Chapter 13 Counting and Probability

Transcription

1 Chapter 3 Counting and Probability Section 3.. union. intersection 3. True; the union of two sets includes those elements that are in one or both of the sets. The intersection consists of the elements that are in both sets. Thus, the intersection is a subset of the union. 4. True; every element in the universal set is either in the set A or the complement of A. 5. subset; 6. finite 7. n( A) + n( B) n( A B) 8. True 9. { a} { b} { c} { d} { ab} { ac} { ad} { bc, },{ bd, },{ cd, },{ abc,, },{ abd,, }, { acd,, },{ bcd,, },{ abcd,,, },,,,,,,,,,, 0. { a} { b} { c} { d} { e} { a b} { a c} { ad, },{ ae, },{ bc, },{ bd, },{ be, }, { cd} { ce} { de} { abc} { abd} { abe,, },{ acd,, },{ ace,, },{ ade,, }, { bcd,, },{ bce,, },{ bde,, },{ cde,, }, { abcd,,, },{ abce,,, },{ abde,,, }, { acde,,, },{ bcde,,, },{ abcde,,,, },,,,,,,,,,,,,,,,,,,,,,. na ( ) 5, nb ( ) 0, na ( B) 0 na ( B) na ( ) + nb ( ) na ( B) na ( ) 30, nb ( ) 40, na ( B) 45 na ( B) na ( ) + nb ( ) na ( B) na ( B) na ( B) na ( B) 50, na ( B) 0, nb ( ) 0 na ( B) na ( ) + nb ( ) na ( B) 50 na ( ) na ( ) 4. na ( B) 60, na ( B) 40, na ( ) nb ( ) na ( B) na ( ) + nb ( ) na ( B) 60 na ( ) + na ( ) na ( ) na ( ) From the figure: na ( ) From the figure: nb ( ) From the figure: na ( or B) na ( B) From the figure: na ( and B) na ( B) From the figure: na ( but not C) na ( ) na ( C) From the figure: na ( ) From the figure: na ( and Band C) na ( B C) 5. From the figure: na ( or Bor C) na ( B C) There are 5 choices of shirts and 3 choices of ties; there are (5)(3) 5 different arrangements. 4. There are 5 choices of blouses and 8 choices of skirts; there are (5)(8) 40 different outfits. 5. There are 9 choices for the first digit, and 0 choices for each of the other three digits. Thus, there are (9)(0)(0)(0) 9000 possible fourdigit numbers. 303 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

2 Chapter 3: Counting and Probability 6. There are 8 choices for the first digit, and 0 choices for each of the other four digits. Thus, there are (8)(0)(0)(0)(0) 80,000 possible five-digit numbers. 7. Let A { those who will purchase a major appliance} and B { those who will buy a car} nu ( ) 500, n( A) 00, nb ( ) 50, na ( B) 5 na ( B) na ( ) + nb ( ) na ( B) n(purchase neither) n( U) n( A B) n(purchase only a car) n B n A B ( ) ( ) 8. Let A { those who will attend Summer Session I} and B { those who will attend Summer Session II} na ( ) 00, nb ( ) 50, na ( B) 75, n( A B) 75 na ( B) na ( ) + nb ( ) na ( B) nu ( ) n( A B) + n( A B) students participated in the survey. 9. Construct a Venn diagram: 5 5 IBM GE AT&T 0 0 (a) 5 (b) 5 (c) 5 (d) 5 (e) Construct a Venn diagram: There are 8 different kinds of blood: A-Rh+, B-Rh+, AB-Rh+, O-Rh+, A-Rh, B-Rh, AB-Rh, O-Rh 3. a. n( widowed or divorced) n( widowed) + n( divorced),73+ 9,00, 93 There were,93 thousand males 8 years old and older who were widowed or divorced. b. n( married, widowed or divorced) n( married) + n( widowed) + n( divorced) 63,38 +,73+ 9,00 75,4 There were 75,4 thousand males 8 years old and older who were married, widowed, or divorced. 3. a. n( widowed or divorced) n( widowed) + n( divorced),05 +,93 4,037 There were 4,037 thousand females 8 years old and older who were widowed or divorced. b. n( married, widowed or divorced) n( married) + n( widowed) + n( divorced) 63, 97+,05 +, 93 88,008 There were 88,008 thousand females 8 years old and older who were married, widowed, or divorced. 33. There are 8 choices for the DOW stocks, 5 choices for the NASDAQ stocks, and 4 choices for the global stocks. Thus, there are (8)(5)(4) 480 different portfolios Answers will vary. Section 3.. ;. False; n! 3. permutation ( n + )! n + 4. combination 304 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

3 Section 3.: Permutations and Combinations 5. n! Pnr (, ) ( n r)!. 6! 6! C (6, 3) 0, 400, 600 (6 3)!3! 3!3! 6. n! Cnr (, ) ( n r)! r!. 8! 8! C (8, 9) 48, 60 (8 9)!9! 9!9! ! 6! 6 5 4! P(6, ) 30 (6 )! 4! 4! 7! 7! 7 6 5! P(7, ) 4 (7 )! 5! 5! 4! 4! 4 3 P(4, 4) 4 (4 4)! 0! 8! 8! P (8, 8) (8 8)! 0! ,30 7! 7! P (7, 0) (7 0)! 7! 9! 9! P (9, 0) (9 0)! 9! 8! 8! ! P(8, 4) 680 (8 4)! 4! 4! 8! 8! ! P(8, 3) 336 (8 3)! 5! 5! 8! 8! 8 7 6! C(8, ) 8 (8 )!! 6!! 6! 8! 8! 8 7 6! C(8, 6) 8 (8 6)! 6!! 6! 6! 7! 7! 7654! C(7, 4) 35 (7 4)! 4! 3! 4! 4! 3 6! 6! 6 5 4! C(6, ) 5 (6 )!! 4!! 4! C 5! 5! 5! (5, 5) (5 5)!5! 0!5! 5! 8! 8! 8 7! C(8, ) 8 (8 )!! 7!! 7! 3. { abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca, bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb, cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec, eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc} 5! 5! 5 4 3! P(5,3) 60 (5 3)!!! 4. { ab, ac, ad, ae, ba, bc, bd, be, ca, cb, cd, ce, da, db, dc, de, ea, eb, ec, ed} 5! 5! 5 4 3! P(5,) 0 (5 )! 3! 3! 5. {3, 4, 3,34,4,43, 3, 4, 3, 34, 4, 43, 3, 34, 3, 34, 34, 34, 4, 43, 4, 43, 43, 43} 4! 4! 4 3 P(4,3) 4 (4 3)!! 6. {3, 4,5, 6,3,34, 35,36, 4, 43, 45, 46, 5,53, 54,56,6, 63, 64,65, 3, 4, 5, 6, 3, 34, 35, 36, 4, 43, 45, 46, 5, 53, 54, 56, 6, 63, 64, 65, 3, 34, 35, 36, 3, 34, 35, 36, 34, 34, 345, 346, 35, 35, 354, 356, 36, 36, 364, 365, 4, 43, 45, 46, 4, 43, 45, 46, 43, 43, 435, 436, 45, 45, 453, 456, 46, 46, 463, 465, 5, 53, 54, 56, 5, 53, 54, 56, 53, 53, 534, 536, 54, 54, 543, 546, 56, 56,563, 564, 6, 63, 64, 65, 6, 63, 64, 65, 63, 63, 634, 635, 64, 64, 643, 645, 65, 65, 653, 654} 6! 6! 6543! P(6,3) 0 (6 3)! 3! 3! 305 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

4 Chapter 3: Counting and Probability 7. { abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde} 5! 5 4 3! C(5, 3) 0 (5 3)!3! 3! 8. { ab, ac, ad, ae, bc, bd, be, cd, ce, de } 5! 5 4 3! C(5, ) 0 (5 )!! 3! 9. {3, 4, 34, 34} 4! 4 3! C(4,3) 4 (4 3)!3!!3! 30. {3, 4,5,6,34, 35,36, 45, 46, 56, 34, 35, 36, 45, 46, 56, 345,346, 356, 456} 6! ! C(6,3) 0 (6 3)!3! 3 3! 3. There are 4 choices for the first letter in the code and 4 choices for the second letter in the code; there are (4)(4) 6 possible two-letter codes. 3. There are 5 choices for the first letter in the code and 5 choices for the second letter in the code; there are (5)(5) 5 possible two-letter codes. 33. There are two choices for each of three positions; there are ()()() 8 possible threedigit numbers. 34. There are ten choices for each of three positions; there are (0)(0)(0) 000 possible three-digit numbers. (Note this is if we allow numbers with initial zeros such as 0.) 35. To line up the four people, there are 4 choices for the first position, 3 choices for the second position, choices for the third position, and choice for the fourth position. Thus there are (4)(3)()() 4 possible ways four people can be lined up. 36. To stack the five boxes, there are 5 choices for the first position, 4 choices for the second position, 3 choices for the third position, choices for the fourth position, and choice for the fifth position. Thus, there are (5)(4)(3)()() 0 possible ways five boxes can be stacked. 37. Since no letter can be repeated, there are 5 choices for the first letter, 4 choices for the second letter, and 3 choices for the third letter. Thus, there are (5)(4)(3) 60 possible threeletter codes. 38. Since no letter can be repeated, there are 6 choices for the first letter, 5 choices for the second letter, 4 choices for the third letter, and 3 choices for the fourth letter. Thus, there are (6)(5)(4)(3) 360 possible three-letter codes. 39. There are 6 possible one-letter names. There are (6)(6) 676 possible two-letter names. There are (6)(6)(6) 7,576 possible threeletter names. Thus, there are ,576 8,78 possible companies that can be listed on the New York Stock Exchange. 40. There are (6)(6)(6)(6) 456,976 possible four-letter names. There are 6)(6)(6)(6)(6),88,376 possible five-letter names. Thus, there are 456,976 +,88,376,338,35 possible companies that can be listed on the NASDAQ. 4. A committee of 4 from a total of 7 students is given by: 7! 7! ! C(7,4) 35 (7 4)!4! 3!4! 3 4! 35 committees are possible. 4. A committee of 3 from a total of 8 professors is given by: 8! 8! ! C(8, 3) 56 (8 3)!3! 5!3! 35! 56 committees are possible. 43. There are possible answers for each question. 0 Therefore, there are 04 different possible arrangements of the answers. 44. There are 4 possible answers for each question. 5 Therefore, there are 4 04 different possible arrangements of the answers. 45. There are 5 choices for the first position, 4 choices for the second position, 3 choices for the third position, choices for the fourth position, and choice for the fifth position. Thus, there are (5)(4)(3)()() 0 possible arrangements of the books. 46. a. There are 6 choices for each of the first two positions, and 0 choices for each of the next four positions. Thus, there are 306 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

5 Section 3.: Permutations and Combinations (6)(6)(0)(0)(0)(0) 6,760,000 possible license plates. b. There are 6 choices for each of the first two positions, 0 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Thus, there are (6)(6)(0)(9)(8)(7) 3,407,040 possible license plates. c. There are 6 choices for the first letter, 5 choices for the second letter, 0 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Thus, there are (6)(5)(0)(9)(8)(7) 3,76,000 possible license plates. 47. The st person can have any of 365 days, the nd person can have any of the remaining 364 days. Thus, there are (365)(364) 3,860 possible ways two people can have different birthdays. 48. The first person can have any of 365 days, the second person can have any of the remaining 364 days, the third person can have any of the remaining 363 days, the fourth person can have any of the remaining 36 days, and the fifth person can have any of the remaining 36 days. Thus, there are (365)(364)(363)(36)(36) 6,30,555,08,760 possible ways five people can have different birthdays. 49. Choosing boys from the 4 boys can be done C(4,) ways. Choosing 3 girls from the 8 girls can be done in C(8,3) ways. Thus, there are a total of: 4! 8! C(4,) C(8,3) (4 )!! (8 3)!3! 4! 8!!! 5!3! 4 3! !! 5!3! The committee is made up of of 4 administrators, 3 of 8 faculty members, and 5 of 0 students. The number of possible committees is: C(4, ) C(8,3) C(0,5) 4! 8! 0! (4 )!! (8 3)!3! (0 5)!5! 4! 8! 0!!! 5!3! 5!5! 4! 8765! ! 54!3! 5!5! 5, 09,344 possible committees 5. This is a permutation with repetition. There are 9! 90,70 different words.!! 5. This is a permutation with repetition. There are! 4,989,600 different words.!!! 53. a. C(7,) C(3,) 3 63 b. C(7,3) C(3,0) c. C(3,3) C(7,0) 54. a. C(5,5) C(0,0) b. C(5, 3) C(0, ) , 475 c. C(5, 4) C(0,) + C(5, 5) C(0, 0) , , There are C(00, ) ways to form the first committee. There are 78 senators left, so there are C(78, 3) ways to form the second committee. There are C(65, 0) ways to form the third committee. There are C(55, 5) ways to form the fourth committee. There are C(50, 6) ways to form the fifth committee. There are C(34, 7) ways to form the sixth committee. There are C(7, 7) ways to form the seventh committee. The total number of committees C(00,) C(78,3) C(65,0) C(55,5) C(50,6) C(34,7) C(7,7) The team is made up of 5 of 0 linemen, 3 of 0 linebackers, and 3 of 5 safeties. The number of possible teams is: Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

6 Chapter 3: Counting and Probability C(0,5) C(0,3) C(5,3) , 400 There are 30,400 possible defensive teams. 57. There are 9 choices for the first position, 8 choices for the second position, 7 for the third position, etc. There are ! 36,880 possible batting orders. 58. There are 8 choices for the first position, 7 choices for the second position, 6 for the third position, etc. and choice for the last position. There are ! 40,30 possible batting orders. 59. The team must have pitcher and 8 position players (non-pitchers). For pitcher, choose player from a group of 4 players, i.e., C(4, ). For position players, choose 8 players from a group of players, i.e., C(, 8). Thus, the number different teams possible is C(4,) C(,8) Consider the ways that the American League can win. Then multiply by to get the total for both leagues. To win the World Series, the last game must be won. There is way to win in four games. To win in 5 games, three of the first four must be won, so there are C(4, 3) 4 ways to win in 5 games. To win in 6 games, three of the first five must be won, so there are C(5, 3) 0 ways to win in 6 games. To win in 7 games, three of the first six must be won, so there are C(6, 3) 0 ways to win in 7 games. Therefore, there are ways the American League can win the World Series. There are also 35 ways the National League can win the World Series. There are a total of 70 different sequences possible. 6. Choose players from a group of 6 players. Thus, there are (6,) 5 C different teams possible. 6. Choose of centers, of 3 guards, and of 7 forwards. There are C(,) C(3,) C(7,) 3 6 different teams possible. 63. a. If numbers can be repeated, there are (50)(50)(50) 5,000 different lock combinations. If no number can be repeated, then there are ,600 different lock combinations. b. Answers will vary. Typical combination locks require two full clockwise rotations to the first number, followed by a full counter-clockwise rotation past the first number to the second number, followed by a clockwise rotation to the third number (not past the second). This is not clear from the given directions. Perhaps a better name for a combination lock would be a permutation lock since the order in which the numbers are entered matters Answers will vary. 66. A permutation is an ordered arrangement of objects while with a combination order does not matter. For example, the number of ways the teams in the Big Ten can come in first, second, and third would be a permutation problem. The number of ways to pick 6 numbers in the Illinois State lottery is a combination problem because the order in which the numbers are selected is irrelevant. Section 3.3. equally likely. complement 3. False; probability may equal 0. In such cases, the corresponding event will never happen. 4. True; in a valid probability model, all probabilities are between 0 and, and the sum of the probabilities is. 5. Probabilities must be between 0 and, inclusive. Thus, 0, 0.0, 0.35, and could be probabilities. 6. Probabilities must be between 0 and, inclusive. Thus,, 3,, and 0 could be probabilities All the probabilities are between 0 and. The sum of the probabilities is This is a probability model. 8. All the probabilities are between 0 and. The sum of the probabilities is This is a probability model. 9. All the probabilities are between 0 and. The sum of the probabilities is 308 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

7 Section 3.3: Probability This is not a probability model. 0. One probability is not between 0 and. This is not a probability model.. The sample space is: S { HH, HT, TH, TT}. Each outcome is equally likely to occur; so ne ( ) PE ( ). The probabilities are: PHH ( ), PHT ( ), PTH ( ), PTT ( ) The sample space is: S { HH, HT, TH, TT}. Each outcome is equally likely to occur; so ne ( ) PE ( ). The probabilities are: PHH ( ), PHT ( ), PTH ( ), PTT ( ) The sample space of tossing two fair coins and a fair die is: S { HH, HH, HH3, HH4, HH5, HH6, HT, HT, HT3, HT 4, HT5, HT 6, TH, TH, TH3, TH 4, TH5, TH 6, TT, TT, TT3, TT 4, TT5, TT6} There are 4 equally likely outcomes and the probability of each is The sample space of tossing a fair coin, a fair die, and a fair coin is: S { H H, H H, H3 H, H4 H, H5 H, H6 H, H T, H T, H3 T, H4 T, H5 T, H6 T, T H, T H, T3 H, T4 H, T5 H, T6 H, T T, T T, T3 T, T4 T, T5 T, T6 T} There are 4 equally likely outcomes and the probability of each is The sample space for tossing three fair coins is: S { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} There are 8 equally likely outcomes and the probability of each is The sample space for tossing one fair coin three times is: S { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }. There are 8 equally likely outcomes and the probability of each is The sample space is: S { Yellow, Red, Green, Yellow, Red, Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green} There are equally likely events and the probability of each is. The probability of getting a or 4 followed by a Red is P( Red) + P(4 Red) The sample space is: S {Forward Yellow, Forward Red, Forward Green, Backward Yellow, Backward Red, Backward Green} There are 6 equally likely events and the probability of each is. The probability of 6 getting Forward followed by Yellow or Green is: P( Forward Yellow) + P( Forward Green) The sample space is: S { Yellow Forward, Yellow Backward, Red Forward, Red Backward, Green Forward, Green Backward, Yellow Forward, Yellow Backward, Red Forward, Red Backward, Green Forward, Green Backward, 3 Yellow Forward, 3 Yellow Backward, 3 Red Forward, 3 Red Backward, 3 Green Forward, 3 Green Backward, 4 Yellow Forward, 4 Yellow Backward, 4 Red Forward, 4 Red Backward, 4 Green Forward, 4 Green Backward} There are 4 equally likely events and the probability of each is 4. The probability of getting a, followed by a Red or Green, followed by a Backward is: P( Red Backward) + P( Green Backward) The sample space is: S {Yellow Forward, Yellow Backward, Red Forward, Red Backward, Green Forward, Green Backward, Yellow Forward, Yellow Backward, Red Forward, Red Backward, Green Forward, Green Backward, Yellow 3 Forward, Yellow 3 Backward, Red Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

8 Chapter 3: Counting and Probability Forward, Red 3 Backward, Green 3 Forward, Green 3 Backward, Yellow 4 Forward, Yellow 4 Backward, Red 4 Forward, Red 4 Backward, Green 4 Forward, Green 4 Backward} There are 4 equally likely events and the probability of each is 4. The probability of getting a Yellow, followed by a or 4, followed by a Forward is P(Yellow Forward) + P(Yellow 4 Forward) The sample space is: S { Yellow, Red, Green, Yellow, Red, Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green, Yellow, Red, Green, Yellow, Red, Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green, 3 Yellow, 3 Red, 3 Green, 3 Yellow, 3 Red, 3 Green, 3 3 Yellow, 3 3 Red, 3 3 Green, 3 4 Yellow, 3 4 Red, 3 4 Green, 4 Yellow, 4 Red, 4 Green, 4 Yellow, 4 Red, 4 Green, 4 3 Yellow, 4 3 Red, 4 3 Green, 4 4 Yellow, 4 4 Red, 4 4 Green} There are 48 equally likely events and the probability of each is. The probability of 48 getting a, followed by a or 4, followed by a Red or Green is P( Red) + P( 4 Red) + P( Green) + P( 4 Green) The sample space is: S {Forward, Forward, Forward 3, Forward 4, Forward, Forward, Forward 3, Forward 4, Forward 3, Forward 3, Forward 33, Forward 34, Forward 4, Forward 4, Forward 43, Forward 44, Backward, Backward, Backward 3, Backward 4, Backward, Backward, Backward 3, Backward 4, Backward 3, Backward 3, Backward 33, Backward 34, Backward 4, Backward 4, Backward 43, Backward 44} There are 3 equally likely events and the probability of each is. The probability of 3 getting a Forward, followed by a or 3, followed by a or 4 is P( Fwd ) + P( Fwd 4) + P( Fwd 3) + P( Fwd 34) A, B, C, F 4. A (equally likely outcomes) 5. B 6. F 7. Let P(tails) x, then P(heads) 4x x+ 4x 5x x 5 4 P(tails), P(heads) Let P(heads) x, then P(tails) x x+ x 3x x 3 P(heads), P(tails) P() P(4) P(6) x P() P(3) P(5) x P() + P() + P(3) + P(4) + P(5) + P(6) x+ x+ x+ x+ x+ x 9x x 9 P() P(4) P(6) 9 P() P(3) P(5) 9 30 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

9 Section 3.3: Probability 30. P() P() P(3) P(4) P(5) x; P(6) 0 P() + P() + P(3) + P(4) + P(5) + P(6) x+ x+ x+ x+ x+ 0 5x x 5 P() P() P(3) P(4) P(5) ; P(6) ne ( ) n{,,3} 3 PE ( ) 0 0 nf ( ) n{3,5,9,0} 4 PF ( ) ne ( ) n{,4,6,8,0} 5 PE ( ) n( girls, boys) 6 3 P( girls, boys) 6 8 n( sum of two dice is 7) P( sum of two dice is 7) n{,6 or,5 or 3,4 or 4,3 or 5, or 6,} n( sum of two dice is ) P( sum of two dice is ) n{5,6 or 6,5} 36 8 n( sum of two dice is 3) P( sum of two dice is 3) n{, or,} nf ( ) n{,3,5,7,9} 5 PF ( ) 0 0 n(white) 5 5 P(white) n(black) 7 7 P(black) The sample space is: S {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} n(3 boys) P(3 boys) The sample space is: S {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} n(3 girls) P(3 girls) The sample space is: S {BBBB, BBBG, BBGB, BGBB, GBBB, BBGG, BGBG, GBBG, BGGB, GBGB, GGBB, BGGG, GBGG, GGBG, GGGB, GGGG} n( girl, 3 boys) 4 P( girl, 3 boys) The sample space is: S {BBBB, BBBG, BBGB, BGBB, GBBB, BBGG, BGBG, GBBG, BGGB, GBGB, GGBB, BGGG, GBGG, GGBG, GGGB, GGGG} 44. n( sum of two dice is ) P( sum of two dice is ) n{6, 6} PA ( B) PA ( ) + PB ( ) PA ( B) PA ( B) PA ( ) + PB ( ) PA ( B) PA ( B) PA ( ) + PB ( ) PA ( B) PA ( B) PA ( ) + PB ( ) PA ( B) PB ( ) 0.05 PB ( ) PA ( B) PA ( ) + PB ( ) PA ( B) 0.65 PA ( ) PA ( ) P(theft not cleared) P( theft cleared) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

10 Chapter 3: Counting and Probability 5. P(does not own a pet) P(owns a pet) P(does not own cat) P(owns cat) P(not in engineering) P(in engineering) P(never gambled online) P(gambled online) P( not shortbread/trefoils) P( shortbread/trefoils) P(white or green) P(white) + P(green) n(white) + n(green) P(white or orange) P(white) + P(orange) n(white) + n(orange) P(not white) P(white) n(white) P(not green) P(green) n(green) P(strike or one) P(strike) + P(one) n(strike) + n(one) P(00 or 30) P(00) + P(30) n(00) + n(30) There are 30 households out of 00 with an income of $30,000 or more. ne ( ) n(30, 000 or more) 30 3 PE ( ) n(total households) There are 65 households out of 00 with an income between $0,000 and $9,999. ne ( ) n(0,000 to 9,999) 65 3 PE ( ) n(total households) There are 40 households out of 00 with an income of less than $0,000. ne ( ) n(less than $0,000) 40 PE ( ) n(total households) There are 60 households out of 00 with an income of $0,000 or more. ne ( ) n($0,000 or more) 60 3 PE ( ) n(total households) a. P( or ) P() + P() b. P or more P( ) ( ) none c. P(3 or fewer) P( 4 or more) d. P(3 or more) P(3) + P(4 or more) e. P(fewer than ) P(0) + P() f. P(fewer than ) P(0) 0.05 g. P(,, or 3) P() + P() + P(3) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

11 Chapter 3 Review Exercises h. P( or more) P() + P(3) + P(4 or more) a. P(at most ) P(0) + P() + P() b. P(at least ) P() + P(3) + P(4 or more) c. P(at least ) P(0) P(at least with same birthday) P(none with same birthday) n(different birthdays) a. b. P(freshman or female) P( freshman) + P( female) P( freshman and female) n( freshman) + n( female) n( freshman and female) P(sophomore or male) P( sophomore) + P( male) P( sophomore and male) n( sophomore) + n( male) n( sophomore and male) The sample space for picking 5 out of 0 numbers in a particular order contains 0! 0! P (0, 5) 30, 40 possible (0 5)! 5! outcomes. One of these is the desired outcome. Thus, the probability of winning is: ne ( ) n(winning) PE ( ) n(total possible outcomes) ,40 Chapter 3 Review Exercises 70. a. b. P(female or under 40) P( female) + P( under 40) P( female and under 40) n( female) + n( under 40) n( female and under 40) P(male or over 40) P(male) + P(over 40) P(male and over 40) n(male) + n(over 40) n(male and over 40) P(at least with same birthday) P(none with same birthday) n(different birthdays) ,{ Dave },{ Joanne },{ Erica },{ Dave,Joanne }, { Dave, Erica }, { Joanne, Erica }, { Dave, Joanne,Erica}.,{ Green },{ Blue },{ Red },{ Green,Blue }, { Green, Red }, { Blue, Red }, { Green, Blue, Red } 3. na ( ) 8, nb ( ), na ( B) 3 na ( B) na ( ) + nb ( ) na ( B) na ( ), na ( B) 30, na ( B) 6 na ( B) na ( ) + nb ( ) na ( B) 30 + nb ( ) 6 nb ( ) From the figure: na ( ) From the figure: na ( or B ) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

12 Chapter 3: Counting and Probability 7. From the figure: na ( and C) na ( C) From the figure: n(not in B ) From the figure: n( neither in A nor in C) n( A C) From the figure: n ( in B but not in C) ! 8! ! P(8,3) 336 (8 3)! 5! 5! 7! 7! ! P(7,3) 0 (7 3)! 4! 4! 8! 8! ! C(8, 3) 56 (8 3)!3! 5!3! 5! 3 7! 7! 7654! C(7,3) 35 (7 3)!3! 4!3! 4! 3 5. There are choices of material, 3 choices of color, and 0 choices of size. The complete assortment would have: suits. 6. This is a permutation of 5 items taken 5 at a 5! 5! time. There are P (5,5) 5! 0 (5 5)! 0! possible wirings. 7. There are two possible outcomes for each game 7 or 8outcomes for 7 games. 8. There are two possible outcomes for each game 6 or 64outcomes for 6 games. 9. Since order is significant, this is a permutation. 9! 9! ! P(9,4) 304 (9 4)! 5! 5! ways to seat 4 people in 9 seats. 0. Since order is significant, this is a permutation. 4! 4! 4 3 P(4,4) 4 (4 4)! 0! arrangements of the letters in ROSE.. Choose 4 runners order is significant: 8! 8! ! P(8, 4) 680 (8 4)! 4! 4! ways a squad can be chosen.. Choose 3 problems order is not significant: 0! 0! ! C(0, 3) 0 (0 3)!3! 7! 3! 3 7! different tests are possible. 3. Choose teams from 4 order is not significant: 4! 4! 4 3! C(4,) 9 (4 )!!!!! ways to choose teams. 4. a. Since order is important, this is a permutation: 5! 5! P(5,5) P(5,5) (5 5)! (5 5! 5! 5! 0 0 4,400 different arrangements. b. There would be ,400 different arrangements. 5. There are ,600,000 possible phone numbers. 6. There are different types of homes that can be built. 7. There are ,000possible license plates. 8. There are two choices for each digit, so there are 8 56 different numbers. (Note this allows numbers with initial zeros, such as 0.) 9. Since there are repeated letters: 7! different words!! can be formed. 30. Since there are repeated colors: 0! ,600 4!3!!! 433 different vertical arrangements. 3. a. C(9,4) C(9,3) C(9,) , 04 committees can be formed. 34 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

13 Chapter 3 Test 3. a. 33. a. b. C(9,4) C(5,3) C(, ) committees can be formed. 5! 8! C(5,) C(8,3) (5 )!! (8 3)!3! committees containing exactly man. b. C(5, ) C(8,) committees containing exactly women. c. C(5,) C(8,3) + C(5, ) C(8, ) + C(5,3) C(8,) committees containing at least man. b P(no one has same birthday) c. P(at least have same birthday) P(no one has same birthday) a. P (heart disease) 0.9 b. P(not heart disease) P(heart disease) a. P (unemployed) b. P(not unemployed) P(unemployed) n(40 watt) 3 P(40 watt) 0.5 n(bulbs) 0 n(75 watt) P(not 75 watt) P(75 watt) n(bulbs) n($ bill) 4 P($ bill) Let S be all possible selections, so 00. Let D be a card that is divisible by 5, so nd ( ) 0. Let PN be a card that is or a prime number, so npn ( ) 6. nd ( ) 0 PD ( ) npn ( ) 6 3 PPN ( ) Let S be all possible selections, let T be a car that needs a tune-up, and let B be a car that needs a brake job. a. P( Tune-up or Brake job) PT ( B) PT ( ) + PB ( ) PT ( B) b. P( Tune-up but not Brake job) P Tune-up P Tune-up and Brake job PT ( ) PT ( B) c. P( Neither Tune-up nor Brake job) P(Tune-up or Brake job) ( PT ( ) + PB ( ) PT ( B) ) ( ) 0.3 Chapter 3 Test ( ) ( ). From the figure: n physics ( ). From the figure: n biology or chemistry or physics Therefore, n none of the three ( ) ( ) 38. P(ROSE) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

14 Chapter 3: Counting and Probability 3. From the figure: n( only biology and chemistry) n biol. and chem. n biol. and chem. and phys. ( 8+ ) 8 ( ) ( ) 4. From the figure: n( physics or chemistry) ! ! 0! P ( 0 6 )! 4! ! 4! ,00 6. ( 0,6)!! C 5! ( 5 )! 5!6! ! 5436! (,5 ) 8. Since the order in which the colors are selected doesn t matter, this is a combination problem. We have n colors and we wish to select r 6 of them.!! C (,6 ) 6! ( 6 )! 6!5! ! 6!5! , 64 There are 54,64 ways to choose 6 colors from the available colors. 9. Because the letters are not distinct and order matters, we use the permutation formula for nondistinct objects. We have four different letters, two of which are repeated (E four times and D two times). n! 8! n! n! n3! n4! 4!!!! 87654! 4! There are 840 distinct arrangements of the letters in the word REDEEMED. 0. Since the order of the horses matters and all the horses are distinct, we use the permutation formula for distinct objects. 8! 8! 8 7 6! P( 8, ) ( 8 )! 6! 6! There are 56 different exacta bets for an 8-horse race.. We are choosing 3 letters from 6 distinct letters and 4 digits from 0 distinct digits. The letters and numbers are placed in order following the format LLL DDDD with repetitions being allowed. Using the Multiplication Principle, we get ,480,000 Note that there are only 3 possibilities for the third letter. There are 55,480,000 possible license plates using the new format.. Let A Kiersten accepted at USC, and B Kiersten accepted at FSU. Then, we get P A 0.60, P( B ) 0.70, and ( ) ( B) 0.35 P A. a. Here we need to use the Addition Rule. P( A B) P( A) + P( B) P( A B) Kiersten has a 95% chance of being admitted to at least one of the universities. b. Here we need the Complement of an event. ( ) P( B) P B Kiersten has a 30% chance of not being admitted to FSU. 3. a. Since the bottle is chosen at random, all bottles are equally likely to be selected. Thus, 5 5 P ( Coke) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

15 Chapter 3 Cumulative Review There is a 5% chance that the selected bottle contains Coke Pepsi IBC There is a 55% chance that the selected bottle contains either Pepsi or IBC. b. P( ) 4. Since the ages cover all possibilities and the age groups are mutually exclusive, the sum of all the probabilities must equal The given probabilities sum to 0.8. This means the missing probability (for 8-0) must be The number of different selections of 6 numbers is the number of ways we can choose 5 white balls and red ball, where the order of the white balls is not important. This requires the use of the Multiplication Principle and the combination formula. Thus, the total number of distinct ways to pick the 6 numbers is given by n( white balls) n( red ball) C( 53, 5) C( 4,) 53! 4! 5! ( 53 5 )!! ( 4 )! 53! 4! 5! 48!! 4! ! 543 4! ,56,770 Since each possible combination is equally likely, the probability of winning on a $ play is P ( win on $ play) 0, 56, The number of elements in the sample space can be obtained by using the Multiplication Principle: ,776 Consider the rolls as a sequence of 5 slots. The number of ways to position fours in 5 slots is C ( 5, ). The remaining three slots can be filled with any of the five remaining numbers from the die. Repetitions are allowed so this can be done in 555 5different ways. Therefore, the total number of ways to get exactly fours is 5! C ( 5,) ! 3! The probability of getting exactly fours on 5 rolls of a die is given by 50 P ( exactly fours) Chapter 3 Cumulative Review.. 3x x 3x x+ 0 b± b 4ac x a ± ( ) ( ) 4( 3)( ) 3 ( ) ± 4 ± ± i ± i 6 3 i + i The solution set is,. 3 3 f( x) x + 4x 5 a, b 4, c 5. Since a > 0, the graph opens up. The x-coordinate of the vertex is b 4 x. a () The y-coordinate of the vertex is b f f ( ) ( ) + 4( ) 5. a Thus, the vertex is (, 9). The axis of symmetry is the line x.the discriminant is: b 4 ac (4) 4()( 5) > 0. So the graph has two x-intercepts. The x-intercepts are found by solving: x + 4x 5 0 ( x+ 5)( x ) 0 x 5 or x The x-intercepts are 5 and. 37 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

16 Chapter 3: Counting and Probability The y-intercept is f (0) (0) + 4 (0) 5 5. Graphing using the bounds: (Second graph has a better window.) y x+ 4 Using the graph of y x, horizontally shift to the left unit, vertically stretch by a factor of, and vertically shift down 4 units. 3. ( ) 4. x x x x 4.0 The solution set is x 3.99 x 4.0 or 3.99, 4.0 { } [ ] f x 5x 9x 7x 3x 6 Step : f( x ) has at most 4 real zeros. Step : Possible rational zeros: p ±, ±, ± 3, ± 6; q ±, ± 5; p 3 6 ±, ±, ±, ±, ± 3, ±, ± 6, ± q Step 3: Using the Bounds on Zeros Theorem: 5. ( ) ( ) f( x) 5 x.8x.4x 6.x. a3.8, a.4, a 6., a0. Max, Max, { } { } { } + Max., 6.,.4, The smaller of the two numbers is 7.. Thus, every zero of f lies between 7. and 7.. Step 4: From the graph we see that there are x-intercepts at 0. and 3. Using synthetic division with 3: Since the remainder is 0, x 3 is a factor. The 3 other factor is the quotient: 5x + 6x + x+. Using synthetic division with on the quotient: x+ 0. is a factor. The other factor is the quotient: 5x + 5x+ 0 5 x + x+. Since the remainder is 0, x ( ) ( ) Factoring, f( x) 5( x + x+ ) ( x 3)( x+ 0.) The real zeros are 3 and 0.. The complex zeros come from solving x + x+ 0. ()( ) () b± b 4ac ± 4 x a ± 8 ± 7 ± 7i Therefore, over the set of complex numbers, f x 5x 9x 7x 3x 6 has zeros ( ) i, i,, Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

17 Chapter 3 Cumulative Review x gx ( ) Using the graph of y 3 x, shift the graph horizontally unit to the right, then shift the graph vertically 5 units upward. Domain: All real numbers or (, ) Range: { y y > 5} or (5, ) Horizontal Asymptote: y 5 log (9) log (3 ) log (3x ) + log x 4 log ( x(3x ) ) 4 4 x(3x ) 3x x 6 3x x 6 0 (3x 8)( x+ ) 0 8 x or x 3 Since x makes the original logarithms 8 undefined, the solution set is Multiply each side of the first equation by 3 and add to the second equation to eliminate x; multiply each side of the first equation by and add to the third equation to eliminate x: x y+ z 5 3x+ y 3z 8 x + 4 y z 7 3x+ 6y 3z 45 3x+ y 3 z 8 7y 6z 53 ( ) z 3 7y y 35 y 5 z 3, y 5 x ( 5) x x The solution is x, y 5, z 3 or (, 5, 3). 0. 3,, 5, 9,... is an arithmetic sequence with a 3, d 4. Using a a+ ( n ) d, n a (33 ) To compute the sum of the first 0 terms, we use 0 S0 ( a+ a0 ). a (0 ) Therefore, 0 S0 ( a+ a0 ) 0 ( 3+ 73) π y 3sin x+ 3sin x+ Amplitude: A 3 3. ( π ) π Period: T π φ π π Phase Shift: ω x y+ z 5 x 4y+ z 30 x+ 4 y z 7 x+ 4y z 7 z 3 Substituting and solving for the other variables: 39 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

18 Chapter 3: Counting and Probability. Use the Law of Cosines: a b + c bccosa a o cos cos 40 a 6.09 b a + c accosb o a + c b cos B ac ( )( ) B cos o o o o o o C 80 A B Area of the triangle 59sin40 o ( ) 4.46 square units. Chapter 3 Projects Project I. Research will vary. See answer to part (c).. P ( win not switched) 3 P ( win switched) 3 Results of simulations will vary. 3. In the Monty Hall Game, a curtain is selected by the contestant and left unopened. The host then reveals the contents behind one of the unselected curtains. In this situation, the host knows the contents behind the curtain being opened. The grand prize will never be revealed by the host. In Deal or No Deal, a suitcase is selected by the contestant and left unopened. The contestant then chooses another unselected suitcase to open. In this situation, the content within the suitcase being opened is not known. Since the contestant selects the case to open, the grand prize may be revealed (and eliminated). 4. P (winning grand prize) 6 5. Answers will vary. 6. Answers will vary. Project II. 0 bit errors: 0 bit errors: bit errors: bit errors: bit errors: 000. P( symbol received correctly) # of received symbols with bit errors: (8, ) 8 C 56 P( received correctly) P received incorrectly P received correctly ( ) ( ) 4. Let k # of errors, n 8 length of symbol. Probability of k errors : 8 30 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall

19 Chapter 3 Projects Project III n k n k Pnk (, ) ( p) ( p) k k 8 k 8 P(8, k) k 3 3 Since this parity code only detects odd numbers of errors, P(error detected) P(8,) + P(8,3) + P(8,5) + P(8,7) To find the probability that an error occurred but is not detected, we need to assume that an even number of errors occurred: P(error occured, but not detected) P(8, ) + P(8, 4) + P(8,6) + P(8,8) Project IV e. Answers will vary, depending on the L generated by the calculator. f. The data accumulates around y 0.5. Project V One simulation might be: Woman Woman told you has about Probability Boy-Boy Older boy 4 Boy-Boy Younger boy 4 Boy-Girl Younger boy 4 Girl-Boy Older boy 4 We leave out the combinations where she would have to tell you about a girl. Thus, the probability that she has boys is Man has Boy-Boy Girl-Boy Man told you about Older boy Older boy Probability Thus the probability he has two boys is. The probabilities are the same. Answers will vary. 3 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall