P( A | B) =
2-121.
(0.38)(0.53) P( A B) P( A) P( B | A) = = = 39.3821% P( B) P( A) P( B | A) + P ( A' ) P( B | A' ) (0.38)(0.53) + (0.62)(0.5)
Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, an

3-96.
Let X denote a geometric random variable with parameter p. Let q = 1-p.
1 1 E ( X ) = x (1 p ) x 1 p = p xq x 1 = p 2 = x =1 x =1 p p
V ( X ) = ( x 1 ) 2 (1 p) x 1 p = px 2 2 x + p
x =1 x =1
1 p
(
1 p
)(1 p)
x 1
= p x 2 q x 1 2 xq x 1 +
x =1
x =1
q

3-79.
The probability is 0.651 that at least one sample from the next five will contain more than one defective Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1.
a) P( X 5) = 1 P( X

3-60.
E (cX ) = cxf ( x) = c xf ( x) = cE ( X ) , V (cX ) = (cx c ) f ( x) = c 2 ( x ) 2 f ( x) = cV ( X )
2 x x x x
Section 3-6 3-61. A binomial distribution is based on independent trials with two outcomes and a constant probability of success on each t

3-38.
The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1 a) P(X1/18) = 0 b) P(X1/4) = 0.9 c) P(X5/16) = 0.9 d) P(X>1/4) = 0.1 e) P(X1/2) = 1
Section 3-4 3-39. Mean and Va

CHAPTER 3 Section 3-1 3-1. 3-2. 3-3. 3-4. 3-5. The range of X is
cfw_0,1,2,.,1000
The range of X is cfw_0,12,.,50 , The range of X is cfw_0,12,.,99999 , The range of X is cfw_0,12,3,4,5 , The range of X is 1,2,.,491 . Because 490 parts are conforming, a n

CHAPTER 3 Section 3-1 3-1. 3-2. 3-3. 3-4. 3-5. The range of X is
cfw_0,1,2,.,1000
The range of X is cfw_0,12,.,50 , The range of X is cfw_0,12,.,99999 , The range of X is cfw_0,12,3,4,5 , The range of X is 1,2,.,491 . Because 490 parts are conforming, a n

Section 3-9 3-107.
e 4 4 0 = e 4 = 0.0183 0! b) P( X 2) = P( X = 0) + P( X = 1) + P( X = 2)
a) P( X = 0) =
e 4 41 e 4 42 + 1! 2! = 0.2381 = e 4 +
e 4 4 4 = 01954 . 4! e 4 4 8 = 0.0298 d) P( X = 8) = 8!
c) P ( X = 4 ) = 3-108 a) P ( X = 0) = e 0.4 = 0.6703

3-118.
a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with = 0.16.
P( X = 0) = e 0.16 = 0.8521
b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with = 0.48.
P(Y 1) = 1 P(Y = 0) = 1 e 48 = 0

(d) P ( A' B ) =
63 + 35 + 15 = 0.113 1000 0.032 P( A B) (e) P ( A | B ) = = = 0.2207 (25 + 63 + 15 + 7 + 35) / 1000 P( B) 5 (f) P = = 0.005 1000
2-167.
(a) Let A denote that a part conforms to specifications and let B denote for a part a simple component

P( A | B) =
2-121.
(0.38)(0.53) P( A B) P( A) P( B | A) = = = 39.3821% P( B) P( A) P( B | A) + P ( A' ) P( B | A' ) (0.38)(0.53) + (0.62)(0.5)
Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, an

2-112.
10 6 = 10 10 16 10 1 (b) P = 0.25 ( ) = 0.020833 12
(a) P =
Let A denote the event that a sample is produced in cavity one of the mold. 1 a) By independence, P( A1 A 2 A 3 A 4 A 5 ) = ( )5 = 0.00003 8 b) Let Bi be the event that all five samples ar

a) P(B) = P(B|A)P(A) + P(B|A')P(A') = (19/99)(20/100) + (20/99)(80/100) = 0.2
b) Let C denote the event that the third chip selected is defective. P( A B C ) = P(C A B) P( A B) = P(C A B) P( B A) P( A)
18 19 20 98 99 100 = 0.00705 =
2-99.
Open surgery suc

c) P( A B ) = 0 , because A B = d) P( ( A B ) C ) = 0, because ( A B ) C = ( A C ) ( B C ) e) P( A B C ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1
=
2-68.
(a) P(Caused by sports) = P(Caused by contact sports or by noncontact sports) = P(Caused by c

(b) There are 36 experiments that use all three steps. The probability the best result uses all three steps is 36/52 = 0.6923. (a) No, it will not change. With k amounts in the first step the number of experiments is k + 3k + 9k = 13k. The number of exper

CHAPTER 2
Section 2-1 2-1. 2-2. Let "a", "b" denote a part above, below the specification S = cfw_aaa, aab, aba, abb, baa, bab, bba, bbb
Let "e" denote a bit in error Let "o" denote a bit not in error ("o" denotes okay)
eeee, eoee, oeee, ooee, eeeo, eoeo,

125 375 5 0 2.3453E8 = = 0.00092 f X (5) = 2.5524 E11 500 5
b)
x f(x)
3-150.
0 1 2 3 4 5 6 7 8 9 10 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000
Let X denote the number of totes in the sample that exceed the moisture conte