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Part IV: Complete Solutions, Chapter 4 296 Chapter 4: Elementary Probability Theory Section 4.1 1. Equally likely outcomes, relative frequency, intuition 2. The complement is “not rain today.” This probability is 100% – 30% = 70%. 3. (a) The probability of a certain event is 1. (b) The probability of an impossible event is 0. 4. The law of large numbers states that in the long run, as the sample size or number of trials increases, the relative frequency of outcomes approaches the theoretical probability of the outcome. Five hundred trials are better because the law of large numbers works better for larger samples. 5. No. The probability of throwing tails on the second toss is 0.50 regardless of the outcome on the first toss. 6. (a) (b) (c) (d) 7. The resulting relative frequency can be used as an estimate of the true probability of all Americans who can wiggle their ears. 8. The resulting relative frequency can be used as an estimate of the true probability of all Americans who can raise one eyebrow. 15 71 124 131 34 , P(1) , P(2) , P(3) , P(4) (a) P(no similar preferences) P(0) 375 375 375 375 375 15 71 124 131 34 375 1, yes (b) 375 375 Personality types were classified into four main preferences; all possible numbers of shared preferences were considered. The sample space is 0, 1, 2, 3, and 4 shared preferences. 9. Probabilities must be between 0 and 1 inclusive. –0.41 < 0 Probabilities must be between 0 and 1 inclusive. 1.21 > 1 120% = 1.20, and probabilities must be between 0 and 1 inclusive. 1.20 > 1 Yes, 0 ≤ 0.56 ≤ 1. 10. (a) The sample space would be 1, 2, 3, 4, 5, and 6 dots. If the die is fair, all outcomes will be equally likely. 1 (b) P(1) P(2) P(3) P(4) P(5) P(6) because the die faces are equally likely, and there are six 6 1 1 1 1 1 1 6 outcomes. The probabilities should and do add to 1 1 because all 6 6 6 6 6 6 6 possible outcomes have been considered. 1 1 1 1 4 2 (c) P(number of dots < 5) = P(1 or 2 or 3 or 4 dots) = P(1) + P (2) + P(3) + P(4) 6 6 6 6 6 3 1 2 or P(dots < 5) = 1 – P(5 or 6 dots) 1 (The applicable probability rule used here will be 3 3 discussed in the next section of the text; rely on your common sense for now.) (d) Complementary event rule: P(A) = 1 – P(not A) P(5 or 6 dots) = 1 – P(1 or 2 or 3 or 4 dots) 1 2 1 1 1 2 1 , or P(5 or 6) = P(5) + P(6) 3 3 6 6 6 3 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 297 11. (a) Note: “Includes the left limit but not the right limit” means 6 A.M. ≤ time t < noon, noon ≤ t < 6 P.M., 6 P.M. ≤ t < midnight, midnight ≤ t < 6 A.M. 290 0.30 966 135 0.14 P(best idea 12 noon–6 P.M.) 966 319 0.33 P(best idea 6 P.M.–12 midnight) 966 222 0.23 P(best idea from 12 midnight to 6 A.M.) 966 (b) The probabilities add up to 1. They should add up to 1 provided that the intervals do not overlap and each inventor chose only one interval. The sample space is the set of four time intervals. P(best idea 6 A.M.–12 noon) 12. (a) P(germinate) = number germinated 2, 430 0.81 number planted 3, 000 3,000 2, 430 570 0.19 3,000 3,000 (c) The sample space is two outcomes, germinate and not germinate. (b) P(not germinate) = P(germinate) + P(not germinate) = 0.81 + 0.19 = 1 The probabilities of all the outcomes in the sample space should and do sum to 1. (d) No because P(germinate) = 0.81 ≠ P(not germinate) = 0.19 1 If they were equally likely, each would have probability 0.5. 2 n 13. (a) Given: Odds in favor of A are n:m i.e., . m n Show: P( A) mn P( A) Proof: Odds in favor of A are by definition P(not A) P(not A) 1 P( A) complementary events n P( A) P( A) substitution m P(not A) 1 P( A) n[1 P( A)] m[ P( A)] cross multiply n n[ P( A)] m[ P( A)] n n[ P( A)] m[ P( A)] n (n m)[ P ( A)] So n P ( A), as was to be shown. nm (b) Odds of a successful call are 2 to 15. Now 2 to 15 can be written as 2:15 or From part (a): if the odds are 2:15 (let n = 2, m = 15), then P(sale) Copyright © Houghton Mifflin Company. All rights reserved. 2 . 15 n 2 2 0.118. n m 2 15 17 Part IV: Complete Solutions, Chapter 4 298 (c) Odds of free throw are 3 to 5, i.e., 3:5. Let n = 3 and m = 5 here; then, from part (a): P(free throw) n 3 3 0.375 n m 35 8 a 14. (a) Given: Odds against W are a:b or . b Show: P(not W) a . ab P (not W ) by definition. P (W ) P(W ) 1 P(not W ) complementary events P(not W ) a substitution P(W ) b P(not W ) a substitution 1 P(not W ) b b[ P(not W )] a[1 P(not W )] cross-multiply b[ P(not W )] a a[ P(not W )] b[ P(not W )] a[ P(not W )] a (a b)[ P (not W )] a a P(not W ) ab a P(not W ) , as was to be shown. ab Proof: Odds against W are (b) Point Given’s betting odds are 9:5. Betting odds are based on the probability that the horse does not P(not PG wins) . win, so odds against Point Given (PG) winning are P(PG wins) Let a = 9 and b = 5 in part (a) formula. From part (a), P(not PG wins) “not PG wins” is the same as “PG loses,” so P(PG loses) 1 a 9 9 , but event a b 9 5 14 9 0.64, and P(PG wins) 14 9 5 0.36. 14 14 (c) Betting odds for Monarchos are 6:1. Betting odds are based on the probability that the horse does not win; i.e., the horse loses. Let W be the event that Monarchos wins. From part (a), if the events against W are given as a:b, the a . Let a = 6 and b = 1 in the part (a) formula, so P(not W) ab Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 P(not W ) 299 6 6 6 1 7 6 0.86 7 P(Monarchos wins) P(W ) 1 P(not W ) 6 1 1 0.14 7 7 P(not W ) P(Monarchos loses)= (d) Invisible Ink was given betting odds of 30 to 1; i.e., odds against Invisible Ink winning were 30 . 1 Let W denote the event that Invisible Ink wins. Let a = 30, b = 1 in formula from part (a). Then, from part (a), P(not W) a 30 30 , P (not Invisible Ink wins) ; i.e., ab 30 1 31 30 0.97 31 P (Invisible Ink wins) 1 P(Invisible Ink loses) 30 1 1 0.03 31 31 P (Invisible Ink loses) 15. One approach is to make a table showing the information about the 127 people who walked by the store. Came into the store Did not come in Column total Buy 25 0 25 Did Not Buy 58 – 25 = 33 69 102 Row Total 58 127 – 58 = 69 127 If 58 came in, 69 didn’t; 25 of the 58 bought something, so 33 came in but didn’t buy anything. Those who did not come in couldn’t buy anything. The row entries must sum to the row totals, the column entries must sum to the column totals, and the row totals, as well as the column totals, must sum to the overall total, i.e., the 127 people who walked by the store. Also, the four inner cells must sum to the overall total: 25 + 33 + 0 + 69 = 127. number outcomes favorable to A . This kind of problem relies on formula (2), P(event A) total number of outcomes (a) P ( A) 58 0.46 127 Here, we divide by 127 people. (b) P( A) 25 0.43 58 Here, we divide by 58 people (only those who entered). (c) P( A) P(Enter and buy) 58 25 25 0.20 127 58 127 Or similarly, read from the table that 25 people both entered and bought something. Divide this by the total number of people, namely, 127. 33 0.57 (d) P( A) P(Buy nothing) Here, we divide by 58 people. 58 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 300 Section 4.2 1. No. Mutually exclusive events cannot occur at the same time. 2. If A and B are independent, then P(A) = P(A | B). Therefore, P(A | B) = 0.3. 3. (a) Event A cannot occur if event B has occurred. Therefore, P(A | B) = 0. (b) Since we are told that P(A) ≠ 0, and we have determined that P(A | B) = 0, we can deduce that P(A) ≠ P(A | B). Therefore, events A and B are not independent. 4. (a) P(A and B) = P(A) × P(B) if events A and B are independent. This product can equal zero only if either P(A) = 0 or P(B) = 0 (or both). We are told that P(A) ≠ 0 and that P(B) ≠ 0. Therefore, P(A and B) ≠ 0. (b) By the preceding line, the definition of mutually exclusive events is violated. Thus A and B are not mutually exclusive. 5. 6. 7. (a) P(A and B) (b) P(B | A) (d) P(A or B) (e) P(A or Bc) (a) P(Ac or B) (b) P(B | A) (d) P(A and Bc) (e) P(A and B) (c) P(Ac | B) (c) P(A | B) (a) Green and blue are mutually exclusive because each M&M candy is only one color. P(green or blue) = P(green) + P(blue) = 10% + 10% = 20% = 0.20. (b) Yellow and red are mutually exclusive once again because each candy is only one color. P(yellow or red) = P(yellow) + P(red) = 20% + 20% = 40% = 0.40. (c) Use the complementary event. P(not purple) = 1 – P(purple) = 1 – 0.20 = 0.80 = 80% 8. The total number of arches tabled is 288. Arch heights are mutually exclusive. 111 (a) P(3 to 9 feet) 288 30 33 18 81 (b) P(30 feet or taller) = P(30 to 49) + P(50 to 74) + P(75 and higher) 288 288 288 288 111 96 30 237 (c) P(3 to 49 feet) = P(3 to 9) + P(10 to 29) + P(30 to 49) 288 288 288 288 96 30 33 159 (d) P(10 to 74 feet) = P(10 to 29) + P(30 to 49) + P(50 to 74) 288 288 288 288 18 (e) P(75 feet or taller) 288 Hint: For Problems 9–12, refer to Figure 4-2 if necessary. Think of the outcomes as an (x, y) ordered pair. 9. (a) Yes, the outcome of the red die does not influence the outcome of the green die. 1 1 1 (b) P(5 on green and 3 on red) = P(5 on green) · P(3 on red) 0.028 . 6 6 36 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 301 1 1 1 (c) P(3 on green and 5 on red) = P(3 on green) · P(5 on red) 0.028 6 6 36 (d) P[(5 on green and 3 on red) or (3 on green and 5 on red)] = P(5 on green and 3 on red) + P(3 on green and 5 on red) 1 1 2 1 0.056 (because they are mutually exclusive outcomes). = 36 36 36 18 10. (a) Yes. 1 1 1 (b) P(1 on green and 2 on red) = P(1 on green) · P(2 on red) 6 6 36 1 1 1 (c) P(2 on green and 1 on red) = P(2 on green) · P(1 on red) 6 6 36 (d) P[(1 on green and 2 on red) or (2 on green and 1 on red)] = P(1 on green and 2 on red) + P(2 on green and 1 on red) = 1 1 2 1 (because they are mutually exclusive outcomes). 36 36 36 18 11. (a) We can obtain a sum of 6 as follows: 1+5=6 2+4=6 3+3=6 4+2=6 5+1=6 P(sum 6) P[(1, 5) or (2, 4) or (3 on red, 3 on green) or (4, 2) or (5, 1)] P(1, 5) P(2, 4) P(3, 3) P(4, 2) P(5, 1) because the (red, green) outcomes are mutually exclusive 1 1 1 1 1 1 1 1 1 1 6 6 6 6 6 6 6 6 6 6 because the red die outcome is independent of the green die outcome 1 1 1 1 1 5 36 36 36 36 36 36 (b) We can obtain a sum of 4 as follows: 1+3=4 2+2=4 3+1=4 P(sum is 4) P[(1, 3) or (2, 2) or (3, 1)] P(1, 3) P(2, 2) P(3, 1) because the (red, green) outcomes are mutually exclusive 1 1 1 1 1 1 6 6 6 6 6 6 because the red die outcome is independent of the green die outcome 1 1 1 3 1 36 36 36 36 12 (c) You cannot roll a sum of 6 and a sum of 4 at the same time. These are mutually exclusive events. 5 3 8 2 P(sum of 6 or 4) = P(sum of 6) + P(sum of 4) = 36 36 36 9 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 302 12. (a) We can obtain a sum of 7 as follows: 1+6=7 2+5=7 3+4=7 4+3=7 5+2=7 6+1=7 P(sum is 7) P[(1, 6) or (2, 5) or (3, 4) or (4, 3) or (5, 2) or (6, 1)] P(1, 6) P(2, 5) P(3, 4) P(4, 3) P(5, 2) P(6, 1) because the (red, green) outcomes are mutually exclusive 1 1 1 1 1 1 1 1 1 1 1 1 6 6 6 6 6 6 6 6 6 6 6 6 because the red die outcome is independent of the green die outcome 1 1 1 1 1 1 6 1 36 36 36 36 36 36 36 6 (b) We can obtain a sum of 11 as follows: 5 + 6 = 11 or 6 + 5 = 11 P(sum is 11) P[(5, 6) or (6, 5)] P(5, 6) P(6, 5) because the (red, green) outcomes are mutually exclusive 1 1 1 1 6 6 6 6 because the red die outcome is independent of the green die outcome 1 1 2 1 36 36 36 18 (c) You cannot roll a sum of 7 and a sum of 11 at the same time. These are mutually exclusive events. 6 2 8 2 P(sum is 7 or 11) = P(sum is 7) + P(sum is 11) = 36 36 36 9 13. (a) No, the draws are not independent. The key idea is “without replacement” because the probability of the second card drawn depends on the first card drawn. Let the card draws be represented by an (x, y) ordered pair. For example, (K, 6) means the first card drawn was a king and the second card drawn was a 6. Here the order of the cards is important. 16 4 4 4 (b) P(ace on first draw and king on second draw) = P(ace, king) 52 51 2,652 663 There are four aces and fpour kings in the deck. Once the first card is drawn and not replaced, there are only 51 cards left to draw from, but all the kings are available. 16 4 4 4 (c) P(king, ace) 52 51 2652 663 (d) P(ace and king in either order) = P[(ace, king) or (king, ace)] = P(ace, king) + P(king, ace) because these two outcomes are mutually exclusive 16 16 32 8 = 2,652 2,652 2,652 663 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 303 14. (a) No, the draws are not independent. The key idea is “without replacement” because the probability of the second card drawn depends on the first card drawn. Let the card draws be represented by an (x, y) ordered pair. For example, (K, 6) means the first card drawn was a king and the second card drawn was a 6. Here the order of the cards is important. (b) P(3, 10) P[(3 on 1st) and (10 on 2nd, given 3 on 1st)] P(3 on 1st) P(10 on 2nd, given 3 on 1st) 16 4 4 4 0.006 52 51 2, 652 663 (c) P(10, 3) P[(10 on 1st) and (3 on 2nd, given 10 on 1st)] P(10 on 1st) P(3 on 2nd, given 10 on 1st) 16 4 4 4 0.006 52 51 2, 652 663 (d) P[(3, 10) or (10, 3)] = P(3, 10) + P(10, 3) because these two outcomes are mutually exclusive. 4 4 8 0.012 = 663 663 663 15. (a) Yes, the draws are independent. The key idea is “with replacement.” When the first card drawn is replaced, the sample space is the same for the second card as it was for the first card. In fact, it is possible to draw the same card twice. Let the card draws be represented by an (x, y) ordered pair; for example, (K, 6) means a king was drawn, replaced, and then the second card, a 6, was drawn. (b) P(A, K) P(A) P(K) because they are independent. 16 1 4 4 52 52 2, 704 169 (c) P(K, A) P(K) P(A) because they are independent. 16 1 4 4 52 52 2, 704 169 (d) P[(A, K) or (K, A)] = P(A, K) + P(K, A) because the two outcomes are mutually exclusive. 1 1 2 = 169 169 169 16. (a) Yes, the draws are independent. The key idea is “with replacement.” When the first card drawn is replaced, the sample space is the same for the second card as it was for the first card. In fact, it is possible to draw the same card twice. Let the card draws be represented by an (x, y) ordered pair; for example, (K, 6) means a king was drawn, replaced, and then the second card, a 6, was drawn. (b) P(3, 10) P(3) P(10) because draws are independent. 16 1 4 4 0.0059 52 52 2, 704 169 (c) P(10, 3) P(10) P(3) because of independence. 16 1 4 4 0.0059 52 52 2, 704 169 (d) P[(3, 10) or (10, 3)] = P(3, 10) + P(10, 3) because the two outcomes are mutually exclusive. 1 1 2 0.0118 = 169 169 169 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 304 17. (a) (b) (c) (d) P(6 years old or older) = 27% + 14% + 22% = 63% P(12 years old or younger) = 1 – P(13 years old or older) = 100% – 22% = 78% P(Between 6 and 12 years old) = 27% + 14% = 41% P(Between 2 and 9 years old) = 22% + 27% = 49% The 13-and-older category may include children up to 17 or 18 years old. This is a larger category. 18. Let S denote “senior.” Let F denote “got the flu.” We are given the following probabilities: P(F | S) = 0.14 P(F | Sc) = 0.24 P(S) = 0.125 P(Sc) = 0.875 (a) P(S and F) = P(S) × P(F | S) = (0.125) × (0.14) = 0.0175 (b) P(Sc and F) = P(Sc) × P(F | Sc) = (0.875) × (0.24) = 0.21 (c) Here, P(S) = 0.95, so P(Sc) = 1 – 0.95 = 0.05 (a) P(S and F) = P(S) × P(F | S) = (0.95) × (0.14) = 0.133 (b) P(Sc and F) = P(Sc) × P(F | Sc) = (0.05) × (0.24) = 0.012 (d) Here, P(S) = P(Sc) = 0.50. (a) P(S and F) = P(S) × P(F | S) = (0.50) × (0.14) = 0.07 (b) P(Sc and F) = P(Sc) × P(F | Sc) = (0.50) × (0.24) = 0.12 19. Let T denote “telling the truth.” Let L denote “machine catches a person lying.” We are given the following probabilities: P(L | Tc) = 0.72 P(L | T) = 0.07 (a) Given P(T) = 0.90. Then P(T and L) = P(T) × P(L | T) = (0.90) × (0.07) = 0.063 (b) Given P(Tc) = 0.10. Then P(Tc and L) = P(Tc) × P(L | Tc) = (0.10) × (0.72) = 0.072 (c) Given P(T) = P(Tc) = 0.50. Then P(T and L) = (0.50) × (0.07) = 0.035 P(Tc and L) = (0.50) × (0.72) = 0.36 (d) Given P(T) = 0.15 and P(Tc) = 0.85. Then P(T and L) = (0.15) × (0.07) = 0.0105 P(Tc and L) = (0.85) × (0.72) = 0.612 20. (a) We want to solve for P(Tc). There are two possibilities when the polygraph says that the person is lying: Either the polygraph is right, or the polygraph is wrong. If the polygraph is right, the polygraph results show “lying,” and the person is not telling the truth; i.e., P(L and not T). If the polygraph is wrong, then the polygraph results show “lying,” but in fact, the person is telling the truth; i.e., P(L and T). P(L) = P(L and Tc) + P(L and T) = [P(Tc) × P(L | Tc)] + [P(T) × P(L | T)] = [P(Tc) × P(L | Tc)] + {[1 – P(Tc)] × P(L | T)} Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 305 We are told that P(L) = 0.30, so 0.30 = [P(Tc) × P(L | Tc)] + {[1 – P(Tc)] × P(L | T)} = [P(Tc) × 0.72 ] + {[1 – P(Tc)] × 0.07} = (0.72) × P(Tc) + {0.07 – [0.07 × P(Tc)]} 0.23 = P(Tc) × (0.72 – 0.07) = P(Tc) × (0.65) (**) 0.23/0.65 = P(Tc) = 0.354 = 35.4% (b) Here, P(L) = 70% = 0.70. Replace the 0.30 with 0.70 in (**) and solve. P(Tc) = 0.63/0.65 = 0.969 686 1,160 270 P(S | A) 580 416 P(S | Pa) 580 (b) No, they are not independent. P(S | Pa) ≠ P(S) based on the previous part. 21. (a) P(S) (c) P(A and S) = 270/1,160 using the table. P(Pa and S) = 416/1,160 using the table. 474 1,160 310 P(N | A) 580 (e) No, they are not independent. P(N | A) ≠ P(N) based on the preceding part. (d) P(N) (f) P( A or S ) P( A) P( S ) P( A and S ) 580 686 270 996 1,160 1,160 1,160 1,160 110 130 20 (b) P(– |condition present) 130 50 (c) P(– | condition absent) = 70 20 (d) P(+ | condition absent) 70 (e) P(condition present and +) = P(condition present) × P(+ | condition present) 130 110 110 200 130 200 (f) P(condition present and –) = P(condition present) × P(– | condition present) 130 20 20 200 130 200 22. (a) P(+ | condition present) Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 306 23. Let C denote the presence of the condition and not C denote absence of the condition. 72 (a) P(+ | C) 154 82 (b) P(– | C) 154 79 (c) P(– | not C) 116 37 (d) P(+ | not C) 116 154 72 72 (e) P(C and +) = P(C) × P(+ | C) 270 154 270 154 82 82 (f) P(C and –) = P(C) × P(– | C) 270 154 270 24. (a) P(10 to 14 years) = 291 2008 (b) P(10 to 14 years | East) = (c) P(at least 10 years) = 77 452 291 535 826 2008 2008 (d) P(at least 10 years | East) = 45 86 131 373 373 (e) P(West | less than 1 year) = 41 157 (f) P(South | less than 1 year) = 53 157 (g) P(1 or more years | East) = 1 – P(less than 1 year | East) = 1 32 420 452 452 (h) P(1 or more years | West) = 1 – P(less than 1 year | West) = 1 41 332 373 373 (i) We can check if P(East) = P(15 or more years | East). If these probabilities are equal, then the events are independent. P(East) 452 0.225 2008 P(15 years | East) 118 0.261 452 Since the probabilities are not equal, the events are not independent. Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 307 25. Given: Let A be the event that a new store grosses > $940,000 in year 1; then Ac is the event the new store grosses ≤ $940,000 the first year. Let B be the event that the store grosses > $940,000 in the second year; then Bc is the event the store grosses ≤ $940,000 in the second year of operation. 2-Year Results A and B A and Bc Ac and B Ac and Bc Translations Profitable both years Profitable first but not second year Profitable second but not first year Not profitable either year P(A) = 65% (show profit in first year) P(Ac) = 35% P(B) = 71% (show profit in second year) P(Bc) = 29% P(close) = P(Ac and Bc) P(B, given A) = 87% (a) (b) (c) (d) (e) (f) P(A) = 65% = 0.65 P(B) = 71% = 0.71 P(B | A) = 87% = 0.87 P(A and B) = P(A) × P(B | A) = (0.65)(0.87) = 0.5655 ≈ 0.57 P(A or B) = P(A) + P(B) – P(A and B) = 0.65 + 0.71 – 0.57 = 0.79 P(not closed) = P(show a profit in year 1 or year 2 or both) = 0.79 P(closed) = 1 – P(not closed) = 1 – 0.79 = 0.21 26. P(female) = 85%, so P(male) = 15% P(BSN | female) = 70% P(BSN | male) = 90% (a) (b) (c) (d) (e) P(BSN | female) = 70% = 0.70 P(BSN and female) = P(female) × P(BSN | female) = (0.85) × (0.70) = 0.595 P(BSN | male) = 90% = 0.90 P(BSN and male) = P(male) × P(BSN | male) = (0.15)(0.90) = 0.135 Of the graduates, some are female and some are male. We can add the mutually exclusive probabilities. P(BSN) = [P(BSN | female) × P(female)] + [P(BSN | male) × P(male)] = [(0.70) × (0.85)] + [(0.90) × (0.15)] = 0.73 (f) The phrase “will graduate and is female” describes the proportion of all students who are female and will graduate. The phrase “will graduate, given female” describes the proportion of the females who will graduate. Observe from parts (a) and (b) that the probabilities are indeed different. 27. Let TB denote that the person has tuberculosis. Let + denote the test for tuberculosis indicates the presence of the disease. Let – denote the test for tuberculosis indicates the absence of the disease. We are given the following probabilities: P(+ | TB) = 0.82 (sensitivity of the test) P(+ | TBc) = 0.09 (false-positive rate) P(TB) = 0.04 (a) P(TB and +) = P(+ | TB) × P(TB) = (0.82) × (0.04) = 0.0328 (b) P(TBc) = 1 – P(TB) = 1 – 0.04 = 0.96 (c) P(TBc and +) = P(+ | TBc) × P(TBc) = (0.09) × (0.96) = 0.0864 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 308 28. Known: Let A be the event the client relapses in phase I. Let B be the event the client relapses in phase II. Let C be the event that the client has no relapse in phase I; i.e., C = not A. Let D be the event that the client has no relapse in phase II; i.e., D = not B. P(A) = 0.27, so P(Ac) = P(C) = 1 – 0.27 = 0.73 P(B) = 0.23, so P(Bc) = P(D) = 1 – 0.23 = 0.77 P(Bc | Ac) = 0.95 = P(D | C) = 0.95 P(B | A) = 0.70 Possible Outcomes A, B Ac, B (= C, B) A, Bc (= A, D) Ac, Bc (= C, D) Translation Relapse in I, relapse in II No relapse in I, relapse in II Relapse in I, no relapse in II No relapse in I no relapse in II (a) P(A) = 0.27, P(B) = 0.23, P(C) = 0.73, P(D) = 0.77 (b) P(B | A) = 0.70, P(D | C) = 0.95 (c) P(A and B) = P(A) × P(B | A) = (0.27) × (0.70) = 0.189 P(C and D) = P(C) × P(D | C) = (0.73) × (0.95) = 0.6935 (d) P(A or B) = P(A) + P(B) – P(A and B) = 0.27 + 0.23 – 0.189 = 0.311 (e) P(C and D) = 0.69 (f) P(A and B) = 0.189 (g) Translate as the inclusive or. P(A or B) = 0.31. Section 4.3 1. The permutations rule counts the number of different arrangements, or r items out of n distinct items. Here, the ordering matters. The combinations rule counts the number of groups of r items out of n distinct items. Here, the ordering does not matter. For a permutation, ABC is different from ACB. For a combination, ABC and ACB are the same item. The number of permutations is larger than the number of combinations. 2. A tree diagram lists all possible events. The user of the diagram can trace the sequential event from the start to the end by following a distinct path along the branches. Counting the number of final branches gives the total number of outcomes. 3. (a) Use the combinations rule because we are concerned only with the groups of size five. (b) Use the permutations rules because we are concerned with the number of different arrangements of size five. 4. Both methods are correct because you are counting the number of possible arrangements of five items taken five at a time. Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 5. 309 (a) (b) HHT, HTH, THH. There are three outcomes. (c) There are eight possible outcomes, and three outcomes have exactly two heads. Copyright © Houghton Mifflin Company. All rights reserved. 3 . 8 Part IV: Complete Solutions, Chapter 4 310 6. (a) (b) H5, H6. There are two outcomes. (c) There are 12 possible outcomes, and 2 outcomes meet the requirements. 2 1 . 12 6 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 7. 311 (a) (b) Let P(x, y) be the probability of choosing an x-colored ball on the first draw and a y-colored ball on the second draw. Notice that the probabilities add to 1. 3 2 6 1 P( B, R) 6 5 30 5 1 3 2 6 1 2 1 2 P( B, B) P( R, R) 6 5 30 5 6 5 30 15 1 3 1 3 2 3 6 1 P( B, Y ) P( R, B) 6 5 30 10 6 5 30 5 1 1 1 2 2 2 1 2 P(Y , R) P( R, Y ) 6 5 30 15 6 5 30 15 1 1 3 3 P(Y , B) 6 5 30 10 Copyright © Houghton Mifflin Company. All rights reserved. 312 8. Part IV: Complete Solutions, Chapter 4 (a) For clarity, only a partial tree diagram is provided. Each of the branches for B, C, and D would continue in the same manner as the fully expanded A branch. 1 1 1 1 (b) If the outcomes are equally likely, then P(all 3 correct) . 4 4 4 64 9. Using the provided hint, we multiply. There are 4 × 3 × 2 × 1 = 4! = 24 possible wiring configurations. 10. Using the multiplication rule, we multiply. There are 4! = 24 possible ways to visit the four cities. This problem is exactly like Problem 9. 11. There are four fertilizers, three temperature zones for each fertilizer, and three water treatments for every fertilizer–temperature zone combination. She needs to test 4 × 3 × 3 = 36 plots. 12. (a) The die rolls are independent, so multiply the six outcomes for the first die and the six outcomes for the second die. There are 6 × 6 = 36 possible outcomes. (b) There are three possible even outcomes per die. There are 3 × 3 = 9 outcomes. Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 313 9 1 0.25 36 4 number of favorable outcomes Using P(event) total number of outcomes (c) P(even, even) Problems 13, 14, 15, and 16 deal with permutations. n! Use Pn, r to count the number of ways r objects can be selected from n objects when ordering (n r )! matters. 13. P5, 2 : n 5, r 2 P5, 2 5! 5 4 3 2 1 20 (5 2)! 3! 14. P8,3 : n 8, r 3 P8,3 8! 8 7 6 5 4 3 2 1 8 7 6 5! 336 (8 3)! 5! 5! 15. P7,7 : n r 7 P7, 7 7! 7! 7! 5, 040 (recall 0! 1) (7 7)! 0! In general, Pn, n n! n! n! n! . (n n)! 0! 1 16. P9,9 : n r 9 P9,9 9! 9! 9! 362,880 (9 9)! 0! 1 Problems 17, 18, 19, and 20 deal with combinations. n! Use Cn, r to count the number of ways r objects can be selected from n objects when ordering r !(n r )! is irrelevant. 17. C5, 2 : n 5, r 2 C5, 2 5! 5! 5 4 3 2 1 20 10 2!(5 2)! 2!3! 2 1 3 2 1 2 18. C8,3 : n 8, r 3 C8,3 8! 8! 8 7 6 5! 56 3!(8 3)! 3!5! 3 2 1 5! Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 314 19. C7,7 : n r 7 7! 7! 7! 1 (recall 0! 1) 7!(7 7)! 7!0! 7!(1) C7, 7 In general, Cn, n n! n! n! 1. There is only one way to choose n objects without regard n !(n n)! n !0! n !(1) to order. 20. C8,8 : n r 8 8! 8! 8! 1 (recall 0! 1) 8!(8 8)! 8!0! 8!(1) C8,8 21. Since the order matters (first is day supervisor, second is night supervisor, and third is coordinator), this is a permutation of 15 nurse candidates to fill three positions. P15,3 15! 15! 15 14 13 12! 2, 730 (15 3)! 12! 12! 22. Order matters here because the order of the finalists selected determines the prize awarded. P10,3 10! 10! 10 9 8 7! 720 (10 3)! 7! 7! 23. Order matters because the resulting sequence determines who wins first, second, and third place. P5,3 5! 5! 120 60 (5 3)! 2! 2 24. The order of the software packages selected is irrelevant, so use the combinations method. C10,3 10! 10! 10 9 8 7! 720 120 3!(10 3)! 3!7! 3!7! 6 25. The order of trainee selection is irrelevant, so use the combinations method. C15,5 15! 15! 15 14 13 12 1110! 15 14 13 12 11 3, 003 5!(15 5)! 5!10! 5!10! 5 4 3 2 1 26. The order of the problems selected is irrelevant, so use the combinations method. 12! 12! 12 11 10 9 8 7! 12 11 10 9 8 792 5!(12 5)! 5!7! 5!7! 5 4 3 2 1 (b) Jerry must have completed the same five problems as the professor selected to grade. 1 0.001 P(Jerry chose the right problems) 792 7! 7! 7 6 5! 7 6 42 21 (c) Silvia did seven problems, so she completed C7,5 5!(7 5)! 5!2! 5!2! 2 1 2 possible subsets. (a) C12,5 Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 P(Silvia picked the correct set of graded problems) 315 21 0.027 792 Silvia increased her chances by a factor of 21 compared with Jerry. 27. (a) Six applicants are selected from among 12 without regard to order. C12, 6 12! 479, 001, 600 924. 6!6! (720)2 (b) This problem is asking, “In how many ways can six women be selected from seven applicants?” C7,6 7! 7 6! 1! number of favorable outcomes total number of outcomes 7 1 0.008 P(all hired are women) 924 132 (c) P(event A) Chapter 4 Review 1. (a) (b) (c ) (d) (e) The individual does not own a cell phone. The individual owns both a cell phone and a laptop computer. The individual owns either a cell phone or a laptop computer or both. A laptop owner who owns a cell phone. A cell phone owner who owns a laptop. 2. (a) Only if events A and B are mutually exclusive. Then P(A and B) = 0 and P(A or B) = P(A) + P(B). (b) Yes, see above. 3. (a) No, unless events A and B are independent. If they are not, we need either P(A | B) or P(B | A) to compute P(A and B). (b) Yes, now we can compute P(A and B) = P(A) × P(B). 4. The information yields P(B | A) = 2. Probabilities must be between 0 and 1 inclusive. Also, P(A and B) cannot be greater than P(A) or P(B) individually. 5. P(asked) = 24% = 0.24 P(received | asked) = 45% = 0.45 P(asked and received) = P(asked) × P(received | asked) = (0.24) × (0.45) = 0.108 = 10.8% 6. P(asked) = 20% = 0.20 P(received | asked) = 59% = 0.59 P(asked and received) = P(asked) × P(received | asked) = (0.20) × (0.59) = 0.118 = 11.8% 7. (a) Throw a large number of similar thumbtacks or one thumbtack a large number of times, and record the relative frequency of the outcomes. Assume that the thumbtack falls either flat side down or tilted. To estimate the probability the tack lands on its flat side, find the relative frequency of this occurrence, dividing the number of times this occurred by the total number of thumbtack tosses. Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 316 (b) The sample space consists of two outcomes: flat side down and tilted. 340 0.68 (c) P(flat side down) 500 8. (a) (b) (c) (d) P(tilted) 1 0.68 0.32 470 P( N ) 0.470 1000 390 P( M ) 0.390 1000 140 P( S ) 0.140 1000 420 0.840 P(N | W) 500 20 0.040 P(S | W) 500 50 0.100 P(N | A) 500 120 0.240 P(S | A) 500 P(N and W) = P(W) × P(N | W) = (0.50) × (0.84) = 0.42 P(M and W) = P(W) × P(M | W) = (0.50) – (0.12) = 0.06 (e) P( N or M ) P( N ) P( M ) if mutually exclusive 470 390 860 0.860 1, 000 1, 000 1, 000 No reaction is mutually exclusive from a mild reaction; they cannot occur at the same time. (f) If N and W were independent, P(N and W) = P(N) · P(W) = (0.470) × (0.500) = 0.235. However, from (d), we have P(N and W) = 0.420. They are not independent. 9. (a) Possible values for x are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. (b) Below, the values for x are listed, along with the combinations required. 2 3 4 5 6 7 8 9 10 11 12 1 and 1 1 and 2, or 2 and 1 1 and 3, 2 and 2, 3 and 1 1 and 4, 2 and 3, 3 and 2, 4 and 1 1 and 5, 2 and 4, 3 and 3, 4 and 2, 5 and 1 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1 2 and 6, 3 and 5, 4 and 4, 5 and 3, 6 and 2 3 and 6, 4 and 5, 5 and 4, 6 and 3 4 and 6, 5 and 5, 6 and 4 5 and 6, 6 and 5 6 and 6 1 way 2 ways 3 ways 4 ways 5 ways 6 ways 5 ways 4 ways 3 ways 2 ways 1 way Copyright © Houghton Mifflin Company. All rights reserved. Part IV: Complete Solutions, Chapter 4 x 2 3 4 5 6 7 8 9 10 11 12 P(x) 1 0.028 36 2 0.056 36 3 0.083 36 4 0.111 36 5 0.139 36 6 0.167 36 5 0.139 36 4 0.111 36 3 0.083 36 2 0.056 36 1 0.028 36 Where there are (6)(6) = 36 possible, equally likely outcomes. (The sums, however, are not equally likely). 10. P(pass 101) = 0.77 P(pass 102 | pass 101) = 0.90 P(pass 101 and pass 102) = P(pass 101) × P(pass 102 | pass 101) = (0.77) × (0.90) = 0.693 11. C8, 2 8! 8 7 6! 56 28 2!6! (2 1)6! 2 12. (a) P7, 2 7! 7! 7(6) 42 (7 2)! 5! 7! 76 21 2!5! 2 3! 3! 6 (c) P3,3 (3 3)! 0! (b) C7, 2 (d) C4, 4 4! 4! 1 4!(4 4)! 4!0! 13. Five multiple choice questions, each with flurossible (A, B, C, or D). There are 4 × 4 × 4 × 4 × 4 = 1,024 possible sequences, such as A, D, B, B. 1 0.00098 P(getting the correct sequence) 1024 Copyright © Houghton Mifflin Company. All rights reserved. 317 318 Part IV: Complete Solutions, Chapter 4 14. 15. There are 10 possible numbers per turn of dial and, we turn the dial three times. There are 10 × 10 × 10 = 1,000 possible combinations. 16. The combination uses the three numbers 2, 9, and 5, in an ordered sequence. 3! 3 2 1 6. The number of sequences is P3,3 (3 3)! 0! The possible combinations are 259, 295, 529, 592, 925, and 952. Copyright © Houghton Mifflin Company. All rights reserved.