# Solution Manual for An Introduction to Mathematical Statistics and Its Applications, 6th Edition

0. (b) No, because P(A ๏ B) = 0.2 ๏น P(A) ๏ P(B) = (0.6)(0.5) = 0.3 (c) P(AC ๏ BC) = P((A ๏ B)C) = 1 ๏ญ P(A ๏ B) = 1 ๏ญ 0.2 = 0.8. 2.5.2 Let C and M be the events that Spike passes chemistry and mathematics, respectively. Since P(C ๏ M) = 0.12 ๏น P(C) ๏ P(M) = (0.35)(0.40) = 0.14, C and M are not independent. Copyright ยฉ 2018 Pearson Education, Inc. 14 Chapter 2: Probability P(Spike fails both) = 1 ๏ญ P(Spike passes at least one) = 1 ๏ญ P(C ๏ M) = 1 ๏ญ [P(C) + P(M) ๏ญ P(C ๏ M)] = 0.37. 6 . 36 2.5.3 P(one face is twice the other face) = P((1, 2), (2, 1), (2, 4), (4, 2), (3, 6), (6, 3)) = 2.5.4 Consider the probability of the complementary event that they have the same blood types: 0.42 ๏ซ 0.012 ๏ซ 0.052 ๏ซ 0.452 ๏ฝ 0.375 . Then the probability they have different blood types is 1 โ 0.375 = 0.625. 2.5.5 P(Dana wins at least 1 game out of 2) = 0.3, which implies that P(Dana loses 2 games out of 2) = 0.7. Therefore, P(Dana wins at least 1 game out of 4) = 1 ๏ญ P(Dana loses all 4 games) = 1 ๏ญ P(Dana loses first 2 games and Dana loses second 2 games) = 1 ๏ญ (0.7)(0.7) = 0.51. 2.5.6 Six equally-likely orderings are possible for any set of three distinct random numbers: x1 < x2 < x3, x1 < x3 < x2, x2 < x1 < x3, x2 < x3 < x1, x3 < x1 < x2, and x3 < x2 0, so A, B, and C are not mutually independent. However, P(A ๏ B) = 36 3 3 9 3 18 9 = P(A) ๏ P(B) = ๏ , P(A ๏ C) = = P(A) ๏ P(C) = ๏ , and P(B ๏ C) = = P(B) ๏ 6 6 36 6 36 36 3 18 P(C) = ๏ , so A, B, and C are pairwise independent. 6 36 2.5.16 Let Ri and Gi be the events that the ith light is red and green, respectively, i = 1, 2, 3, 4. Then 1 1 P(R1) = P(R2) = and P(R3) = P(R4) = . Because of the considerable distance between the 3 2 intersections, what happens from light to light can be considered independent events. P(driver stops at least 3 times) = P(driver stops exactly 3 times) + P(driver stops all 4 times) = P((R1 ๏ R2 ๏ R3 ๏ G4) ๏ (R1 ๏ R2 ๏ G3 ๏ R4) ๏ (R1 ๏ G2 ๏ R3 ๏ R4) ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ ๏ (G1 ๏ R2 ๏ R3 ๏ R4) ๏ (R1 ๏ R2 ๏ R3 ๏ R4)) = ๏ง ๏ท๏ง ๏ท๏ง ๏ท๏ง ๏ท ๏ซ ๏ง ๏ท๏ง ๏ท๏ง ๏ท๏ง ๏ท ๏จ 3 ๏ธ๏จ 3 ๏ธ๏จ 2 ๏ธ๏จ 2 ๏ธ ๏จ 3 ๏ธ๏จ 3 ๏ธ๏จ 2 ๏ธ๏จ 2 ๏ธ ๏ฆ 1 ๏ถ๏ฆ 2 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ ๏ฆ 2 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ๏ฆ 1 ๏ถ 7 . + ๏ง ๏ท๏ง ๏ท๏ง ๏ท๏ง ๏ท ๏ซ ๏ง ๏ท๏ง ๏ท๏ง ๏ท๏ง ๏ท ๏ซ ๏ง ๏ท๏ง ๏ท๏ง ๏ท๏ง ๏ท ๏ฝ ๏จ 3 ๏ธ๏จ 3 ๏ธ๏จ 2 ๏ธ๏จ 2 ๏ธ ๏จ 3 ๏ธ๏จ 3 ๏ธ๏จ 2 ๏ธ๏จ 2 ๏ธ ๏จ 3 ๏ธ๏จ 3 ๏ธ๏จ 2 ๏ธ๏จ 2 ๏ธ 36 2.5.17 Let M, L, and G be the events that a student passes the mathematics, language, and general 6175 7600 8075 knowledge tests, respectively. Then P(M) = , P(L) = , and P(G) = . 9500 9500 9500 P(student fails to qualify) = P(student fails at least one exam) = 1 ๏ญ P(student passes all three exams) = 1 ๏ญ P(M ๏ L ๏ G) = 1 ๏ญ P(M) ๏ P(L) ๏ P(G) = 0.56. 2.5.18 Let Ai denote the event that switch Ai closes, i = 1, 2, 3, 4. Since the Aiโs are independent events, P(circuit is completed) = P((A1 ๏ A2) ๏ (A3 ๏ A4)) = P(A1 ๏ A2) + P(A3 ๏ A4) ๏ญ P((A1 ๏ A2) ๏ (A3 ๏ A4)) = 2p2 ๏ญ p4. 2.5.19 Let p be the probability of having a winning game card. Then 0.32 = P(winning at least once in 5 tries) = 1 ๏ญ P(not winning in 5 tries) = 1 ๏ญ (1 ๏ญ p)5, so p = 0.074 Copyright ยฉ 2018 Pearson Education, Inc. 16 Chapter 2: Probability 2.5.20 Let AH, AT, BH, BT, CH, and CT denote the events that players A, B, and C throw heads and tails on individual tosses. Then P(A throws first head) = P(AH ๏ (AT ๏ BT ๏ CT ๏ AH) ๏ ๏๏๏) 2 ๏ถ 4 1 1๏ฆ1๏ถ 1๏ฆ1๏ถ 1๏ฆ 1 = ๏ซ ๏ง ๏ท ๏ซ ๏ง ๏ท ๏ซ๏ ๏ฝ ๏ง ๏ท๏ฝ . 2 2๏จ8๏ธ 2๏จ8๏ธ 2 ๏จ1 ๏ญ 1/ 8 ๏ธ 7 Similarly, P(B throws first head) = P((AT ๏ BH) ๏ (AT ๏ BT ๏ CT ๏ AT ๏ BH) ๏ โฆ) 2 ๏ถ 2 1 1๏ฆ1๏ถ 1๏ฆ1๏ถ 1๏ฆ 1 = ๏ซ ๏ง ๏ท ๏ซ ๏ง ๏ท ๏ซ … ๏ฝ ๏ง ๏ท๏ฝ . 4 4๏จ8๏ธ 4๏จ8๏ธ 4 ๏จ1 ๏ญ 1/ 8 ๏ธ 7 P(C throws first head) = 1 ๏ญ 4 2 1 ๏ญ ๏ฝ . 7 7 7 2.5.21 P(at least one child becomes adult) = 1 ๏ญ P(no child becomes adult) = 1 ๏ญ 0.8n . ln 0.25 Then 1 ๏ญ 0.8n ๏ณ 0.75 implies n ๏ณ or n โฅ 6.2 , so take n = 7. ln 0.8 2.5.22 P(at least one viewer can name actor) = 1 ๏ญ P(no viewer can name actor) = 1 ๏ญ (0.85)10 = 0.80. 2.5.23 Let B be the event that no heads appear, and let Ai be the event that i coins are tossed, 2 6 6 1๏ฆ1๏ถ ๏ฆ1๏ถ ๏ฆ1๏ถ ๏ฆ 1 ๏ถ ๏ฆ 1 ๏ถ 63 . i = 1, 2, โฆ, 6. Then P(B) = ๏ฅ P ( B | Ai ) P( Ai ) ๏ฝ ๏ง ๏ท ๏ซ ๏ง ๏ท ๏ง ๏ท ๏ซ … ๏ซ ๏ง ๏ท ๏ง ๏ท ๏ฝ 2๏จ6๏ธ ๏จ 2๏ธ ๏จ6๏ธ ๏จ 2 ๏ธ ๏จ 6 ๏ธ 384 i ๏ฝ1 2.5.24 P(at least one red chip is drawn from at least one urn) = 1 ๏ญ P(all chips drawn are white) r r r rm ๏ฆ4๏ถ ๏ฆ4๏ถ ๏ฆ4๏ถ ๏ฆ4๏ถ = 1 ๏ญ ๏ง ๏ท ๏ ๏ง ๏ท ๏๏ง ๏ท ๏ฝ 1 ๏ญ ๏ง ๏ท . ๏จ7๏ธ ๏จ7๏ธ ๏จ7๏ธ ๏จ7๏ธ n 2.5.25 ๏ฆ 35 ๏ถ P(at least one double six in n throws) = 1 ๏ญ P(no double sixes in n throws) = 1 ๏ญ ๏ง ๏ท . By ๏จ 36 ๏ธ trial and error, the smallest n for which P(at least one double six in n throws) exceeds 0.50 is 24 25 ๏ฆ 35 ๏ถ ๏ฆ 35 ๏ถ 25 [1 ๏ญ ๏ง ๏ท = 0.49; 1 ๏ญ ๏ง ๏ท = 0.51]. ๏จ 36 ๏ธ ๏จ 36 ๏ธ 2.5.26 Let A be the event that a sum of 8 appears before a sum of 7. Let B be the event that a sum of 8 appears on a given roll and let C be the event that the sum appearing on a given roll is 5 25 , P(C) = , and P(A) = P(B) + P(C)P(B) + P(C)P(C)P(B) neither 7 nor 8. Then P(B) = 36 36 2 k ๏ถ 5 5 25 5 ๏ฆ 25 ๏ถ 5 5 ๏ฅ ๏ฆ 25 ๏ถ 5๏ฆ 1 + ๏๏๏ = ๏ซ ๏ซ๏ง ๏ท ๏ซ๏ ๏ฝ ๏ฅ ๏ท๏ฝ . ๏ง ๏ท ๏ฝ ๏ง 36 36 36 ๏จ 36 ๏ธ 36 36 k ๏ฝ 0 ๏จ 36 ๏ธ 36 ๏จ 1 ๏ญ 25 / 36 ๏ธ 11 2.5.27 Let W, B, and R denote the events of getting a white, black and red chip, respectively, on a given draw. Then P(white appears before red) = P(W ๏ (B ๏ W) ๏ (B ๏ B ๏ W) ๏ ๏๏๏) 2 w b w b w ๏ฆ ๏ถ = ๏ซ ๏ ๏ซ๏ง ๏ซ๏ ๏ท ๏ w๏ซb๏ซr w๏ซb๏ซr w๏ซb๏ซr ๏จ w๏ซb๏ซr ๏ธ w๏ซb๏ซr = ๏ฆ ๏ถ w 1 w ๏๏ง . ๏ท๏ฝ w ๏ซ b ๏ซ r ๏จ 1 ๏ญ b / (w ๏ซ b ๏ซ r ) ๏ธ w ๏ซ r Copyright ยฉ 2018 Pearson Education, Inc. Section 2.6: Combinatorics 17 2.5.28 P(B|A1) = 1 ๏ญ P(all m I-teams fail) = 1 ๏ญ (1 ๏ญ r)m; similarly, P(B|A2) = 1 ๏ญ (1 ๏ญ r)n๏ญm. From Theorem 2.4.1, P(B) = [1 ๏ญ (1 ๏ญ r)m]p + [1 ๏ญ (1 ๏ญ r)n๏ญm](1 ๏ญ p). Treating m as a continuous variable and differentiating P(B) gives dP ( B ) dP ( B ) = ๏ญp(1 ๏ญ r)m๏ln(1 ๏ญ r) + (1 ๏ญ p)(1 ๏ญ r)n๏ญm ๏ln(1 ๏ญ r). Setting = 0 implies that dm dm n ln[(1 ๏ญ p ) / p ] m= ๏ซ . 2 2ln(1 ๏ญ r ) 2.5.29 P(at least one four) = 1 ๏ญ P(no fours) = 1 ๏ญ (0.9)n. 1 ๏ญ (0.9)n ๏ณ 0.7 implies n = 12 2.5.30 Let B be the event that all n tosses come up heads. Let A1 be the event that the coin has two heads, and let A2 be the event the coin is fair. Then (1 / 2) n (8 / 9) 8(1 / 2) n ๏ฝ 1(1 / 9) ๏ซ (1 / 2) n (8 / 9) 1 ๏ซ 8(1 / 2) n By inspection, the limit of P ( A2 | B ) as n goes to infinity is 0. P ( A2 | B ) ๏ฝ 2.5.31 Assume the events that Stanley answers correctly from question to question are independent. (Is this a reasonable assumption?) The probability of answering at least one question on Final A is (0.45)(0.55) + (0.55)(0.45) + (0.45)(0.45) = 0.6975 The probability of answering at least one question on Final B is (0.30)(0.70)(0.70) + ((0.70)(0.30)(0.70) +(0.30)(0.30)(0.70) + (0.30)(0.30)(0.70) + (0.30)(0.70)(0.30) + (0.70)(0.30)(0.30) + (0.30)(0.30)(0.30) = 0.657 A simpler way to answer the question is to take the complement of the event that he answers none correctly. For Final A this is 1 ๏ญ (0.55) 2 = 0.6975. For Final B this is 1 ๏ญ (0.70)3 = 0.657. With either method of solution, Stanley will be a bit better off taking Final A. 2.5.32 Take the complementary event that none of n switches opens. This probability is 0.4n . Then 0.04n ๏ฃ 0.02 implies nln(0.04) โค ln(0.02) or n โฅ ln(0.02)/ln(0.04) = 4.27. So the smallest n is 5. Section 2.6: Combinatorics 2.6.1 2 ๏ 3 ๏ 2 ๏ 2 = 24 2.6.2 20 ๏ 9 ๏ 6 ๏ 20 = 21,600 2.6.3 3 ๏ 3 ๏ 5 = 45. Included will be aeu and cdx. 2.6.4 (a) 262 ๏ 104 = 6,760,000 (b) 262 ๏ 10 ๏ 9 ๏ 9 ๏ 8 ๏ 7 = 3,407,040 (c) The total number of plates with four zeros is 26 ๏ 26, so the total number not having four zeros must be 262 ๏ 104 ๏ญ 262 = 6,759,324. Copyright ยฉ 2018 Pearson Education, Inc. 18 Chapter 2: Probability 2.6.5 There are 9 choices for the first digit (1 through 9), 9 choices for the second digit (0 + whichever eight digits are not appearing in the hundreds place), and 8 choices for the last digit. The number of admissible integers, then, is 9 ๏ 9 ๏ 8 = 648. For the integer to be odd, the last digit must be either 1, 3, 5, 7, or 9. That leaves 8 choices for the first digit and 8 choices for the second digit, making a total of 320 (= 8 ๏ 8 ๏ 5) odd integers. 2.6.6 For each topping, the customer has 2 choices: โaddโ or โdo not add.โ The eight available toppings, then, can produce a total of 28 = 256 different hamburgers. 2.6.7 The bases can be occupied in any of 27 ways (each of the seven can be either โemptyโ or โoccupiedโ). Moreover, the batter can come to the plate facing any of five possible โoutโ situations (0 through 4). It follows that the number of base-out configurations is 5 ๏ 27, or 640. 2.6.8 (a) There are 3 choices for the leading digitโ7, 8,9, There are 10 choices for each of the remaining eight places, for a total of 3 ๏ 108 However, this count includes 7,000,000,000, so the answer is 3 ๏ 108 ๏ญ 1 . (b) Suppose the sequence starts with an even number. Counting 0 as even, there are 5 choices for each of the five even places and five choices for the four odd places, giving a total of 59 . But the same number occurs if the sequence starts with an odd number, so the answer is 2 ๏ 59 (c) There are 5! choices for the first five digits. Then there are 6 choices for where the 2โs go, so the answer is 5!๏ 6 . 2.6.9 4 ๏ 14 ๏ 6 + 4 ๏ 6 ๏ 5 + 14 ๏ 6 ๏ 5 + 4 ๏ 14 ๏ 5 = 1156 2.6.10 There are two mutually exclusive sets of ways for the black and white keys to alternateโthe black keys can be 1st, 3rd, 5th, and 7th notes in the melody, or the 2nd, 4th 6th, and 8th. Since there are 5 black keys and 7 white keys, there are 5 ๏ 7 ๏ 5 ๏ 7 ๏ 5 ๏ 7 ๏ 5 ๏ 7 variations in the first set and 7 ๏ 5 ๏ 7 ๏ 5 ๏ 7 ๏ 5 ๏ 7 ๏ 5 in the second set. The total number of alternating melodies is the sum 54 7 4 ๏ซ 7 4 54 = 3,001,250. 2.6.11 The number of usable garage codes is 28 ๏ญ 1 = 255, because the โcombinationโ where none of the buttons is pushed is inadmissible (recall Example 2.6.3). Five additional families can be added before the eight-button system becomes inadequate. 2.6.12 4, because 21 + 22 + 23 < 26 but 21 + 22 + 23 + 24 ๏ณ 26. 2.6.13 In order to exceed 256, the binary sequence of coins must have a head in the ninth position and at least one head somewhere in the first eight tosses. The number of sequences satisfying those conditions is 28 ๏ญ 1, or 255. (The โ1โ corresponds to the sequences TTTTTTTTH, whose value would not exceed 256.) 2.6.14 There are 3 choices for the vowel and 4 choices for the consonant, so there are 3 ๏ 4 = 12 choices, if order doesnโt matter. If we are taking ordered arrangements, then there are 24 ways, since each unordered selection can be written vowel first or consonant first. 2.6.15 There are 1 ๏ 3 ways if the ace of clubs is the first card and 12 ๏ 4 ways if it is not. The total is then 3 + 12 ๏ 4 = 51 Copyright ยฉ 2018 Pearson Education, Inc.

## Document Preview (25 of 249 Pages)

### Solution Manual for An Introduction to Mathematical Statistics and Its Applications, 6th Edition

$18.99 ~~$29.99~~Save:$11.00(37%)

##### Alexander Robinson

0 (0 Reviews)