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# Solucionario Walpole 8 ED

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```letters follow g, we get a probability of 19/26.
2.58 (a) Let A = Defect in brake system; B = Defect in fuel system; P (A\u222aB) = P (A) +
P (B)\u2212 P (A \u2229 B) = 0.25 + 0.17\u2212 0.15 = 0.27.
(b) P (No defect) = 1\u2212 P (A \u222aB) = 1\u2212 0.27 = 0.73.
2.59 By Theorem 2.2, there are N = (26)(25)(24)(9)(8)(7)(6) = 47, 174, 400 possible ways
to code the items of which n = (5)(25)(24)(8)(7)(6)(4) = 4, 032, 000 begin with a vowel
and end with an even digit. Therefore, n
N
= 10
117
.
2.60 (a) Of the (6)(6) = 36 elements in the sample space, only 5 elements (2,6), (3,5),
(4,4), (5,3), and (6,2) add to 8. Hence the probability of obtaining a total of 8 is
then 5/36.
(b) Ten of the 36 elements total at most 5. Hence the probability of obtaining a total
of at most is 10/36=5/18.
Solutions for Exercises in Chapter 2 19
2.61 Since there are 20 cards greater than 2 and less than 8, the probability of selecting two
of these in succession is (
20
52
)(
19
51
)
=
95
663
.
2.62 (a)
(11)(
8
2)
(93)
= 1
3
.
(b)
(52)(
3
1)
(93)
= 5
14
.
2.63 (a)
(43)(
48
2 )
(525 )
= 94
54145
.
(b)
(134 )(
13
1 )
(525 )
= 143
39984
.
2.64 Any four of a kind, say four 2\u2019s and one 5 occur in
(
5
1
)
= 5 ways each with probability
(1/6)(1/6)(1/6)(1/6)(1/6) = (1/6)5. Since there are 6P2 = 30 ways to choose various
pairs of numbers to constitute four of one kind and one of the other (we use permutation
instead of combination is because that four 2\u2019s and one 5, and four 5\u2019s and one 2 are
two different ways), the probability is (5)(30)(1/6)5 = 25/1296.
2.65 (a) P (M \u222aH) = 88/100 = 22/25;
(b) P (M \u2032 \u2229H \u2032) = 12/100 = 3/25;
(c) P (H \u2229M \u2032) = 34/100 = 17/50.
2.66 (a) 9;
(b) 1/9.
2.67 (a) 0.32;
(b) 0.68;
(c) office or den.
2.68 (a) 1\u2212 0.42 = 0.58;
(b) 1\u2212 0.04 = 0.96.
2.69 P (A) = 0.2 and P (B) = 0.35
(a) P (A\u2032) = 1\u2212 0.2 = 0.8;
(b) P (A\u2032 \u2229 B\u2032) = 1\u2212 P (A \u222aB) = 1\u2212 0.2\u2212 0.35 = 0.45;
(c) P (A \u222a B) = 0.2 + 0.35 = 0.55.
2.70 (a) 0.02 + 0.30 = 0.32 = 32%;
(b) 0.32 + 0.25 + 0.30 = 0.87 = 87%;
20 Chapter 2 Probability
(c) 0.05 + 0.06 + 0.02 = 0.13 = 13%;
(d) 1\u2212 0.05\u2212 0.32 = 0.63 = 63%.
2.71 (a) 0.12 + 0.19 = 0.31;
(b) 1\u2212 0.07 = 0.93;
(c) 0.12 + 0.19 = 0.31.
2.72 (a) 1\u2212 0.40 = 0.60.
(b) The probability that all six purchasing the electric oven or all six purchasing the
gas oven is 0.007 + 0.104 = 0.111. So the probability that at least one of each
type is purchased is 1\u2212 0.111 = 0.889.
2.73 (a) P (C) = 1\u2212 P (A)\u2212 P (B) = 1\u2212 0.990\u2212 0.001 = 0.009;
(b) P (B\u2032) = 1\u2212 P (B) = 1\u2212 0.001 = 0.999;
(c) P (B) + P (C) = 0.01.
2.74 (a) (\$4.50\u2212 \$4.00)× 50, 000 = \$25, 000;
(b) Since the probability of underfilling is 0.001, we would expect 50, 000×0.001 = 50
boxes to be underfilled. So, instead of having (\$4.50 \u2212 \$4.00) × 50 = \$25 profit
for those 50 boxes, there are a loss of \$4.00× 50 = \$200 due to the cost. So, the
loss in profit expected due to underfilling is \$25 + \$200 = \$250.
2.75 (a) 1\u2212 0.95\u2212 0.002 = 0.048;
(b) (\$25.00\u2212 \$20.00)× 10, 000 = \$50, 000;
(c) (0.05)(10, 000)× \$5.00 + (0.05)(10, 000)× \$20 = \$12, 500.
2.76 P (A\u2032\u2229B\u2032) = 1\u2212P (A\u222aB) = 1\u2212(P (A)+P (B)\u2212P (A\u2229B) = 1+P (A\u2229B)\u2212P (A)\u2212P (B).
2.77 (a) The probability that a convict who pushed dope, also committed armed robbery.
(b) The probability that a convict who committed armed robbery, did not push dope.
(c) The probability that a convict who did not push dope also did not commit armed
robbery.
2.78 P (S | A) = 10/18 = 5/9.
2.79 Consider the events:
M : a person is a male;
S: a person has a secondary education;
C: a person has a college degree.
(a) P (M | S) = 28/78 = 14/39;
(b) P (C \u2032 | M \u2032) = 95/112.
Solutions for Exercises in Chapter 2 21
2.80 Consider the events:
A: a person is experiencing hypertension,
B: a person is a heavy smoker,
C: a person is a nonsmoker.
(a) P (A | B) = 30/49;
(b) P (C | A\u2032) = 48/93 = 16/31.
2.81 (a) P (M \u2229 P \u2229H) = 10
68
= 5
34
;
(b) P (H \u2229M | P \u2032) = P (H\u2229M\u2229P \u2032)
P (P \u2032)
= 22\u221210
100\u221268 =
12
32
= 3
8
.
2.82 (a) (0.90)(0.08) = 0.072;
(b) (0.90)(0.92)(0.12) = 0.099.
2.83 (a) 0.018;
(b) 0.22 + 0.002 + 0.160 + 0.102 + 0.046 + 0.084 = 0.614;
(c) 0.102/0.614 = 0.166;
(d) 0.102+0.046
0.175+0.134
= 0.479.
2.84 Consider the events:
C: an oil change is needed,
F : an oil filter is needed.
(a) P (F | C) = P (F\u2229C)
P (C)
= 0.14
0.25
= 0.56.
(b) P (C | F ) = P (C\u2229F )
P (F )
= 0.14
0.40
= 0.35.
2.85 Consider the events:
H : husband watches a certain show,
W : wife watches the same show.
(a) P (W \u2229H) = P (W )P (H | W ) = (0.5)(0.7) = 0.35.
(b) P (W | H) = P (W\u2229H)
P (H)
= 0.35
0.4
= 0.875.
(c) P (W \u222aH) = P (W ) + P (H)\u2212 P (W \u2229H) = 0.5 + 0.4\u2212 0.35 = 0.55.
2.86 Consider the events:
H : the husband will vote on the bond referendum,
W : the wife will vote on the bond referendum.
Then P (H) = 0.21, P (W ) = 0.28, and P (H \u2229W ) = 0.15.
(a) P (H \u222aW ) = P (H) + P (W )\u2212 P (H \u2229W ) = 0.21 + 0.28\u2212 0.15 = 0.34.
(b) P (W | H) = P (H\u2229W )
P (H)
= 0.15
0.21
= 5
7
.
(c) P (H | W \u2032) = P (H\u2229W \u2032)
P (W \u2032)
= 0.06
0.72
= 1
12
.
22 Chapter 2 Probability
2.87 Consider the events:
A: the vehicle is a camper,
(a) P (B | A) = P (A\u2229B)
P (A)
= 0.09
0.28
= 9
28
.
(b) P (A | B) = P (A\u2229B)
P (B)
= 0.09
0.12
= 3
4
.
(c) P (B\u2032 \u222a A\u2032) = 1\u2212 P (A \u2229B) = 1\u2212 0.09 = 0.91.
2.88 Define
H : head of household is home,
C: a change is made in long distance carriers.
P (H \u2229 C) = P (H)P (C | H) = (0.4)(0.3) = 0.12.
2.89 Consider the events:
A: the doctor makes a correct diagnosis,
B: the patient sues.
P (A\u2032 \u2229B) = P (A\u2032)P (B | A\u2032) = (0.3)(0.9) = 0.27.
2.90 (a) 0.43;
(b) (0.53)(0.22) = 0.12;
(c) 1\u2212 (0.47)(0.22) = 0.90.
2.91 Consider the events:
A: the house is open,
B: the correct key is selected.
P (A) = 0.4, P (A\u2032) = 0.6, and P (B) = (
1
1)(
7
2)
(83)
= 3
8
= 0.375.
So, P [A \u222a (A\u2032 \u2229 B)] = P (A) + P (A\u2032)P (B) = 0.4 + (0.6)(0.375) = 0.625.
2.92 Consider the events:
F : failed the test,
P : passed the test.
(a) P (failed at least one tests) = 1 \u2212 P (P1P2P3P4) = 1 \u2212 (0.99)(0.97)(0.98)(0.99) =
1\u2212 0.93 = 0.07,
(b) P (failed 2 or 3) = P (P1)P (P4)(1 \u2212 P (P2P3)) = (0.99)(0.99)(1 \u2212 (0.97)(0.98)) =
0.0484.
(c) 100× 0.07 = 7.
(d) 0.25.
2.93 Let A and B represent the availability of each fire engine.
(a) P (A\u2032 \u2229 B\u2032) = P (A\u2032)P (B\u2032) = (0.04)(0.04) = 0.0016.
(b) P (A \u222a B) = 1\u2212 P (A\u2032 \u2229 B\u2032) = 1\u2212 0.0016 = 0.9984.
Solutions for Exercises in Chapter 2 23
2.94 P (T \u2032 \u2229N \u2032) = P (T \u2032)P (N \u2032) = (1\u2212 P (T ))(1\u2212 P (N)) = (0.3)(0.1) = 0.03.
2.95 Consider the events:
A1: aspirin tablets are selected from the overnight case,
A2: aspirin tablets are selected from the tote bag,
L2: laxative tablets are selected from the tote bag,
T1: thyroid tablets are selected from the overnight case,
T2: thyroid tablets are selected from the tote bag.
(a) P (T1 \u2229 T2) = P (T1)P (T2) = (3/5)(2/6) = 1/5.
(b) P (T
\u2032
1 \u2229 T \u20322) = P (T \u20321)P (T \u20322) = (2/5)(4/6) = 4/15.
(c) 1\u2212P (A1 \u2229A2)\u2212P (T1\u2229T2) = 1\u2212P (A1)P (A2)\u2212P (T1)P (T2) = 1\u2212 (2/5)(3/6)\u2212
(3/5)(2/6) = 3/5.
2.96 Consider the events:
X: a person has an X-ray,
C: a cavity is filled,
T : a tooth is extracted.
P (X \u2229 C \u2229 T ) = P (X)P (C | X)P (T | X \u2229 C) = (0.6)(0.3)(0.1) = 0.018.
2.97 (a) P (Q1\u2229Q2\u2229Q3\u2229Q4) = P (Q1)P (Q2 | Q1)P (Q3 | Q1\u2229Q2)P (Q4 | Q1\u2229Q2\u2229Q3) =
(15/20)(14/19)(13/18)(12/17) = 91/323.
(b) Let A be the event that 4 good quarts of milk are selected. Then
P (A) =
(
15
4
)(
20
4
) = 91
323
.
2.98 P = (0.95)[1\u2212 (1\u2212 0.7)(1\u2212 0.8)](0.9) = 0.8037.
2.99 This is a parallel system of two series subsystems.
(a) P = 1\u2212 [1\u2212 (0.7)(0.7)][1\u2212 (0.8)(0.8)(0.8)] = 0.75112.
(b) P = P (A
\u2032\u2229C\u2229D\u2229E)
P system works
= (0.3)(0.8)(0.8)(0.8)
0.75112
= 0.2045.
2.100 Define S: the```