 5 5 5 5 hapter 4 Section 2 Question 9 Page 58 a) There are four possible outcomes: P confesses but Q denies, P and Q deny, P and Q confess, and Q confesses but P denies, corresponding to 0,, 2, and 3 years, respectively. Number of Years in P(x) Prison, x 0 2 3 4 4 4 4 24 MHR Data Management 2 Solutions
b) Answers may vary. The only chance of going free is if prisoner P confesses. If Q confesses, then P should also confess and he will get prison time of 2 years, whereas if P denies, P will go to jail for 3 years. If Q denies, then P should confess because he will be set free; otherwise P will go to jail for year. hapter 4 Section 2 Question 0 Page 58 a) Answers may vary. My predictions for the expected outcomes of the dice are 2.5, 3.5, 4.5, 6.5, and 0.5 (the average of the face values for each platonic solid). b) Use this common fraction to determine the expected value. For 4 faces, For 6 faces, Ex ( ) ( 2 3 4) Ex ( ) ( 2 3 4 5 6) 4 6 0 6(6 ) 4 6 2 2.5 3.5 The expected outcome is 2.5. The expected outcome is 3.5. For 8 faces, Ex ( ) ( 2 3 4 5 6 7 8) 8 8(8 ) 8 2 4.5 The expected outcome is 4.5. For 2 faces, Ex ( ) ( 2 3 4 5 6 7 8 9 0 2) 2 2(2 ) 2 2 6.5 The expected outcome is 6.5. c) Answers may vary. My findings in part b) confirm my prediction for the icosahedron. hapter 4 Section 2 Question Page 59 alculate the expectation. Each sector is equally likely with a probability of. The total of the 0 winning options is $27 000. The total of the losing options is $20 000. Use the sum of the values of x P(x). MHR Data Management 2 Solutions 25
Ex ( ) (27 000 20 000) 0 7000 0 700 The expected outcome is winning $700. The amount of risk money that would make this a fair game is $700. hapter 4 Section 2 Question 2 Page 59 a) Find the area of each ring. For region A, 2 2 A π(20 2 ) For region B, 2 2 A π(2 8 ) 256π 80π 64π Since the areas of the three regions are not equal, this is not a uniform distribution. 26 MHR Data Management 2 Solutions For region, 2 A π8 b) Answers may vary. To make this a fair game, let region A be worth 5 points, region B be worth 6 points, and region be worth 20 points. c) Answers may vary. A similar target with a uniform distribution would have regions of equal area. hoose radii of 8 cm, 28 cm, and 92 cm. hapter 4 Section 2 Question 3 Page 59 The total prizes are worth $.50(50 000) + $200(2) + $5 000 + $35 000, or $27 400. total value of all prizes EX ( ) price per coffee number of coffees sold 27 400.50 000 000 0.274.5.3726 The expected value of a cup of coffee is $.3726. hapter 4 Section 2 Question 4 Page 59 Determine the expected payout per ticket for total prizes worth $ 0 000. total value of all prizes Expected payout price per ticket number of tickets sold 0 000 500 000 2.22 The expected payout per ticket is $2.22. Let c represent the cost of a ticket. The desired profit is 0.6c. Solve for c. c2.22 0.6c 0.4c 2.22 c 5.55 A ticket should cost $5.55 in order for the charity to make a 60% profit.
hapter 4 Section 2 Question 5 Page 59 Answers may vary. a) A spinner with 9 equal sectors, labelled to 9, can be used in a uniform distribution that has P(x) = 9. b) A spinner with 5 equal sectors, labelled to 5, can be used in a uniform distribution that has an expected outcome of 8. The expected outcome is the average of the values on the spinner. c) A scenario similar to the game presented in Example 2 will provide an outcome for an unfair game. The game involves rolling a die. A player who rolls an even number receives points equal to three times the face value of the die. If the player rolls an odd number, the player loses five times the face value of the die. Roll Point Value, x P(x) x P(x) 5 2 6 3 5 4 2 5 25 6 8 6 5 6 6 6 6 6 5 6 2 6 6 6 25 6 8 6 6 Sum 9 6 9 Ex ( ) 6.5 In this case, the player will lose.5 points on each turn, on average. hapter 4 Section 2 Question 6 Page 59 Answers may vary. hapter 4 Section 2 Question 7 Page 59 A uniform distribution has possible outcomes from to n. The probability of each outcome is n. Use this common fraction to determine the expected value. E( x) ( 2 3 n) n nn ( ) n 2 n 2 n The expected outcome is. 2 MHR Data Management 2 Solutions 27
hapter 4 Section 2 Question 8 Page 59 Answers may vary. A table of possible winnings is shown. Number ard Winnings ($) ard 2 Winnings ($) Midpoint (Expectation) 40 + 2() = 42 00 2 = 99 70.5 2 40 + 2(2) = 44 00 2 2 = 96 70.0 3 40 + 2(3) = 46 00 3 2 = 9 68.5 4 40 + 2(4) = 48 00 4 2 = 84 66.0 5 40 + 2(5) = 50 00 5 2 = 75 62.5 6 40 + 2(6) = 52 00 6 2 = 64 58.0 7 40 + 2(7) = 54 00 7 2 = 5 52.5 8 40 + 2(8) = 56 00 8 2 = 36 46.0 9 40 + 2(9) = 58 00 9 2 = 9 38.5 0 40 + 2(0) = 60 00 0 2 = 0 30.0 If a number from to 6 is chosen, more money is won with card 2. If a number from 7 to 0 is chosen, more money is won with card. hoosing gives the greatest expected outcome at $70.50. hapter 4 Section 3 Binomial Distributions hapter 4 Section 3 Example Your Turn Page 62 a) The probability of success (an ace) on any individual cut of a deck of cards 3. The probability of failure is 2. There will be one success and nine failures in 0 tires. So, there 3 2 is some combination of 3 3. The ace can occur on any of the 0 tries, in 0 ways. The nine non-aces can occur in the remaining 9 9 ways. The probability of success on one try and failure on the other nine is 0 9 9 28 MHR Data Management 2 Solutions 9 9 2, or approximately 0.3743. 3 3 b) There will be three successes and seven failures in 0 tires. So, there is some combination of 3 7 2 3 3. The ace can occur on any three of the 0 tries, in 0 3 ways. The seven non-aces can occur in the remaining 7 7 ways. The probability of success on three tries and failure on the other seven is 3 7 2, or approximately 0.032. 3 3 0 3 7 7 hapter 4 Section 3 Example 2 Your Turn Page 65 a) X = the number of occurrences of a girl b) Each trial is independent, and on each trial the probability of a girl is 0.5.
Number of Girls, x Probability, P(x) x P(x) 0 6 0 (0.5) (0.5) 0.05 625 0 2 3 4 5 6 c) The probabilities sum to. d) Use a graphing calculator. 6 0 5 6 (0.5) (0.5) 0.093 75 2 4 6 2 (0.5) (0.5) 0.234 375 3 3 6 3 (0.5) (0.5) 0.325 4 2 6 4 (0.5) (0.5) 0.234 375 5 6 5 (0.5) (0.5) 0.093 75 6 0 6 6 (0.5) (0.5) 0.05 625 0.093 75 0.468 75 0.9375 0.9375 0.468 75 0.093 75 ompared to the shape of the histogram in Example 2, the probability is more closely bell-shaped with the mode at x = 3 girls. It is slightly skewed to the left. e) Use the values from the table in part b). 6 x P( x) 0 0.093 75 0.468 75 0.9375 0.093 75 0.468 75 0.05 625 x0 3 On average, you can expect 3 girls in a family of six children. hapter 4 Section 3 Example 3 Your Turn Page 66 a) In this case, the probability of success means conditions for rain, so p = 40% q = 0.4 = 0.4 = 0.6 Use the indirect method to determine P(x < 6) = P(x 6). P( x 6) P(6) P(7) P(8) (0.4) (0.6) (0.4) (0.6) (0.4) (0.6) 0.9502 6 2 7 8 0 8 6 8 7 8 8 95.02% There is about 95.02% chance that it rained on fewer than six days. b) E( X ) np 8( 0.4) 3.2 On average, there will be 3.2 days of rain out of eight days with the correct atmospheric conditions. MHR Data Management 2 Solutions 29
hapter 4 Section 3 R Page 67 In the binomial distribution, the coefficient n x represents the number of ways each of the number of successes can happen. hapter 4 Section 3 R2 Page 67 Answers may vary. In a binomial distribution, p and q represent the probability of success and failure, respectively. They are related by p = q. For example, the probability of drawing a heart (success) from a deck of cards is 0.25, while the probability of not drawing a heart (failure) is 0.75. hapter 4 Section 3 R3 Page 67 Answers may vary. The statement that about % of anadians are left-handed means that in a group of 25 people, the expected number of left-handed people is 2.75. However, the probability 24 of one person in 25 being left-handed is (0.) (0.89), or about 6.78%. 25 hapter 4 Section 3 Question Page 67 The probability of three successes in seven independent trials in a binomial distribution can be 3 4 represented by p q. Answer A. 7 3 hapter 4 Section 3 Question 2 Page 67 For n = 8 and p = 0.5, E( X ) np 8( 0.5) 4 Answer B. hapter 4 Section 3 Question 3 Page 67 Look for independent trials whose outcomes are either success or failure. This is not the scenario in the case of the number of queens in a five-card hand, the sum when two dice are rolled, or each lane for a 00-m race. An example of a binomial distribution is the probabilities of the number of times a 5 occurs when spinning a spinner six times. Answer D. 30 MHR Data Management 2 Solutions
hapter 4 Section 3 Question 4 Page 67 a) and b) MHR Data Management 2 Solutions 3
c) Assume a success is a rolling a one. Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 4, x = 0,, 2, 3, 4, p = 0.75, and q = 0.25. d) Use p = 0.75 and q = 0.25. (0.75 0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) 4 4 0 3 2 2 3 4 0 4 4 2 4 3 (0.75) (0.25) 4 4 0 4 4 0 3 2 2 3 0 4 (0.75) (0.25) 4(0.75) (0.25) 6(0.75) (0.25) 4(0.75) (0.25) (0.75) (0.25) The expansion contains the same terms as determined using the binomial distribution formula, but in reverse order. hapter 4 Section 3 Question 5 Page 68 a) Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 6, x = 0,, 2, 3, 4, 5, 6, p = 0.3, and q = 0.7. b) Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 8, x = 0,, 2, 3, 4, 5, 6, 7, 8, p = 9, and q = 8 9. 32 MHR Data Management 2 Solutions
hapter 4 Section 3 Question 6 Page 68 For n = 2000 and p = 6, E( X ) np 2000 6 333.333... The expected number of times a 6 appears when rolling a die 2000 times is about 333.33. hapter 4 Section 3 Question 7 Page 68 a) The probability of a family of five children having exactly two girls is 0.325. 2 3 (0.5) (0.5), or 5 2 b) The probability of a family of five children having exactly three boys is 0.325. This is the same as part a). 3 2 (0.5) (0.5), or 5 3 hapter 4 Section 3 Question 8 Page 68 a) The probability that two people out of six choose the number 9 is 0.0305. 2 4 (.05) (0.95), or about 6 2 b) Use the indirect method to determine P(x 2) = P(x < 2). P( x 2) P(0) P() (0.05) (0.95) (0.05) (0.95) 0 6 5 6 0 6 0.0328 There is about a 0.0328 chance that at least two people out of six choose the number 9. hapter 4 Section 3 Question 9 Page 68 a) Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 5, x = 0,, 2, 3, 4, 5, p = 6, and q = 5 6. MHR Data Management 2 Solutions 33
b) Use a graphing calculator. c) Use a graphing calculator to calculate the sum of the x P(x) values. The expected number of sums of 7 in five rolls is about 0.8333. For n = 5 and p = 6, E( X ) np 5 6 5 6 The expected number of of sums of 7 in five rolls is 5, or about 0.8333. 6 The two methods agree. hapter 4 Section 3 Question 0 Page 68 a) Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 8, x = 0,, 2, 3, 4, 5, 6, 7, 8, p = 0.45, and q = 0.55. 34 MHR Data Management 2 Solutions
b) Use a graphing calculator to calculate the sum of the x P(x) values. The expected number of bull s-eyes in eight attempts is 3.6. c) Read the value of P(8) from the table created in part b). Since the probability of eight bull seyes in eight attempts is 0.0068, it is highly unlikely. hapter 4 Section 3 Question Page 68 a) The probability that you roll two 3s when rolling five dice is 5 2 2 3 5, or about 0.608. 6 6 b) Use 6 and 5 6. 5 5 0 4 3 2 2 3 4 0 5 5 5 5 5 5 5 5 55 54 53 52 5 50 6 6 6 6 6 6 6 6 6 6 6 6 6 6 5 0 4 3 2 2 3 4 0 5 5 5 5 5 5 5 5 0 0 5 6 6 6 6 6 6 6 6 6 6 6 6 c) The fourth term: 5 2 2 3 5 6 6. x n x d) The general term, nxp q, in the binomial expansion represents the probability of x successes in n independent trials, where p is the probability of success on an individual trial and q is the probability of failure on that same individual trial (q = p). hapter 4 Section 3 Question 2 Page 68 a) The probability that three bulbs out of 0 do not meet specification is about 0.068. 3 7 (0.06) (0.94), or 0 3 b) The probability that seven bulbs out of 0 do not meet specification is about 2.790 0 7. 7 3 (0.06) (0.94), or 0 7 c) Interpret between 3 and 7 to mean 3, 4, 5, 6, or 7 do not meet specifications. The probability that between three and seven bulbs out of 0 do not meet specification is 3 7 (0.06) (0.94) + 4 6 (0.06) (0.94) + 5 5 (0.06) (0.94) + 6 4 (0.06) (0.94) + 0 3 0 4 7 3 0 7 (0.06) (0.94), or about 0.08 84. 0 5 0 6 MHR Data Management 2 Solutions 35
d) Answers may vary. I used the direct method in part c). An alternate method is to use the indirect method: P(0) P() P(2) P(8) P(9) P(0). e) Answers may vary. The expected number of bulbs that do not meet specification out of 0 bulbs is 0(0.06), or 0.6. Yes, the inspector should be concerned if two bulbs do not meet specifications. hapter 4 Section 3 Question 3 Page 68 alculate the expectation. Points received indicate a positive value for the random variable, x. Each point represents a car given away. Number of ars That Start Point Value, x P(x) x P(x) 0 0 0 2 2 3 3 4 4 5 5 5 0 0 5 9 0 0 5 4 9 0 0 5 2 2 3 9 0 0 5 3 3 2 9 0 0 5 4 4 9 0 0 5 5 5 0 9 0 0 0 0 0.458 0.0243 0.008 0.000 05 Sum 0.7 95 A fair game will have an expectation equal to 0. This game favours the contestant, on average. hapter 4 Section 3 Question 4 Page 68 8 2 a) The probability that Jamal is successful on eight of ten free-throws is 0 8 (0.8) (0.2), or about 0.3020. He is 30.2% likely to be successful on 8 of 0 free-throw attempts. b) Use the direct method to determine P(x 8). P( x 8) P(8) P(9) P(0) (0.8) (0.2) (0.8) (0.2) (0.8) (0.2) 8 2 9 0 0 0 8 0 9 0 0 0.6777... He is about 67.8% likely to be successful on at least 8 of 0 free-throw attempts. hapter 4 Section 3 Question 5 Page 69 a) The probability that Jean will be successful on two out of eight multiple choice questions is 2 6 (0.25) (0.75), or about 0.35. 8 2 b) Assume that a pass means more than four correct questions. Use the direct method to determine P(x 4). 36 MHR Data Management 2 Solutions
P( x 4) P(4) P(5) P(6) P(7) P(8) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75) (0.25) (0.75) 4 4 5 3 6 2 7 8 0 8 4 8 5 8 6 8 7 8 8 0.3807... The probability that Jean will pass is about 0.3 807. c) The expected number of correct answers on the quiz is 8(.25), or 2. d) Answers may vary. The probability distribution will be approximately bell-shaped, with the mode at 2. It will be skewed to the right. e) Answers may vary. Since the probability of success increases, the probability distribution will more closely resemble a bell-shape, with the mode at 8(0.4), or about 3. f) Probability histogram for part d) Probability histogram for part e) hapter 4 Section 3 Question 6 Page 69 a) The binomial distribution is not a suitable model in the situation because the probability of success changes with each trial. b) Answers may vary. Example: A jar contains 2 red balls and eight green balls. Six balls are removed with replacement. What is the probability that four of the balls are red? c) Answers may vary. The probability that four of the six balls are red is 0.3 04. 4 2 (0.6) (0.4), or 6 4 hapter 4 Section 3 Question 7 Page 69 a) alculate the probabilities of a majority of people approving the transit system. Sample Size Probability 7 P(x > 3) = P(0) + P() + P(2) + P(3) = binomcdf(7, 0.35, 3) 0.998 00 P(x > 50) = P(0) + P() + P(2) + + P(50) = binomcdf(00, 0.35, 50) 7.3783 0 4 000 P(x > 500) = P(0) + P() + P(2) + + P(500) = binomcdf(000, 0.35, 500) = 0 MHR Data Management 2 Solutions 37
b) The probability of the majority of people approving the transit system decreases with the increase in sample size. hapter 4 Section 3 Question 8 Page 69 6 3 The expected payout per scratch card is 0 25 50, or $8.50. 0 0 0 Then, the expected cost to the store if it has 200 customers one particular day is 200(8.5), or $3700. hapter 4 Section 3 Question 9 Page 69 Answers may vary. hapter 4 Section 3 Question 20 Page 69 Answers may vary. Example: 9% of the anadian population live in rural areas. A group of 30 anadians is selected at Pearson International Airport. a) What is the probability that more than two of them live in a rural area? b) What is the expected number of rural citizens? hapter 4 Section 3 Question 2 Page 69 n! a b c a) P( outcome, outcome 2, outcome 3) p q r where abc!!! P is the probability of outcome, outcome 2, and outcome 3 a is the number of times outcome occurs b is the number of times outcome 2 occurs c is the number of times outcome 3 occurs p is the probability of outcome q is the probability of outcome 2 r is the probability of outcome 3 b) In this situation, n = 0, a = 3, b = 2, c = 5, p = q = r = 3. n! P( X a, Y b, Z c) p q r abc!!! a b c 0 3 2 5 0! P( X 3, Y 2, Z 5) 325!!! 3 3 3 P( X 3, Y 2, Z 5) 2520 3 P( X 3, Y 2, Z 5) 0.042 676... The probability of rolling three reds, two greens, and five yellows in 0 rolls of the die is about 0.0427. 38 MHR Data Management 2 Solutions
hapter 4 Section 3 Question 22 Page 69 n E( X ) x P( x ) i n x0 n x0 n x n x i x n! x x p ( p) ( n x)! x! n! x x p ( p) ( n x)! x( x )! n! x p ( p) ( n x)!( x )! n x nx np p ( p) x i x nx nxp q ( n )! ( n x)!( x )! nx nx nx n ( n )! x ( n) ( x) np p ( p) x (( n ) ( x ))!( x )! Let m = n and y = x. m m! y m y np p ( p) y0 ( m y)! y! This is a form of the binomial theorem (x + y) n with x = p, y = p, and n = m. m np( p ( p)) np() np m hapter 4 Section 4 Hypergeometric Distributions hapter 4 Section 4 Example Your Turn Page 72 There are 5 astronauts with 6 women and 9 men. Four are to be selected: n(s) = 5 4 For the probability of 2 men and 2 women, n(a) = 6 2 9 2. 62 92 PA ( ) 5 4 540 365 0.3956 The probability that 2 men and 2 women are selected is about 39.56%. hapter 4 Section 4 Example 2 Your Turn Page 75 ax nar x a) Use a graphing calculator to determine all Px ( ), where n = 0, r = 4, a = 6, nr x = 0,, 2, 3, 4. MHR Data Management 2 Solutions 39
b) Use a graphing calculator. c) ompared to the shape of the histogram in Example 2, the probability is less bell-shaped with the mode at x = 2. It is skewed to the left. d) P(0) = 0.00476 or 0.476%, which means that getting no green jelly beans is very unlikely. e) From the graphing calculator, E(X) = 2.4. On average, there will be 2.4 green jelly beans in four selections. hapter 4 Section 4 Example 3 Your Turn Page 77 a) In this situation, n = 24, r = 7, and a = 0. Use the indirect method to determine P(x 3) = P(x < 3). P( x 3) P(0) P() P(2) 0 0 4 7 0 4 6 0 2 4 5 24 7 24 7 24 7 0.6430 The probability that at least three are male is about 0.6430. b) For r = 7, a = 0, and n = 24, 40 MHR Data Management 2 Solutions
ra E( X) n 7( 0) 24 2.967 On average, about 2.967 males will be chosen in a selection of seven students. hapter 4 Section 4 Example 4 Your Turn Page 77 The 80 foxes caught were all different from each other. So, there were no repetitions and the trials were dependent. If they were tagged, the trial was deemed a success; if not, the trial was deemed a failure. This is represented by a hypergeometric distribution. n = size of fox population a = number originally tagged r = number later caught (sample size) = 500 = 80 E(X) = number of foxes that had been tagged (expectation from the sample) = 34 Use the expectation formula. ra E( X) n 80( 500) 34 n 80(500) n 34 n 76.4706 The fox population is about 76 foxes. hapter 4 Section 4 R Page 78 a) The random variable is the number of red marbles selected. The size of the sample space is 5 marbles. The size of the population is 0 marbles. The range of the random variable is 0 to 5. b) The random variable is the number of hearts selected. The size of the sample space is 7 cards. The size of the population is 52 cards. The range of the random variable is 0 to 7. hapter 4 Section 4 R2 Page 78 Since the trials are independent, this is not a hypergeometric probability situation. hapter 4 Section 4 Question Page 78 From a bag containing five red and six blue blocks, the probability of getting three red blocks in 53 6 four selections is. Answer D. 4 MHR Data Management 2 Solutions 4
hapter 4 Section 4 Question 2 Page 78 Look for dependent trials whose outcomes are either success or failure. This is not the scenario in the case of the probability of each sum when two dice are rolled, the probability of a given number being randomly chosen from the numbers to 0, or the probability of the number of times a 3 occurs when rolling a die six times. An example of a hypergeometric distribution is the probability of the number of aces in a seven-card hand. Answer A. hapter 4 Section 4 Question 3 Page 78 a) For 6 3 9 2, n = 6 + 9, or 5, and r = 3 + 2, or 5. n r a5 7 b) For 0 6 b, a = 0 7, or 3, and b = 6 5, or. c) For 6 3 c 2, c = 25 6, or 9, and d = 3 + 2, or 5. 25 d hapter 4 Section 4 Question 4 Page 78 ax nar x a) Use a graphing calculator to determine all Px ( ), where n = 5, r = 4, a = 7, nr x = 0,, 2, 3, 4. ax nar x b) Use a graphing calculator to determine all Px ( ), where n = 8, r = 4, a = 4, nr x = 0,, 2, 3, 4. 42 MHR Data Management 2 Solutions
hapter 4 Section 4 Question 5 Page 79 ax nar x a) Use a graphing calculator to determine all Px ( ), where n = 20, r = 5, a = 5, nr x = 0,, 2, 3, 4, 5. b) Use the expectation formula with r = 5, a = 5, and n = 20. ra E( X) n 55) ( 20.25 On average,.25 hearts will be in a five-card hand of honour cards. Use a graphing calculator to calculate the sum of the x P(x) values. On average,.25 hearts will be in a five-card hand of honour cards. hapter 4 Section 4 Question 6 Page 79 a) In this situation, n = 20, r = 3, and a = 5. Use the indirect method to determine P(x ) = P(x < ). P( x ) P(0) 5 0 5 3 20 3 0.6009 The probability that at least one light bulb is defective is about 0.6009. MHR Data Management 2 Solutions 43
b) For r = 3, a = 5, and n = 20, ra E( X) n 35) ( 20 0.75 On average, 0.75 defective light bulbs will be chosen in a selection of 3 light bulbs. 5 3 5 0 c) P(3) 20 3 0.0088 The probability that all three light bulbs are defective is about 0.88%, which is very unlikely. hapter 4 Section 4 Question 7 Page 79 a) There are 4 aces in a deck of cards. 44 489 P(4 aces) 52 3 0.0026 The probability of 4 aces in a 3-card hand is about 0.0026. b) There are 4 kings in a deck of cards. Use the indirect method to determine P(x ) = P(x < ). P( x ) P(0) 4 0 48 3 52 3 0.6962 The probability of at least one king in a 3-card hand is about 0.6962. c) There are 3 cards of each suit in a deck of cards. 35 38 260 P(5 clubs, 8 diamonds) 52 3 6 2.6084 0 The probability of 5 clubs and 8 diamonds in a 3-card hand is about 2.6084 0 6. hapter 4 Section 4 Question 8 Page 79 There are 30 mice, 8 with a genetic mutation, 0 are selected. The probability of fewer than three means P(0) + P() + P(2). The probability of more than seven typically means P(8) + P(9) + P(0). However, there are only 8 mice with a defective mutation. So, the probability of more than seven means P(8). Knowing the shape of the distribution, the ends are highly unlikely. There is a greater probability of the mice having the genetic mutation in fewer than three. 44 MHR Data Management 2 Solutions
hapter 4 Section 4 Question 9 Page 79 n = size of fox population a = number originally tagged r = number later caught (sample size) = 200 = 00 E(X) = number of foxes that had been tagged (expectation from the sample) = 4 Use the expectation formula. ra E( X) n 00( 200) 4 n 00(200) n 4 n 428.57 429 The fox population is about 429 foxes. hapter 4 Section 4 Question 0 Page 79 a) n = size of fox population = 20 a = number originally tagged r = number later caught (sample size) = 80 = 25 E(X) = number of deer that had been tagged (expectation from the sample) Use the expectation formula. ra EX ( ) n 25( 80) 20 6.666... They would expect to find about 7 deer tagged. b) Yes. The expected number of tagged deer is 7. hapter 4 Section 4 Question Page 79 a) Answers may vary. I predict that a five-card hand will have a greater probability of no spades. b) For a seven-card hand, For a five-card hand, 30 397 P(no spades) P(no spades) 52 7 3 0 39 5 0.4 967... 0.22533... My prediction was correct. As the number of cards in a hand increases, so does the possibility that it contains a spade. 52 5 MHR Data Management 2 Solutions 45
hapter 4 Section 4 Question 2 Page 79 a) There are 20 cars. It is not possible to send four cars from the 920s or 930s. Four from 940s, Four from 950s, Four from 960s, 44 60 64 40 P(940s) P(950s) P(960s) 20 4 20 4 7 4 3 0 0.000 206... 0.003 095... 0.007 223... Sending four cars from the 960s has the greatest probability. b) There are 3 cars from the 950s or 960s. Use a graphing calculator to determine all ax nar x Px ( ), where n = 20, r = 4, a = 3, x = 0,, 2, 3, 4. n r 20 4 c) Use the expectation formula with r = 4, a = 4, and n = 20. ra E( X) n 4(4) 20 0.8 On average, 0.8 cars from the 940s will be selected. d) Increasing the number of cars from the 920s changes the value of n to 23 and n a becomes 0. With more choices from other decades, the mode should shift to the left. hapter 4 Section 4 Question 3 Page 79 a) There are 9 men and women from which 2 are selected. Note: It is not possible to have a ax nar x jury with no men. Use a graphing calculator to determine all Px ( ), where n = 20, nr r = 2, a = 9, x =, 2, 3, 4, 5, 6, 7, 8, 9. 46 MHR Data Management 2 Solutions
b) Answers may vary. c) Answers may vary. hapter 4 Section 4 Question 4 Page 79 There are 3 cards of each suit in a deck of cards. The bridge hand has 3 cards. 34 33 33 33 P(4 one suit, 3 of each other suit) 52 3 0.0263 The probability of four cards of one suit and three of each other suit in a 3-card hand is about 0.0263. hapter 4 Section 4 Question 5 Page 79 There is only one way for the coins to total 75 cents, selecting three quarters. 43 80 P(75 cents) 2 3 0.082 The probability of that the value of three coins is 75 cents is about 0.082. hapter 4 Section 5 omparing and Selecting Discrete Probability Distributions hapter 4 Section 5 Example Your Turn Page 84 a) MHR Data Management 2 Solutions 47
b) There are 5 males and 6 female names from which 4 are selected. Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 4, x = 0,, 2, 3, 4, p = 6, and q = 5. ax nar x Use a graphing calculator to determine all Px ( ), where n =, r = 4, a = 6, x = 0, nr, 2, 3, 4. c) The graphs have the same bell-like shape, with the x = 2 female names being the most likely outcome. The hypergeometric graph is slightly taller than the binomial graph at x = 2 (0.455 vs 0.369) and x = 3 (0.303 vs. 0.295), and shorter at the other values of x. This occurs due to the dependent nature of the hypergeometric distribution, causing probabilities to increase when fewer choices are available. 48 MHR Data Management 2 Solutions
hapter 4 Section 5 R Page 84 Answers may vary. hapter 4 Section 5 R2 Page 84 No. For a binomial distribution, the probability of each success is the same, but with hypergeometric the probability of success changes with each trial. hapter 4 Section 5 Question Page 85 Look for independent versus dependent trials. a) Hypergeometric: Trials are dependent. b) Binomial: Trials are independent and given the probability of failure. c) Hypergeometric: Trials are dependent. d) Uniform: Trials are independent and all outcomes are equally likely. hapter 4 Section 5 Question 2 Page 85 If, in a probability distribution, the number of successes is counted, then the distribution may be binomial or hypergeometric. Answer. hapter 4 Section 5 Question 3 Page 85 The probability distribution for the number of squares chosen that contain money is hypergeometric. Answer. MHR Data Management 2 Solutions 49
hapter 4 Section 5 Question 4 Page 85 Answers may vary. a) Possible random variable is the number of cards of a particular suit or denomination, x = 0 to 5. b) Possible random variable is the number of grade students or grade 2 students, x = 0 to 4. c) Possible random variable is the value of the outcome (a particular suit or denomination). d) Possible random variable is the number of a particular digit, x = 0 to r. e) Possible random variable is the number of defective bottles, x = 0 to r. f) Possible random variable is the value of the outcome (winning square). hapter 4 Section 5 Question 5 Page 85 Answers may vary. hapter 4 Section 5 Question 6 Page 85 a) Since each outcome is equally likely, this is a uniform distribution. b) Answers may vary. Example: Six people are asked to choose a number between and 5. What is the probability that two people choose the number 3? hapter 4 Section 5 Question 7 Page 86 a) Since these are independent success/failure trials, this is a binomial distribution. b) Answers may vary. Example: At Bill s Burger Barn, there is a hat with five free hamburger tickets and ten free fries tickets. If three tickets are drawn, what is the probability of winning a free hamburger? 50 MHR Data Management 2 Solutions
hapter 4 Section 5 Question 8 Page 86 a) Since these are dependent success/failure trials, this is a hypergeometric distribution. b) Answers may vary. Example: For a random draw, 20 slips of paper containing people s names are placed into a bin, 20% of which are her friends. What is the probability that at least one friend s name will be drawn? c) The probability of at least one friend winning a prize is P(x ) = P(x < ). In the first scenario, In the second scenario, P( x ) P(0) P( x ) P(0) 0.783 4 0 6 5 20 5 (0.2) (0.8) 5 0 0.6723 0 5 d) Since there is a higher probability of a friend winning in the first scenario, Barb would be happier with the hypergeometric distribution. hapter 4 Section 5 Question 9 Page 86 Answers may vary. hapter 4 Section 5 Question 0 Page 86 For the expectation for cutting a card from a deck four times, use np. For dealing four cards from a deck, use ra n. However, n = r and p = a. So, the results are the same. n a) The expected number of aces in four cuts of the cards is 4, or about 0.3077. The 3 expected number of aces in four dealt cards is 4(4), or about 0.3077. 52 b) The expected number of red cards in four cuts of the cards is number of red cards in four dealt cards is 4(26) 52, or 2. 4, or 2. The expected 2 c) The expected number of hearts in four cuts of the cards is of hearts in four dealt cards is 4(3) 52, or. 4, or. The expected number 4 hapter 4 Section 5 Question Page 86 a) This can be modelled using a hypergeometric distribution because the trials are dependent. MHR Data Management 2 Solutions 5
ax nar x b) Use a graphing calculator to determine all Px ( ), where n = 20, r = 6, a = 2, nr x = 0,, 2, 3, 4, 5, 6. c) Use the expectation formula with r = 6, a = 2, and n = 20. ra E( X) n 6( 2) 20 3.6 On average, 3.6 girls will win a prize. Use a graphing calculator to calculate the sum of the x P(x) values. On average, 3.6 girls will win a prize. d) Answers may vary. A basket contains 20 slips of paper, each with a student s name on it. Eight of the names are males and 2 are females. Six names are selected at random, with replacement, and the number of female names is counted. hapter 4 Section 5 Question 2 Page 86 Answers may vary. a) Show the probability distribution for the number of successful ring-tosses in eight attempts. b) There are 5 bottles, with five green and ten clear. Show the probability distribution for the number of successful ring-tosses onto green bottles in four attempts. c) There are eight bottles numbered to 8, and the probability of a successful ring-toss on any one is equally likely. Show the probability distribution for the bottle number. 52 MHR Data Management 2 Solutions
hapter 4 Section 5 Question 3 Page 87 ax nar x a) Use a graphing calculator to determine all Px ( ), where n = 0, r = 6, a = 6, nr x = 0,, 2, 3, 4, 5, 6. ax nar x Use a graphing calculator to determine all Px ( ), where n = 200, r = 6, a = 6, nr x = 0,, 2, 3, 4, 5, 6. b) Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 6, x = 0,, 2, 3, 4, 5, 6, p = 0.6, and q = 0.4. Let p = 0.6. Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 6, x = 0,, 2, 3, 4, 5, 6, p = 0.03, and q = 0.97. c) The two distributions for n = 0 both have somewhat similar bell shapes, with a mode of x = 4. The hypergeometric graph is slightly taller than the binomial graph at x = 3 and x = 4, and shorter at the other values of x. MHR Data Management 2 Solutions 53
The two distributions for n = 200 both have almost identical half-bell shapes, with a mode of x = 0. The hypergeometric graph is slightly taller than the binomial graph at x =, and slightly shorter at the other values of x. d) The binomial distribution gives a close approximation of a hypergeometric distribution when r is small in relation to n. hapter 4 Section 5 Question 4 Page 87 np x e ( np) Use a graphing calculator to determine the sum of all Px ( ), where n = 2000, x! p = 0.05, x = 0,, 2, 3, 4, 5, 6, 7, 8, 9. Shown in L2. Use a graphing calculator to determine the sum of all P(x) = n x p x q n x, where n = 2000, x = 0,, 2, 3, 4, 5, 6, 7, 8, 9, p = 0.05, and q = 0.985. Shown in L3. The Poisson distribution gives approximately 7. 0 6, where as the binomial distribution gives approximately 6.4 0 6. hapter 4 Section 5 Question 5 Page 87 a) i) E( X ) n n i x nn ( ) n 2 n 2 i ii) E( X ) np a n n a ra iii) E( X) n na n a b) The expected value for the binomial distribution equals that of the hypergeometric distribution. 54 MHR Data Management 2 Solutions
hapter 4 Section 5 Question 6 Page 87 a) i) The probability that your first red light will be on your first trip through the intersection is 0.4. ii) The probability that your first red light will be on your second trip through the intersection is 0.6(0.4), or 0.24. iii) The probability that your first red light will be on your third trip through the intersection is 0.6(0.6)(0.4), or 0.44. iv) The probability that your first red light will be on your nth trip through the intersection is (0.6) n (0.4). b) The probability of an event occurring for the first time after n initial failures, involving independent trials is (probability of failure) n (probability of success). c) The probability of success after a waiting time of x failures is P(x) = q x p, where p is the probability of success in each single trial and q is the probability of failure. d) Use a graphing calculator to determine all P(x) = q x p, where x = 0,, 2, 3, 4, 5, 6, 7, p = 6, and q = 5 6. e) All three distributions have two possible outcomes, success or failure. The geometric and binomial distributions involve independent trials where the probability of success is the same in every trial. This is in contrast to the hypergeometric distribution, where trials are dependent and the probability of success is not the same in every trial. Binomial and hypergeometric distributions involve a fixed number of trials, whereas geometric distributions involve continued trials until success. hapter 4 Review hapter 4 Review Question Page 88 a) ontinuous: The time spent playing a hockey game can be measured in fractions of a second. b) Discrete: The number of times you make a basket in basketball must be a whole number. c) Discrete: The number of candies in a bag must be a whole number. MHR Data Management 2 Solutions 55
d) ontinuous: The mass of candies can be measured in fractions of a kilogram. hapter 4 Review Question 2 Page 88 a) For ease of graphing, convert probabilities to decimal values. b) hapter 4 Review Question 3 Page 88 a) 3 x0 5 x P( x) 0 2 3 2 2 3 6 5 2 3 0 2 3 6 5 8 6 2.583... The expected value of x is about.58. b) Ex ( ) 2(0.05) 4(0.3) 6(0.24) 8(0.38) 0(0.2) 2(0.05) 4(.03) 7.32 The expected value of x is 7.32. hapter 4 Review Question 4 Page 88 A uniform distribution occurs when, in a single trial, all outcomes are equally likely. 56 MHR Data Management 2 Solutions
hapter 4 Review Question 5 Page 88 a) Random Number, x P(x) x P(x) 2 3 4 5 6 6 6 2 6 6 3 6 6 4 6 6 5 6 6 6 6 6 b) Use the sum of the values of x P(x) in the table in part a). 2 3 4 5 6 Ex ( ) 6 6 6 6 6 6 2 6 3.5 The expectation is 3.5. The predicted average value of the random number will be 3.5. hapter 4 Review Question 6 Page 88 a) alculate the expectation. Amounts received indicate a positive value for the random variable, x. Amounts lost indicate a negative value for x. Number of Green Balls Amount ($), x P(x) x P(x) 0 3 0 360 (0.4) (0.6) 77.76 40 2 280 3 0 2 3 (0.4) (0.6) 7.28 2 3 2 (0.4) (0.6) 80.64 3 0 3 600 3 3 (0.4) (0.6) 38.40 Sum 24 A fair game will have an expectation equal to 0. This is not a fair game because the player will win $24 on each turn, on average. MHR Data Management 2 Solutions 57
b) alculate the expectation. Amounts received indicate a positive value for the random variable, x. Amounts lost indicate a negative value for x. Number of Green Balls Amount ($), x P(x) x P(x) 0 360 5 4 3 30 25 24 23 7.22 40 0 5 4 3 25 24 23 8.26 2 280 0 9 5 32 25 24 23 82.7 3 600 0 9 8 33 25 24 23 3.30 Sum 23.99 A fair game will have an expectation equal to 0. This is not a fair game because the player will win $23.99 on each turn, on average. hapter 4 Review Question 7 Page 89 Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 6, x = 0,, 2, 3, 4, 5, 6, p = 6, and q = 5 6. hapter 4 Review Question 8 Page 89 a) For type O, n = 20 and p = 0.46. E( X ) np 20 0. 46 552. The expected number of people with type O blood is 55.2. A binomial distribution has a specific number of identical independent trials in which the result is success or failure. The binomial distribution is a suitable model in the situation because the probability of success (type O) does not change with each trial. b) For type A, n = 20 and p = 0.42. The expected number of people with type A blood is 20(0.42), or 50.4. For type B, n = 20 and p = 0.09. The expected number of people with type B blood is 20(0.09), or 0.8. For type AB, n = 20 and p = 0.03. The expected number of people with type AB blood is 20(0.03), or 3.6. 58 MHR Data Management 2 Solutions
hapter 4 Review Question 9 Page 89 a) The probability of winning a prize once in 0 tries is b) Use the indirect method to determine P(x 3) = P(x < 3). P( x 3) P(0) P() P(2) (0.2) (0.88) (0.2) (0.88) (0.2) (0.88) 0 0 9 2 8 0 0 0 0 2 9 0 (0.2) (0.88), or about 0.3798. 0.087 The probability of winning a prize at least three times in 0 tries is about 0.087. c) The expected number of winning cards in 0 tries is 0(0.2), or.2. hapter 4 Review Question 0 Page 89 ax nar x a) Use a graphing calculator to determine all Px ( ), where n = 25, r = 7, a = 0, nr x = 0,, 2, 3, 4, 5, 6, 7. b) Use the expectation formula with r = 7, a = 0, and n = 25. ra E( X) n 7( 0) 25 2.8 The expected outcome is 2.8. Use a graphing calculator to calculate the sum of the x P(x) values. The expected outcome is 2.8. MHR Data Management 2 Solutions 59
hapter 4 Review Question Page 89 a) There are 56 coins, with 8 being rare. Select 0 coins. 80 380 P(all rare) 56 0 6.2280 The probability that all 0 coins are rare is about.228 0 6. b) There are 56 coins, with 8 being rare. Select 0 coins. 80 380 P(0 rare) 56 0 0.033 The probability that no coins are rare is about 0.033. c) Use the indirect method to determine P(x 2) = P(x < 2). P( x 2) P(0) P() 8 0 38 0 8 38 9 56 0 56 0 0.9043 The probability that at least two coins are rare is about 0.9043. hapter 4 Review Question 2 Page 89 n = size of seal; population a = number originally tagged r = number later caught (sample size) = 420 = 00 E(X) = number of seals that had been tagged (expectation from the sample) = 42 Use the expectation formula. ra E( X) n 00( 420) 42 n 00(420) n 42 n 000 The seal population is 000 seals. hapter 4 Review Question 3 Page 89 Look for independent versus dependent trials. a) Binomial: Trials are independent and given the probability of failure. b) Hypergeometric: Trials are dependent. c) None of these 60 MHR Data Management 2 Solutions
d) Uniform: Trials are independent and all outcomes are equally likely. hapter 4 Review Question 4 Page 89 a) There are 2 faces cards in a deck of 52 cards. The probability that five of seven cards dealt 25 402 are face card is, or about 0.0046. 52 7 b) There are 2 faces cards in a deck of 52 cards: p = 3 3 of seven cards chosen, with replacement are face card is 0 and q =. The probability that five 3 7 5 5 2 3 0, or about 0.008. 3 3 c) There is a greater probability of five face cards with replacement. This is because the trials are independent and the probability of a face card does not change. hapter 4 Test Yourself hapter 4 Test Yourself Question Page 90 This is a uniform distribution. The expected outcome on a typical roll is (2 4 6 8 0 2 4 6), or 9. Answer. 8 hapter 4 Test Yourself Question 2 Page 90 The binomial and hypergeometric distributions are similar in that they both involve counting success. Answer D. hapter 4 Test Yourself Question 3 Page 90 The expectation for a uniform distribution is calculated using n xi i n. Answer A. hapter 4 Test Yourself Question 4 Page 90 ounting the number of tails when a coin is flipped 20 times is an example of a binomial distribution. Answer A. hapter 4 Test Yourself Question 5 Page 90 The probability that exactly two students will be selected when five people are selected from four students and three teachers is 4 2 3 3. Answer B. 7 5 MHR Data Management 2 Solutions 6
hapter 4 Test Yourself Question 6 Page 90 This is a binomial distribution with p = 0.4 and n = 2(5), or 0. The expected number of red lights is 0(0.4), or 4. hapter 4 Test Yourself Question 7 Page 90 There are 2 face cards in an honour deck of cards. 22 8 P(2 face) 20 3 0.4632 The probability that two are face cards is about 0.4632. hapter 4 Test Yourself Question 8 Page 90 a) The situation in #7 is modelled by a hypergeometric distribution because the trials are dependent (without replacement). b) The situation can be changed to a binomial distribution by making it with replacement. hapter 4 Test Yourself Question 9 Page 90 n = size of beaver population = 452 a = number originally tagged r = number later caught (sample size) = 200 = 65 E(X) = number of beaver that had been tagged (expectation from the sample) Use the expectation formula. ra EX ( ) n 65( 200) 452 28.760... They would expect to find about 29 beaver tagged. hapter 4 Test Yourself Question 0 Page 90 a) Use a graphing calculator to determine all P(x) = n x p x q n x, where n = 8, x = 0,, 2, 3, 4, 5, 6, 7, 8, p = 6, and q = 5 6. 62 MHR Data Management 2 Solutions
b) Use a graphing calculator. c) Use a graphing calculator to calculate the sum of the x P(x) values. The expected number of sums of 7 in eight rolls is about.3333. hapter 4 Test Yourself Question Page 9 a) The probability that the line is always busy is Use the direct method to determine P(x 2). P( x 2) P(2) P(3) P(4) P(5) 5 0 5 5 (0.95) (0.05), or about 0.4633. (0.95) (0.05) (0.95) (0.05) (0.95) (0.05) (0.95) (0.05) 2 3 3 2 4 5 0 5 2 5 3 5 4 5 5 0.9945 The probability that the line is busy at least 2 times is about 0.9945. b) The expected number of times the line should be busy is 5(0.95), or 4.25. hapter 4 Test Yourself Question 2 Page 9 There are 0 males and 5 females that have applied for 4 job promotions. 50 04 P(0 women) P(at least women) P(0) 54 0.538 5 0 0 4 0.8462 The probability that no females were promoted is about 0.538, while the probability that at least one female was promoted is about 0.8462. There is a much greater chance that at least one female was promoted, so the committee s claim is unfounded. hapter 4 Test Yourself Question 3 Page 9 a) There are 6 people and p = 0.2. Use the indirect method to determine P(x 2). 5 4 MHR Data Management 2 Solutions 63