285 pág.

# Solucionario Walpole 8 ED

Pré-visualização50 páginas

(M | T )P (T )+P (M | T \u2032)P (T \u2032) = (0.9)(0.5) (0.9)(0.5)+(0.65)(0.5) = 0.5807. 28 Chapter 2 Probability 2.133 Consider the events: A: purchased from vendor A, D: a customer is dissatisfied. Then P (A) = 0.2, P (A | D) = 0.5, and P (D) = 0.1. So, P (D | A) = P (A | D)P (D) P (A) = (0.5)(0.1) 0.2 = 0.25. 2.134 (a) P (Union member | New company (same field)) = 13 13+10 = 13 23 = 0.5652. (b) P (Unemployed | Union member) = 2 40+13+4+2 = 2 59 = 0.034. 2.135 Consider the events: C: the queen is a carrier, P (C) = 0.5, D: a prince has the disease, P (D | C) = 0.5. P (C | D\u20321D\u20322D\u20323) = P (D \u2032 1D \u2032 2D \u2032 3 | C)P (C) P (D \u2032 1D \u2032 2D \u2032 3 | C)P (C)+P (D \u2032 1D \u2032 2D \u2032 3 | C\u2032)P (C\u2032) = (0.5) 3(0.5) (0.5)3(0.5)+1(0.5) = 1 9 . 2.136 Using the solutions to Exercise 2.50, we know that there are total 365P60 ways that no two students have the same birth date. Since the total number of ways of the birth dates that 60 students can have is 36560, the probability that at least two students will have the same birth date in a class of 60 is P = 1 \u2212 365P60 36560 . To compute this number, regular calculator may not be able to handle it. Using approximation (such as Stirling\u2019s approximation formula), we obtain P = 0.9941, which is quite high. Chapter 3 Random Variables and Probability Distributions 3.1 Discrete; continuous; continuous; discrete; discrete; continuous. 3.2 A table of sample space and assigned values of the random variable is shown next. Sample Space x NNN NNB NBN BNN NBB BNB BBN BBB 0 1 1 1 2 2 2 3 3.3 A table of sample space and assigned values of the random variable is shown next. Sample Space w HHH HHT HTH THH HTT THT TTH TTT 3 1 1 1 \u22121 \u22121 \u22121 \u22123 3.4 S = {HHH, THHH,HTHHH, TTHHH, TTTHHH,HTTHHH, THTHHH, HHTHHH, . . .}; The sample space is discrete containing as many elements as there are positive integers. 29 30 Chapter 3 Random Variables and Probability Distributions 3.5 (a) c = 1/30 since 1 = 3\u2211 x=0 c(x2 + 4) = 30c. (b) c = 1/10 since 1 = 2\u2211 x=0 c ( 2 x )( 3 3\u2212 x ) = c [( 2 0 )( 3 3 ) + ( 2 1 )( 3 2 ) + ( 2 2 )( 3 1 )] = 10c. 3.6 (a) P (X > 200) = \u222b\u221e 200 20000 (x+100)3 dx = \u2212 10000 (x+100)2 \u2223\u2223\u2223\u221e 200 = 1 9 . (b) P (80 < X < 200) = \u222b 120 80 20000 (x+100)3 dx = \u2212 10000 (x+100)2 \u2223\u2223\u2223120 80 = 1000 9801 = 0.1020. 3.7 (a) P (X < 1.2) = \u222b 1 0 x dx+ \u222b 1.2 1 (2\u2212 x) dx = x2 2 \u2223\u2223\u22231 0 + ( 2x\u2212 x2 2 )\u2223\u2223\u22231.2 1 = 0.68. (b) P (0.5 < X < 1) = \u222b 1 0.5 x dx = x 2 2 \u2223\u2223\u22231 0.5 = 0.375. 3.8 Referring to the sample space in Exercise 3.3 and making use of the fact that P (H) = 2/3 and P (T ) = 1/3, we have P (W = \u22123) = P (TTT ) = (1/3)3 = 1/27; P (W = \u22121) = P (HTT ) + P (THT ) + P (TTH) = 3(2/3)(1/3)2 = 2/9; P (W = 1) = P (HHT ) + P (HTH) + P (THH) = 3(2/3)2(1/3) = 2/9; P (W = 3) = P (HHH) = (2/3)3 = 8/27; The probability distribution for W is then w \u22123 \u22121 1 3 P (W = w) 1/27 2/9 2/9 8/27 3.9 (a) P (0 < X < 1) = \u222b 1 0 2(x+2) 5 dx = (x+2) 2 5 \u2223\u2223\u22231 0 = 1. (b) P (1/4 < X < 1/2) = \u222b 1/2 1/4 2(x+2) 5 dx = (x+2) 2 5 \u2223\u2223\u22231/2 1/4 = 19/80. 3.10 The die can land in 6 different ways each with probability 1/6. Therefore, f(x) = 1 6 , for x = 1, 2, . . . , 6. 3.11 We can select x defective sets from 2, and 3\u2212 x good sets from 5 in (2 x )( 5 3\u2212x ) ways. A random selection of 3 from 7 sets can be made in ( 7 3 ) ways. Therefore, f(x) = ( 2 x )( 5 3\u2212x )( 7 3 ) , x = 0, 1, 2. In tabular form x 0 1 2 f(x) 2/7 4/7 1/7 Solutions for Exercises in Chapter 3 31 The following is a probability histogram: 1 2 3 xx f(x ) f(x ) 1/7 2/7 3/7 4/7 3.12 (a) P (T = 5) = F (5)\u2212 F (4) = 3/4\u2212 1/2 = 1/4. (b) P (T > 3) = 1\u2212 F (3) = 1\u2212 1/2 = 1/2. (c) P (1.4 < T < 6) = F (6)\u2212 F (1.4) = 3/4\u2212 1/4 = 1/2. 3.13 The c.d.f. of X is F (x) = \uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3 0, for x < 0, 0.41, for 0 \u2264 x < 1, 0.78, for 1 \u2264 x < 2, 0.94, for 2 \u2264 x < 3, 0.99, for 3 \u2264 x < 4, 1, for x \u2265 4. 3.14 (a) P (X < 0.2) = F (0.2) = 1\u2212 e\u22121.6 = 0.7981; (b) f(x) = F \u2032(x) = 8e\u22128x. Therefore, P (X < 0.2) = 8 \u222b 0.2 0 e\u22128x dx = \u2212e\u22128x|0.20 = 0.7981. 3.15 The c.d.f. of X is F (x) = \uf8f1\uf8f4\uf8f4\uf8f4\uf8f2\uf8f4\uf8f4\uf8f4\uf8f3 0, for x < 0, 2/7, for 0 \u2264 x < 1, 6/7, for 1 \u2264 x < 2, 1, for x \u2265 2. (a) P (X = 1) = P (X \u2264 1)\u2212 P (X \u2264 0) = 6/7\u2212 2/7 = 4/7; (b) P (0 < X \u2264 2) = P (X \u2264 2)\u2212 P (X \u2264 0) = 1\u2212 2/7 = 5/7. 32 Chapter 3 Random Variables and Probability Distributions 3.16 A graph of the c.d.f. is shown next. F( x) F( x) xx 1/7 2/7 3/7 4/7 5/7 6/7 1 0 1 2 3.17 (a) Area = \u222b 3 1 (1/2) dx = x 2 \u2223\u22233 1 = 1. (b) P (2 < X < 2.5) \u222b 2.5 2 (1/2) dx = x 2 \u2223\u22232.5 2 = 1 4 . (c) P (X \u2264 1.6) = \u222b 1.6 1 (1/2) dx = x 2 \u2223\u22231.6 1 = 0.3. 3.18 (a) P (X < 4) = \u222b 4 2 2(1+x) 27 dx = (1+x) 2 27 \u2223\u2223\u22234 2 = 16/27. (b) P (3 \u2264 X < 4) = \u222b 4 3 2(1+x) 27 dx = (1+x) 2 27 \u2223\u2223\u22234 3 = 1/3. 3.19 F (x) = \u222b x 1 (1/2) dt = x\u22121 2 , P (2 < X < 2.5) = F (2.5)\u2212 F (2) = 1.5 2 \u2212 1 2 = 1 4 . 3.20 F (x) = 2 27 \u222b x 2 (1 + t) dt = 2 27 ( t+ t 2 2 )\u2223\u2223\u2223x 2 = (x+4)(x\u22122) 27 , P (3 \u2264 X < 4) = F (4)\u2212 F (3) = (8)(2) 27 \u2212 (7)(1) 27 = 1 3 . 3.21 (a) 1 = k \u222b 1 0 \u221a x dx = 2k 3 x3/2 \u2223\u22231 0 = 2k 3 . Therefore, k = 3 2 . (b) F (x) = 3 2 \u222b x 0 \u221a t dt = t3/2 \u2223\u2223x 0 = x3/2. P (0.3 < X < 0.6) = F (0.6)\u2212 F (0.3) = (0.6)3/2 \u2212 (0.3)3/2 = 0.3004. 3.22 Denote by X the number of spades int he three draws. Let S and N stand for a spade and not a spade, respectively. Then P (X = 0) = P (NNN) = (39/52)(38/51)(37/50) = 703/1700, P (X = 1) = P (SNN) + P (NSN) + P (NNS) = 3(13/52)(39/51)(38/50) = 741/1700, P (X = 3) = P (SSS) = (13/52)(12/51)(11/50) = 11/850, and P (X = 2) = 1\u2212 703/1700\u2212 741/1700\u2212 11/850 = 117/850. The probability mass function for X is then x 0 1 2 3 f(x) 703/1700 741/1700 117/850 11/850 Solutions for Exercises in Chapter 3 33 3.23 The c.d.f. of X is F (x) = \uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3 0, for w < \u22123, 1/27, for \u2212 3 \u2264 w < \u22121, 7/27, for \u2212 1 \u2264 w < 1, 19/27, for 1 \u2264 w < 3, 1, for w \u2265 3, (a) P (W > 0 = 1\u2212 P (W \u2264 0) = 1\u2212 7/27 = 20/27. (b) P (\u22121 \u2264W < 3) = F (2)\u2212 F (\u22123) = 19/27\u2212 1/27 = 2/3. 3.24 There are ( 10 4 ) ways of selecting any 4 CDs from 10. We can select x jazz CDs from 5 and 4\u2212 x from the remaining CDs in (5 x )( 5 4\u2212x ) ways. Hence f(x) = ( 5 x )( 5 4\u2212x )( 10 4 ) , x = 0, 1, 2, 3, 4. 3.25 Let T be the total value of the three coins. Let D and N stand for a dime and nickel, respectively. Since we are selecting without replacement, the sample space containing elements for which t = 20, 25, and 30 cents corresponding to the selecting of 2 nickels and 1 dime, 1 nickel and 2 dimes, and 3 dimes. Therefore, P (T = 20) = (22)( 4 1) (63) = 1 5 , P (T = 25) = (21)( 4 2) (63) = 3 5 , P (T = 30) = (43) (63) = 1 5 , and the probability distribution in tabular form is t 20 25 30 P (T = t) 1/5 3/5 1/5 As a probability histogram 20 25 30 xx f(x ) f(x ) 1/5 2/5 3/5 34 Chapter 3 Random Variables and Probability Distributions 3.26 Denote by X the number of green balls in the three draws. Let G and B stand for the colors of green and black, respectively. Simple Event x P (X = x) BBB GBB BGB BBG BGG GBG GGB GGG 0 1 1 1 2 2 2 3 (2/3)3 = 8/27 (1/3)(2/3)2 = 4/27 (1/3)(2/3)2 = 4/27 (1/3)(2/3)2 = 4/27 (1/3)2(2/3) = 2/27 (1/3)2(2/3) = 2/27 (1/3)2(2/3) = 2/27 (1/3)3 = 1/27 The probability mass function for X is then x 0 1 2 3 P (X = x) 8/27 4/9