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# Solucionario Walpole 8 ED

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```(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```