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Chapter 1: What is Statistics?
1.1 a. Population: all generation X age US citizens (specifically, assign a ‘1’ to those who
want to start their own business and a ‘0’ to those who do not, so that the population is the set of 1’s and 0’s). Objective: to estimate the proportion of generation X age US citizens who want to start their own business. b. Population: all healthy adults in the US. Objective: to estimate the true mean body temperature c. Population: single family dwelling units in the city. Objective: to estimate the true mean water consumption d. Population: all tires manufactured by the company for the specific year. Objective: to estimate the proportion of tires with unsafe tread. e. Population: all adult residents of the particular state. Objective: to estimate the proportion who favor a unicameral legislature. f. Population: times until recurrence for all people who have had a particular disease. Objective: to estimate the true average time until recurrence. g. Population: lifetime measurements for all resistors of this type. Objective: to estimate the true mean lifetime (in hours).
1.2 a. This histogram is above.
Histogram of wind
wind
Density
5 10 15 20 25 30 35
b. Yes, it is quite windy there. c. 11/45, or approx. 24.4% d. it is not especially windy in the overall sample.
2 Chapter 1: What is Statistics? Instructor’s Solutions Manual
1.3 The histogram is above.
Histogram of U
U
Density
0 2 4 6 8 10 12
1.4 a. The histogram is above.
Histogram of stocks
stocks
Density
2 4 6 8 10 12
b. 18/40 = 45% c. 29/40 = 72.5%
1.5 a. The categories with the largest grouping of students are 2.45 to 2.65 and 2.65 to 2.85.
(both have 7 students). b. 7/ c. 7/30 + 3/30 + 3/30 + 3/30 = 16/
1.6 a. The modal category is 2 (quarts of milk). About 36% (9 people) of the 25 are in this
category. b. .2 + .12 + .04 =. c. Note that 8% purchased 0 while 4% purchased 5. Thus, 1 – .08 – .04 = .88 purchased between 1 and 4 quarts.
4 Chapter 1: What is Statistics? Instructor’s Solutions Manual
Using the above, the numerator of s^2 is (^) ∑
n
i
yi y 1
( )^2 =
∑
n
i
yi yiy y 1
∑
n
i
yi 1
2
∑
n
i
y yi ny 1
2 2 Since ∑ =
n
i
ny yi 1
, we have (^) ∑
n
i
yi y 1
( )^2 =
∑
n
i
yi ny 1
(^2 2). Let ∑ =
n
i
yi n
y 1
to get the result.
1.12 Using the data, (^) ∑
6
i 1
yi = 14 and (^) ∑
6
1
2 i
yi = 40. So, s 2 = (40 - 14 2 /6)/5 = 1.47. So, s = 1.21.
1.13 a. With (^) ∑
45
i 1
y (^) i = 440.6 and (^) ∑
45
1
2 i
yi = 5067.38, we have that y = 9.79 and s = 4.14.
b. k interval frequency Exp. frequency 1 5.65, 13.93 44 30. 2 1.51, 18.07 44 42. 3 -2.63, 22.21 44 45
1.14 a. With (^) ∑
25
i 1
y (^) i = 80.63 and (^) ∑
25
1
2 i
yi = 500.7459, we have that y = 3.23 and s = 3.17.
b.
1.15 a. With (^) ∑
40
i 1
y (^) i = 175.48 and (^) ∑
40
1
2 i
yi = 906.4118, we have that y = 4.39 and s = 1.87.
b.
1.16 a. Without the extreme value, y = 4.19 and s = 1.44.
b. These counts compare more favorably:
k interval frequency Exp. frequency 1 0.063, 6.397 21 17 2 -3.104, 9.564 23 23. 3 -6.271, 12.731 25 25
k interval frequency Exp. frequency 1 2.52, 6.26 35 27. 2 0.65, 8.13 39 38 3 -1.22, 10 39 40
k interval frequency Exp. frequency 1 2.75, 5.63 25 26. 2 1.31, 7.07 36 37. 3 -0.13, 8.51 39 39
Chapter 1: What is Statistics? 5 Instructor’s Solutions Manual
1.17 For Ex. 1.2, range/4 = 7.35, while s = 4.14. For Ex. 1.3, range/4 = 3.04, while = s = 3.17.
For Ex. 1.4, range/4 = 2.32, while s = 1.87.
1.18 The approximation is (800–200)/4 = 150.
1.19 One standard deviation below the mean is 34 – 53 = –19. The empirical rule suggests
that 16% of all measurements should lie one standard deviation below the mean. Since chloroform measurements cannot be negative, this population cannot be normally distributed.
1.20 Since approximately 68% will fall between $390 ($420 – $30) to $450 ($420 + $30), the
proportion above $450 is approximately 16%.
1.21 (Similar to exercise 1.20) Having a gain of more than 20 pounds represents all
measurements greater than one standard deviation below the mean. By the empirical rule, the proportion above this value is approximately 84%, so the manufacturer is probably correct.
1.22 (See exercise 1.11) (^) ∑
n
i
yi y 1
( ) = (^) ∑
n
i
yi 1
1 1
= (^) ∑ −∑ = = =
n
i
i
n
i
ny yi y.
1.23 a. (Similar to exercise 1.20) 95 sec = 1 standard deviation above 75 sec, so this
percentage is 16% by the empirical rule. b. (35 sec., 115 sec) represents an interval of 2 standard deviations about the mean, so approximately 95% c. 2 minutes = 120 sec = 2.5 standard deviations above the mean. This is unlikely.
1.24 a. (112-78)/4 = 8.
b. The histogram is above.
Histogram of hr
hr
Frequency
80 90 100 110
0
1
2
3
4
5
c. With (^) ∑
20
i 1
yi = 1874.0 and (^) ∑
20
1
2 i
yi = 117,328.0, we have that y = 93.7 and s = 9.55.
Chapter 1: What is Statistics? 7 Instructor’s Solutions Manual
1.33 With k =2, at least 1 – 1/4 = 75% should lie within 2 standard deviations of the mean.
The interval is (0.5, 10.5).
1.34 The point 13 is 13 – 5.5 = 7.5 units above the mean, or 7.5/2.5 = 3 standard deviations
above the mean. By Tchebysheff’s theorem, at least 1 – 1/3^2 = 8/9 will lie within 3 standard deviations of the mean. Thus, at most 1/9 of the values will exceed 13.
1.35 a. (172 – 108)/4 =
b. With (^) ∑
15
i 1
yi = 2041 and (^) ∑
15
1
2 i
y (^) i = 281,807 we have that y = 136.1 and s = 17.
c. a = 136.1 – 2(17.1) = 101.9, b = 136.1 + 2(17.1) = 170.3. d. There are 14 observations contained in this interval, and 14/15 = 93.3%. 75% is a lower bound.
1.36 a. The histogram is above.
0
10
20
30
40
50
60
70
ex1.
0 1 2 3 4 5 6 8
b. With (^) ∑
100
i 1
yi = 66 and (^) ∑
100
1
2 i
y (^) i = 234 we have that y = 0.66 and s = 1.39.
c. Within two standard deviations: 95, within three standard deviations: 96. The calculations agree with Tchebysheff’s theorem.
1.37 Since the lead readings must be non negative, 0 (the smallest possible value) is only 0.
standard deviations from the mean. This indicates that the distribution is skewed.
1.38 By Tchebysheff’s theorem, at least 3/4 = 75% lie between (0, 140), at least 8/9 lie
between (0, 193), and at least 15/16 lie between (0, 246). The lower bounds are all truncated a 0 since the measurement cannot be negative.
Chapter 2: Probability
2.1 A = {FF}, B = {MM}, C = {MF, FM, MM}. Then, A ∩ B = 0 / , B ∩ C = {MM}, C ∩ B =
{MF, FM}, A ∪ B ={FF,MM}, A ∪ C = S , B ∪ C = C.
2.2 a. A ∩ B b. A ∪ B c. A ∪ B d. ( A ∩ B )∪( A ∩ B )
2.4 a.
b.
10 Chapter 2: Probability Instructor’s Solutions Manual
2.16 a. 1/3 b. 1/3 + 1/15 = 6/15 c. 1/3 + 1/16 = 19/48 d. 49/
2.17 Let B = bushing defect, SH = shaft defect. a. P (B) = .06 + .02 =. b. P (B or SH) = .06 + .08 + .02 =. c. P (exactly one defect) = .06 + .08 =. d. P (neither defect) = 1 – P(B or SH) = 1 – .16 =.
2.18 a. S = {HH, TH, HT, TT}
b. if the coin is fair, all events have probability .25. c. A = {HT, TH}, B = {HT, TH, HH} d. P ( A ) = .5, P ( B ) = .75, P ( A ∩ B ) = P ( A ) = .5, P ( A ∪ B ) = P ( B ) = .75, P ( A ∪ B ) = 1.
2.19 a. (V 1 , V 1 ), (V 1 , V 2 ), (V 1 , V 3 ), (V 2 , V 1 ), (V 2 , V 2 ), (V 2 , V 3 ), (V 3 , V 1 ), (V 3 , V 2 ), (V 3 , V 3 )
b. if equally likely, all have probability of 1/9. c. A = {same vendor gets both} = {(V 1 , V 1 ), (V 2 , V 2 ), (V 3 , V 3 )} B = {at least one V2} = {(V 1 , V 2 ), (V 2 , V 1 ), (V 2 , V 2 ), (V 2 , V 3 ), (V 3 , V 2 )} So, P ( A ) = 1/3, P ( B ) = 5/9, P ( A ∪ B ) = 7/9, P ( A ∩ B ) = 1/9.
2.20 a. P ( G ) = P ( D 1 ) = P ( D 2 ) = 1/3.
b. i. The probability of selecting the good prize is 1/3. ii. She will get the other dud. iii. She will get the good prize. iv. Her probability of winning is now 2/3. v. The best strategy is to switch.
2.21 P ( A ) = P ( ( A ∩ B )∪( A ∩ B )) = P (^ A^ ∩^ B )+ P ( A ∩ B )since these are M.E. by Ex. 2.5.
2.22 P ( A ) = P ( B ∪ ( A ∩ B )) = P (B) + P ( A ∩ B )since these are M.E. by Ex. 2.5.
2.23 All elements in B are in A , so that when B occurs, A must also occur. However, it is
possible for A to occur and B not to occur.
2.24 From the relation in Ex. 2.22, P ( A ∩ B )≥ 0, so P ( B ) ≤ P ( A ).
2.25 Unless exactly 1/2 of all cars in the lot are Volkswagens, the claim is not true.
2.26 a. Let N 1 , N 2 denote the empty cans and W 1 , W 2 denote the cans filled with water. Thus,
S = {N 1 N 2 , N 1 W 2 , N 2 W 2 , N 1 W 1 , N 2 W 1 , W 1 W 2 } b. If this a merely a guess, the events are equally likely. So, P (W 1 W 2 ) = 1/6.
2.27 a. S = {CC, CR, CL, RC, RR, RL, LC, LR, LL}
b. 5/ c. 5/
Chapter 2: Probability 11 Instructor’s Solutions Manual
2.28 a. Denote the four candidates as A 1 , A 2 , A 3 , and M. Since order is not important, the
outcomes are {A 1 A 2 , A 1 A 3 , A 1 M, A 2 A 3 , A 2 M, A 3 M}. b. assuming equally likely outcomes, all have probability 1/6. c. P (minority hired) = P (A 1 M) + P (A 2 M) + P (A 3 M) =.
2.29 a. The experiment consists of randomly selecting two jurors from a group of two women
and four men. b. Denoting the women as w1, w2 and the men as m1, m2, m3, m4, the sample space is w1,m1 w2,m1 m1,m2 m2,m3 m3,m w1,m2 w2,m2 m1,m3 m2,m w1,m3 w2,m3 m1,m w1,m4 w2,m4 w1,w c. P (w1,w2) = 1/
2.30 a. Let w1 denote the first wine, w2 the second, and w3 the third. Each sample point is an
ordered triple indicating the ranking. b. triples: (w1,w2,w3), (w1,w3,w2), (w2,w1,w3), (w2,w3,w1), (w3,w1,w2), (w3,w2,w1) c. For each wine, there are 4 ordered triples where it is not last. So, the probability is 2/3.
2.31 a. There are four “good” systems and two “defactive” systems. If two out of the six systems are chosen randomly, there are 15 possible unique pairs. Denoting the systems as g1, g2, g3, g4, d1, d2, the sample space is given by S = {g1g2, g1g3, g1g4, g1d1, g1d2, g2g3, g2g4, g2d1, g2d2, g3g4, g3d1, g3d2, g4g1, g4d1, d1d2}. Thus: P (at least one defective) = 9/15 P (both defective) = P (d1d2) = 1/ b. If four are defective, P (at least one defective) = 14/15. P (both defective) = 6/15.
2.32 a. Let “1” represent a customer seeking style 1, and “2” represent a customer seeking
style 2. The sample space consists of the following 16 four-tuples: 1111, 1112, 1121, 1211, 2111, 1122, 1212, 2112, 1221, 2121, 2211, 2221, 2212, 2122, 1222, 2222 b. If the styles are equally in demand, the ordering should be equally likely. So, the probability is 1/16. c. P ( A ) = P (1111) + P (2222) = 2/16.
2.33 a. Define the events: G = family income is greater than $43,318, N otherwise. The
points are E1: GGGG E2: GGGN E3: GGNG E4: GNGG E5: NGGG E6: GGNN E7: GNGN E8: NGGN E9: GNNG E10: NGNG E11: NNGG E12: GNNN E13: NGNN E14: NNGN E15: NNNG E16: NNNN b. A = {E1, E2, …, E11} B = {E6, E7, …, E11} C = {E2, E3, E4, E5} c. If P (E) = P (N) = .5, each element in the sample space has probability 1/16. Thus, P ( A ) = 11/16, P ( B ) = 6/16, and P ( C ) = 4/16.
Chapter 2: Probability 13 Instructor’s Solutions Manual
2.46 There are (^) ⎟⎟ ⎠
ways to chose two teams for the first game, (^) ⎟⎟ ⎠
for second, etc. So,
there are (^5) 2
= 113,400 ways to assign the ten teams to five games.
2.47 There are (^) ⎟⎟ ⎠
2 n ways to chose two teams for the first game, (^) ⎟⎟ ⎠
2 n 2 for second, etc. So,
following Ex. 2.46, there are (^) n
n 2
ways to assign 2 n teams to n games.
2.48 Same answer: (^) ⎟⎟ ⎠
2.49 a. (^) ⎟⎟ ⎠
b. There are 26*26 = 676 two-letter codes and 26(26)(26) = 17,576 three-letter codes. Thus, 18,252 total major codes. c. 8385 + 130 = 8515 required. d. Yes.
2.50 Two numbers, 4 and 6, are possible for each of the three digits. So, there are 2(2)(2) = 8
potential winning three-digit numbers.
2.51 There are (^) ⎟⎟ ⎠
= 19,600 ways to choose the 3 winners. Each of these is equally likely.
a. There are (^) ⎟⎟ ⎠
= 4 ways for the organizers to win all of the prizes. The probability is
b. There are (^) ⎟⎟ ⎠
= 276 ways the organizers can win two prizes and one of the other
46 people to win the third prize. So, the probability is 276/19600.
c. (^) ⎟⎟ ⎠
= 4140. The probability is 4140/19600.
d. (^) ⎟⎟ ⎠
= 15,180. The probability is 15180/19600.
2.52 The mn rule is used. The total number of experiments is 3(3)(2) = 18.
14 Chapter 2: Probability Instructor’s Solutions Manual
2.53 a. In choosing three of the five firms, order is important. So P 3^5 = 60 sample points.
b. If F 3 is awarded a contract, there are P 2^4 = 12 ways the other contracts can be assigned. Since there are 3 possible contracts, there are 3(12) = 36 total number of ways to award F 3 a contract. So, the probability is 36/60 = 0.6.
2.54 There are (^) ⎟⎟ ⎠
= 70 ways to chose four students from eight. There are (^) ⎟⎟ ⎠
= 30 ways
to chose exactly 2 (of the 3) undergraduates and 2 (of the 5) graduates. If each sample point is equally likely, the probability is 30/70 = 0.7.
2.55 a. (^) ⎟⎟ ⎠
b. (^) ⎟⎟ ⎠
2.56 The student can solve all of the problems if the teacher selects 5 of the 6 problems that
the student can do. The probability is (^) ⎟⎟ ⎠
2.57 There are (^) ⎟⎟ ⎠
= 1326 ways to draw two cards from the deck. The probability is
2.58 There are (^) ⎟⎟ ⎠
= 2,598,960 ways to draw five cards from the deck.
a. There are (^) ⎟⎟ ⎠
= 24 ways to draw three Aces and two Kings. So, the probability is
b. There are 13(12) = 156 types of “full house” hands. From part a. above there are 24 different ways each type of full house hand can be made. So, the probability is 156*24/2598960 = 0.00144.
2.59 There are (^) ⎟⎟ ⎠
= 2,598,960 ways to draw five cards from the deck.
a. (^) ⎟⎟ ⎠
5 = 1024. So, the probability is 1024/2598960 = 0.000394.
b. There are 9 different types of “straight” hands. So, the probability is 9( 5 )/2598960 = 0.00355. Note that this also includes “straight flush” and “royal straight flush” hands.
2.60 a. (^) n
n 365
b. With n = 23, (^23) 365
16 Chapter 2: Probability Instructor’s Solutions Manual
c. Let A be the event that Distributors I, II, and III get exactly 2, 3, and 1 order(s) respectively. Then, there is one remaining unassigned order. Thus, A contains
= 2940 sample points and P ( A ) = 2940/ 7 = 0.00029.
2.68 a. (^) ⎟⎟ ⎠
n
n
!( )!
n n n
n −
= 1. There is only one way to choose all of the items.
b. (^) ⎟⎟ ⎠
n
0 !( 0 )!
n −
n = 1. There is only one way to chose none of the items.
c. (^) ⎟⎟ ⎠
r
n
!( )!
r n r
n −
n r n n r
n − − −
n − r
n
. There are the same number of
ways to choose r out of n objects as there are to choose n – r out of n objects.
d. (^) ∑ ∑
−
n
i
nii
n
i
n n i
n i
n
1 1
1 k n k
n k n k
nk k n k
n n k k n k
n k n k
n k
n k
n
2.70 From Theorem 2.3, let y 1 = y 2 = … = y (^) k = 1.
2.71 a. P ( A | B ) = .1/.3 = 1/3. b. P ( B | A ) = .1/.5 = 1/5.
c. P ( A | A ∪ B ) = .5/(.5+.3-.1) = 5/7 d. P ( A | A ∩ B ) = 1, since A has occurred. e. P ( A ∩ B | A ∪ B ) = .1(.5+.3-.1) = 1/7.
2.72 Note that P ( A ) = 0.6 and P ( A | M ) = .24/.4 = 0.6. So, A and M are independent. Similarly,
P ( A | F ) = .24/.6 = 0.4 = P ( A ), so A and F are independent.
2.73 a. P (at least one R) = P (Red) 3/4. b. P (at least one r) = 3/4.
c. P (one r | Red) = .5/.75 = 2/3.
2.74 a. P ( A ) = 0.61, P ( D ) = .30. P ( A ∩ D ) = .20. Dependent. b. P ( B ) = 0.30, P ( D ) = .30. P ( B ∩ D ) = 0.09. Independent. c. P ( C ) = 0.09, P ( D ) = .30. P ( C ∩ D ) = 0.01. Dependent.
Chapter 2: Probability 17 Instructor’s Solutions Manual
2.75 a. Given the first two cards drawn are spades, there are 11 spades left in the deck. Thus,
the probability is
⎟⎟ ⎠
= 0.0084. Note: this is also equal to P (S 3 S 4 S 5 |S 1 S 2 ).
b. Given the first three cards drawn are spades, there are 10 spades left in the deck. Thus,
the probability is
⎟⎟ ⎠
= 0.0383. Note: this is also equal to P (S 4 S 5 |S 1 S 2 S 3 ).
c. Given the first four cards drawn are spades, there are 9 spades left in the deck. Thus,
the probability is
⎟⎟ ⎠
= 0.1875. Note: this is also equal to P (S 5 |S 1 S 2 S 3 S 4 )
2.76 Define the events: U : job is unsatisfactory A : plumber A does the job
a. P ( U | A ) = P ( A ∩ U )/ P ( A ) = P ( A | U ) P ( U )/ P ( A ) = .5*.1/.4 = 0. b. From part a. above, 1 – P ( U | A ) = 0.875.
2.77 a. 0.40 b. 0.37 c. 0.10 d. 0.40 + 0.37 – 0.10 = 0.
e. 1 – 0.4 = 0.6 f. 1 – 0.67 = 0.33 g. 1 – 0.10 = 0. h. .1/.37 = 0.27 i. 1/.4 = 0.
2.78 1. Assume P ( A | B ) = P ( A ). Then:
P ( A ∩ B ) = P ( A | B ) P ( B ) = P ( A ) P ( B ). P ( B | A ) = P ( B ∩ A )/ P ( A ) = P ( A ) P ( B )/ P ( A ) = P ( B ).
2. Assume P ( B | A ) = P ( B ). Then: P ( A ∩ B ) = P ( B | A ) P ( A ) = P ( B ) P ( A ). P ( A | B ) = P ( A ∩ B )/ P ( B ) = P ( A ) P ( B )/ P ( B ) = P ( A ). 3. Assume P ( A ∩ B ) = P ( B ) P ( A ). The results follow from above.
2.79 If A and B are M.E., P ( A ∩ B ) = 0. But, P ( A )P( B ) > 0. So they are not independent.
2.80 If A ⊂ B , P ( A ∩ B ) = P ( A ) ≠ P ( A ) P ( B ), unless B = S (in which case P ( B ) = 1).
2.81 Given P (A ) < P ( A | B ) = P ( A ∩ B )/ P ( B ) = P ( B | A ) P ( A )/ P ( B ), solve for P ( B | A ) in the
inequality.
2.82 P ( B | A ) = P ( B ∩ A )/ P ( A ) = P ( A )/ P ( A ) = 1
P ( A | B ) = P ( A ∩ B )/ P ( B ) = P ( A )/ P ( B ).
Chapter 2: Probability 19 Instructor’s Solutions Manual
2.93 Let H denote a hit and let M denote a miss. Then, she wins the game in three trials with
the events HHH, HHM, and MHH. If she begins with her right hand, the probability she wins the game, assuming independence, is (.7)(.4)(.7) + (.7)(.4)(.3) + (.3)(.4)(.7) = 0.364.
2.94 Define the events A : device A detects smoke B : device B detects smoke
a. P ( A ∪ B )= .95 + .90 - .88 = 0.97. b. P (smoke is undetected) = 1 - P ( A ∪ B )= 1 – 0.97 = 0.03.
2.95 Part a is found using the Addition Rule. Parts b and c use DeMorgan’s Laws.
a. 0.2 + 0.3 – 0.4 = 0. b. 1 – 0.1 = 0. c. 1 – 0.4 = 0.
d.
. 3
PB
PB P A B
PB
P A B
P A B = 2/3.
2.96 Using the results of Ex. 2.95:
a. 0.5 + 0.2 – (0.5)(0.2) = 0.6. b. 1 – 0.6 = 0.4. c. 1 – 0.1 = 0.9.
2.97 a. P (current flows) = 1 – P (all three relays are open) = 1 – (.1) 3 = 0.999. b. Let A be the event that current flows and B be the event that relay 1 closed properly. Then, P ( B | A ) = P ( B ∩ A )/ P ( A ) = P ( B )/ P ( A ) = .9/.999 = 0.9009. Note that B ⊂ A.
2.98 Series system: P (both relays are closed) = (.9)(.9) = 0.
Parallel system: P (at least one relay is closed) = .9 + .9 – .81 = 0.99.
2.99 Given that P ( A ∪ B )= a , P ( B ) = b , and that A and B are independent. Thus P ( A ∪ B )=
1 – a and P ( B ∩ A ) = bP ( A ). Thus, P ( A ∪ B )= P ( A ) + b - bP ( A ) = 1 – a. Solve for P ( A ).
PC
P A C B C
PC
P A B C
P A B C
P C
P A ∩ C + P B ∩ C − P A ∩ B ∩ C
= P ( A | C ) + P ( B | C ) + P ( A ∩ B | C ).
2.101 Let A be the event the item gets past the first inspector and B the event it gets past the
second inspector. Then, P ( A ) = 0.1 and P ( B | A ) = 0.5. Then P ( A ∩ B ) = .1(.5) = 0.05.
2.102 Define the events: I: disease I us contracted II: disease II is contracted. Then, P (I) = 0.1, P (II) = 0.15, and P (I∩II) = 0.03. a. P (I ∪ II) = .1 + .15 – .03 = 0. b. P (I∩II|I ∪ II) = .03/.22 = 3/22.
20 Chapter 2: Probability Instructor’s Solutions Manual
2.103 Assume that the two state lotteries are independent.
a. P (666 in CT|666 in PA) = P (666 in CT) = 0. b. P (666 in CT∩666 in PA) = P (666 in CT) P (666 in PA) = .001(1/8) = 0.000125.
2.104 By DeMorgan’s Law, P ( A ∩ B )= 1 − P ( A ∩ B )= 1 − P ( A ∪ B ). Since P ( A ∪ B )≤
P ( A )+ P ( B ), P ( A ∩ B )≥ 1 – P ( A )− P ( B ).
2.105 P(landing safely on both jumps) ≥ – 0.05 – 0.05 = 0.90.
2.106 Note that it must be also true that P ( A )= P ( B ). Using the result in Ex. 2.104,
P ( A ∩ B )≥ 1 – 2 P ( A )≥ 0.98, so P ( A ) ≥ 0.99.
2.107 (Answers vary) Consider flipping a coin twice. Define the events:
A : observe at least one tail B : observe two heads or two tails C : observe two heads
2.108 Let U and V be two events. Then, by Ex. 2.104, P ( U ∩ V )≥ 1 – P ( U )− P ( V ). Let U =
A ∩ B and V = C. Then, P ( A ∩ B ∩ C )≥ 1 – P ( A ∩ B )− P ( C ). Apply Ex. 2.104 to
P ( A ∩ B )to obtain the result.
2.109 This is similar to Ex. 2.106. Apply Ex. 2.108: 0.95 ≤ 1 – P ( A )− P ( B )− P ( C )≤
P ( A ∩ B ∩ C ). Since the events have the same probability, 0.95 ≤ 1 − 3 P ( A ). Thus, P ( A ) ≥ 0.9833.
2.110 Define the events:
I : item is from line I II : item is from line II N : item is not defective Then, P ( N ) = P ( N ∩ ( I ∪ II ))= P ( N ∩ I ) + P ( N ∩ II ) = .92(.4) + .90(.6) = 0.908.
2.111 Define the following events:
A : buyer sees the magazine ad B : buyer sees the TV ad C : buyer purchases the product The following are known: P ( A ) = .02, P ( B ) = .20, P ( A ∩ B ) = .01. Thus P ( A ∩ B )= .21. Also, P (buyer sees no ad) = P ( A ∩ B )= 1 − P ( A ∪ B )= 1 – 0.21 = 0.79. Finally, it is known that P ( C | A ∪ B ) = 0.1 and P ( C | A ∩ B )= 1/3. So, we can find P ( C ) as P ( C ) = P ( C ∩ ( A ∪ B ))+ P ( C ∩( A ∩ B ))= (1/3)(.21) + (.1)(.79) = 0.149.
2.112 a. P(aircraft undetected) = P(all three fail to detect) = (.02)(.02)(.02) = (.02) 3 . b. P(all three detect aircraft) = (.98)^3.
2.113 By independence, (.98)(.98)(.98)(.02) = (.98)^3 (.02).
