%3p − 3p2+ p3&
+ (1 − p)3
3 − 3p + p2 = 1
3 − 3p + p2 and
pY(2) = p(1 − p) + (1 − p)2pY(0) = p(1 − p) + (1 − p)2 (1 − p)2 3 − 3p + p2
= p(1 − p)%
3 − 3p + p2&
+ (1 − p)4
3 − 3p + p2 =
(1 − p)8%
3p − 3p2+ p3&
+ (1 − p)39 3 − 3p + p2
= 1 − p
3 − 3p + p2. Hence,
pY(y)=
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
(1 − p)2/(3 − 3p + p2), if y = 0;
1/(3 − 3p + p2), if y = 1;
(1 − p)/(3 − 3p + p2), if y = 2;
0, otherwise.
Review Exercises for Chapter 5
Basic Exercises 5.142
a) The possible values of X are 1, 2, 3, and 4.
b) The event that the student selected is a junior can be represented as {X = 3}.
c) From the table, the total number of undergraduate students is
6,159 + 6,790 + 8,141 + 11,220 = 32,310.
Hence, because the selection is done randomly, we have P (X= 3) = N{X = 3}
N (!) = 8,141
32,310 =0.252.
Thus, 25.2% of the undergraduate students are juniors.
d) Proceeding as in part (c), we obtain the following table for the PMF of X:
x 1 2 3 4
pX(x ) 0.191 0.210 0.252 0.347
e) Following is a probability histogram for X. In the graph, we use p(x) instead of pX(x).
5.143
a) The event that the number of busy lines is exactly four can be expressed as {Y = 4}.
b) The event that the number of busy lines is at least four can be expressed as {Y ≥ 4}.
c) The event that the number of busy lines is between two and four, inclusive, can be expressed as {2 ≤ Y ≤ 4}.
d) We have
P (Y = 4) = pY(4) = 0.174, P (Y ≥ 4) =/
y≥4
pY(y)= /6 k=4
pY(k)= 0.174 + 0.105 + 0.043 = 0.322, and
P (2 ≤ Y ≤ 4) = /
2≤y≤4
pY(y)= /4 k=2
pY(k)= 0.232 + 0.240 + 0.174 = 0.646.
5.144 We note that X ∼ G(0.5). Hence, pX(x)= 2−xif x ∈ N , and pX(x)= 0 otherwise.
a) Following is a partial probability histogram for X. In the graph, we use p(x) instead of pX(x).
b) We can’t draw the entire probability histogram because the range of X is infinite, namely, N .
c) By the FPF,
P (X >4) =/
x>4
pX(x)= /∞ k=5
(1/2)k= (1/2)5
1 − (1/2) = (1/2)4 = 1 16. d) From the complementation rule and the FPF,
P (X >4) = 1 − P (X ≤ 4) = 1 −/
x≤4
pX(x)= 1 − /4 k=1
(1/2)k
= 1 − (1/2) − (1/2)5
1 − (1/2) = 1 −%
1 − (1/2)4&
= (1/2)4 = 1 16. e) Proceeding as in parts (c) and (d), respectively, we get
P (X > x)=/
y>x
pX(y)= /∞ k=x+1
(1/2)k = (1/2)x+1
1 − (1/2) = (1/2)x = 2−x and
P (X > x)= 1 − P (X ≤ x) = 1 −/
y≤x
pX(y)= 1 − /x k=1
(1/2)k
= 1 − (1/2) − (1/2)x+1
1 − (1/2) = 1 −%
1 − (1/2)x&
= (1/2)x = 2−x. f) Let N denote the set of prime numbers. Then
P (X ∈ N) = /
x∈N
pX(x)≥ pX(2) + pX(3) + pX(5) + pX(7) + pX(11)
= 1 22 + 1
23 + 1 25 + 1
27 + 1
211 = 0.41455, to five decimal places. Also, from part (e),
P (X∈ N) = /
x∈N
pX(x)≤ pX(2) + pX(3) + pX(5) + pX(7) + pX(11) + P (X > 12)
= 1 22 + 1
23 + 1 25 + 1
27 + 1 211 + 1
212 = 0.41479,
to five decimal places. Thus, 0.41455 ≤ P (X ∈ N) ≤ 0.41479 and, hence, P (X ∈ N) = 0.415 to three significant digits.
5.145
a) We have p = 0.4 because there is a 40% chance that a drinker is involved.
b) With independence, an outcome consisting of exactly k successes has probability (0.4)k(0.6)3−k. Thus, we have the following table:
Outcome Probability Outcome Probability (s, s, s) (0.4)3(0.6)0= 0.064 (f, s, s) (0.4)2(0.6)1= 0.096 (s, s, f ) (0.4)2(0.6)1= 0.096 (f, s, f ) (0.4)1(0.6)2= 0.144 (s, f, s) (0.4)2(0.6)1= 0.096 (f, f, s) (0.4)1(0.6)2= 0.144 (s, f, f ) (0.4)1(0.6)2= 0.144 (f, f, f ) (0.4)0(0.6)3= 0.216
c) From the table in part (b), we see that the outcomes in which exactly two of three traffic fatalities involve a drinker are (s, s, f ), (s, f, s), and (f, s, s).
d) From the table in part (b), we see that each of the three outcomes in part (c) has probability 0.096.
The three probabilities are the same because each probability is obtained by multiplying two success probabilities of 0.4 and one failure probability of 0.6.
e) From parts (c) and (d), we see that the probability that exactly two of three traffic fatalities involve a drinker is equal to 3 · 0.096 = 0.288.
f) X ∼ B(3, 0.4); that is, X has the binomial distribution with parameters 3 and 0.4.
g) Referring to the table in part (b), we get P (X= 0) = P%
{(f, f, f )}&
= 0.216, P (X= 1) = P%
{(s, f, f ), (f, s, f ), (f, f, s)}&
= P%
{(s, f, f )}&
+ P%
{(f, s, f )}&
+ P%
{(f, f, s)}&
= 0.144 + 0.144 + 0.144 = 0.432, P (X= 2) = P%
{(s, s, f ), (s, f, s), (f, s, s)}&
= P%
{(s, s, f )}&
+ P%
{(s, f, s)}&
+ P%
{(f, s, s)}&
= 0.096 + 0.096 + 0.096 = 0.288, P (X= 3) = P%
{(s, s, s)}&
= 0.064.
Thus, we have the following table for the PMF of X:
x 0 1 2 3
pX(x ) 0.216 0.432 0.288 0.064 h) Applying Procedure 5.1 with n = 3 and p = 0.4, we get
pX(x)='3 x (
(0.4)x(0.6)3−x, x = 0, 1, 2, 3,
and pX(x) = 0 otherwise. Substituting successively x = 0, 1, 2, and 3 into the preceding display and doing the necessary algebra gives the same values as shown in the table in part (g).
5.146 Let Y denote the number of traffic fatalities examined up to and including the first one that involves a drinker. Then Y ∼ G(0.4). From Proposition 5.9 on page 229,
pY(y)= (0.4)(0.6)y−1, y = 1, 2, . . . , and pY(y)= 0 otherwise. Also, from Proposition 5.10 on page 231,
P (Y > n)= (0.6)n, n∈ N . a) We have
P (Y = 4) = pY(4) = (0.4)(0.6)4−1 = (0.4)(0.6)3= 0.0864.
b) From the complementation rule,
P (Y ≤ 4) = 1 − P (Y > 4) = 1 − (0.6)4= 0.870.
c) We have
P (Y ≥ 4) = P (Y > 3) = (0.6)3 = 0.216.
d) We want to find the largest n such that P (Y > n) > 0.2. Thus, (0.6)n>0.2 or n < ln 0.2
ln 0.6 ≈3.15.
Hence, n = 3.
e) See the beginning of the solution to this exercise.
5.147 Let X denote the number of the eight tires tested that get at least 35,000 miles. We know that X ∼ B(8, 0.9). Therefore, by the FPF,
P (X ≥ 0.75 · 8) = P (X ≥ 6) =/
x≥6
pX(x)= /8 k=6
'8 k (
(0.9)k(0.1)8−k= 0.962.
5.148 Let X denote the number of “no-shows” when n reservations are made. Then X ∼ B(n, 0.04).
We want to find the largest n so that P (X = 0) ≥ 0.8. Noting that P (X= 0) =
'n 0 (
(0.04)0(0.96)n = (0.96)n, we have the requirement that
(0.96)n≥ 0.8 or n≤ ln 0.8
ln 0.96 ≈5.5.
Hence, n = 5.
5.149 Let X denote the number of moderate-income families of five sampled that spend more than half their income on housing. Then X ∼ H(N, 5, 0.76), where N is the number of moderate-income families in the United States. Because the sample size (5) is small relative to the population size (N), we know, from Proposition 5.6 on page 215, that X ≈ B(5, 0.76). Thus,
pX(x)≈'5 x (
(0.76)x(0.24)5−x, x = 0, 1, . . . , 5.
a) We have
P (X = 3) = pX(3) ≈'5 3 (
(0.76)3(0.24)2 = 0.253.
b) From the FPF,
P (X ≤ 3) =/
x≤3
pX(x)= /3 k=0
pX(k)≈ /3 k=0
'5 k (
(0.76)k(0.24)5−k = 0.346.
c) From the FPF,
P (X ≥ 4) =/
x≥4
pX(x)= /5 k=4
pX(k)≈ /5 k=4
'5 k (
(0.76)k(0.24)5−k = 0.654.
d) From the complementation rule and the FPF,
P (1 ≤ X ≤ 4) = 1 − P (X = 0 or 5) = 1 −%
pX(0) + pX(5)&
≈ 1 − 1'5
0 (
(0.76)0(0.24)5+'5 5 (
(0.76)5(0.24)0 2
= 1 − (0.24)5− (0.76)5= 0.746.
e) Applying the formula presented at the beginning of this solution for the PDF of a B(5, 0.76) random variable, we obtain the following table:
x 0 1 2 3 4 5
pX(x ) 0.000796 0.012607 0.079847 0.252850 0.400346 0.253553
f) The PMF obtained in part (e) assumes sampling with replacement, whereas the sampling is actually without replacement. Thus, the binomial approximation to the hypergeometric distribution (Proposi-tion 5.6) has been used.
g) As we noted, X ∼ H(N, 5, 0.76). Hence,
a) Let X denote the number of successes in n Bernoulli trials with success probability p. We know that X ∼ B(n, p). Let Xj be 0 or 1, respectively, depending on whether a failure or success occurs on trial j. We want to find P (Xj = 1 | X = n). The event {Xj = 1, X = n} occurs if and only if trial j results in a success and among the other n − 1 trials exactly k − 1 successes occur. The former event has probability p and the latter event has probability
'n− 1
Moreover, by the independence of Bernoulli trials, the two events are independent. Applying the condi-tional probability rule, the special multiplication rule, and the binomial PMF, we get
P (Xj = 1 | X = n) = P (Xj = 1, X = n) b) Given that exactly k successes occur in n Bernoulli trials, we can think of the positions (trial numbers) of the k successes and n − k failures as obtained by randomly placing k “s”s and n − k “f ”s in n boxes, one in each box. By symmetry, the letter (“s” or “f ”) placed in box j is equally likely to be any of the nletters. As k of the n letters are “s”s, the probability is k/n that an “s” goes into box j—that is, the probability is k/n that trial j results in success.
5.151
a) According to the frequentist interpretation of probability (see page 5), we have P (E) ≈ n(E)/n.
Intuitively, then, we estimate P (E) to be n(E)/n.
b) We note that n(E) ∼ B(n, p), where p = P (E). Suppose that E occurs k times in the n trials, that is, that event {n(E) = k} occurs. The probability of that happening is %n
k
&
pk(1 − p)n−k. We want to choose p to maximize that quantity, which we denote f (p). We have
f′(p)= words, the maximum likelihood estimate of P (E) is the random variable ˆp = n(E)/n.
c) The maximum likelihood estimate of P (E) coincides with the intuitive estimate based on the frequen-tist interpretation of probability.
5.152 Referring to Definition 5.5 on page 211, we obtain the following results.
a) Because 4 > 0,
'9 4 (
= 9 · 8 · 7 · 6
4! = 9 · 8 · 7 · 6
24 = 126.
b) Because −4 < 0, ' 9
−4 (
= 0.
c) Because 4 > 0, '−9
4 (
= (−9) · (−10) · (−11) · (−12)
4! = 9 · 10 · 11 · 12
24 = 495.
d) Because −4 < 0, '
−9
−4 (
= 0.
e) Because 9 > 0, '4
9 (
= 4 · 3 · 2 · 1 · 0 · (−1) · (−2) · (−3) · (−4)
9! = 0
9! =0.
f) Because −9 < 0, ' 4
−9 (
= 0.
g) Because 9 > 0, '−4
9 (
= (−4) · (−5) · (−6) · (−7) · (−8) · (−9) · (−10) · (−11) · (−12) 9!
= −9 · 8 · 7 · 6 · 5 · 4 · 10 · 11 · 12
9! = −10 · 11 · 12
3 · 2 · 1 = −220.
h) Because −9 < 0, '
−4
−9 (
= 0.
i) Because 4 > 0, '1/2
4 (
= (1/2) · (−1/2) · (−3/2) · (−5/2)
4! = − 15
16 · 24 = − 5 128. j) Because −4 < 0,
'1/2
−4 (
= 0.
k) Because 4 > 0, '−1/2
4 (
= (−1/2) · (−3/2) · (−5/2) · (−7/2)
4! = 1 · 3 · 5 · 7
16 · 24 = 35 128.
l) Because −4 < 0, '
−1/2
−4 (
= 0.
5.153 Let Y denote the number of tests required under the alternative scheme. Also, let X denote the number of people in the sample who have the disease. We know that X ∼ H(N, n, p), where N is the
number of people in the population. As N is very large, we can use the binomial approximation to the hypergeometric distribution. Hence, X ≈ B(n, p). We have
Y =
-1, if X = 0;
n+ 1, if X ≥ 1.
Now,
pY(1) = P (Y = 1) = P (X = 0) ≈ 'n
0 (
p0(1 − p)n−0= (1 − p)n and, by the complementation rule,
pY(n+ 1) = 1 − pY(0) = 1 − (1 − p)n. Hence,
pY(y)=
"(1 − p)n, if y = 1;
1 − (1 − p)n, if y = n + 1;
0, otherwise.
5.154
a) The number of patients, X, that arrive within the first hour of the doctor’s arrival has, by assumption, the Poisson distribution with parameter λ = 6.9. The first patient arrives more than 1 hour after the doctor arrives if and only if no patients arrive within the first hour of the doctor’s arrival, which means that X = 0. Hence, the required probability is
P (X= 0) = e−6.9(6.9)0
0! = e−6.9= 0.00101.
b) According to Proposition 5.8 on page 225, the most probable number of patients that arrive within the first hour of the doctor’s arrival is ⌊6.9⌋ = 6. From the complementation rule and the FPF,
P (|X − 6| > 1) = 1 − P (|X − 6| ≤ 1) = 1 − /
|x−6|≤1
pX(x)= 1 − /7 k=5
e−6.9 (6.9)k
k! = 0.569.
c) From the complementation rule and the FPF,
P (X ≥ 5) = 1 − P (X < 5) = 1 −/
x<5
pX(x)= 1 − /4 k=0
e−6.9 (6.9)k
k! = 0.818.
d) Let Y denote the number of patients observed by the inspector during a 1-hour period. Also, let L= event the inspector inspects the large hospital, and
S= event the inspector inspects the small hospital.
As the inspector tosses a balanced coin to decide which hospital to inspect, we have P (L) = P (S) = 1/2.
Furthermore, given that event L occurs, Y ∼ P(6.9), whereas, given that event S occurs, Y ∼ P(2.6).
Therefore, by the law of total probability,
P (Y = 3) = P (L)P (Y = 3 | L) + P (S)P (Y = 3 | S) = 1
2 ·e−6.9(6.9)3 3! +1
2 ·e−2.6(2.6)3 3!
= 0.136.
e) From Bayes’s rule,
P (S| Y = 3) = P (S)P (Y = 3 | S)
P (L)P (Y = 3 | L) + P (S)P (Y = 3 | S) =
1
2 ·e−2.6(2.6)3 3!
1
2 ·e−6.9(6.9)3 3! +1
2 ·e−2.6(2.6)3 3!
= 1
e−4.3(6.9/2.6)3+ 1 = 0.798.
5.155
a) By the assumptions, we see that X ∼ B(2500, 0.001). Hence, pX(x)='2500
x (
(0.001)x(0.999)2500−x, x= 0, 1, . . . , 2500, and pX(x)= 0 otherwise.
b) Noting that 2500 · 0.001 = 2.5, we see, from Proposition 5.7 on page 220, that pX(x)≈ e−2.5(2.5)x
x! , x= 0, 1, . . . , 2500.
c) From the complementation rule, the FPF, and part (a),
P (X ≥ 5) = 1 − P (X < 5) = 1 −/
x<5
pX(x)= 1 − /4 k=0
'2500 k
(
(0.001)k(0.999)2500−k = 0.1087.
And, from the complementation rule, the FPF, and part (b), P (X ≥ 5) = 1 − P (X < 5) = 1 −/
x<5
pX(x)≈ 1 − /4 k=0
e−2.5(2.5)k
k! = 0.1088.
The Poisson approximation to the binomial probability is excellent here.
d) Answers will vary. We might, however, be justified in claiming that typographical errors are not independent. For instance, if a typographical error is the result of the typist misplacing his fingers on the keyboard, then subsequent characters are likely to be in error.
5.156 We have
pX(x)= e−1.75 (1.75)x
x! , x = 0, 1, . . . , and pX(x)= 0 otherwise.
a) We have
P (X = 2) = pX(2) = e−1.75 (1.75)2
2! = 0.266.
b) From the FPF,
P (4 ≤ X ≤ 6) = /
4≤x≤6
pX(x)= /6 k=4
e−1.75 (1.75)k
k! = e−1.75 /6 k=4
(1.75)k
k! = 0.0986.
c) From the complementation rule,
P (X ≥ 1) = 1 − P (X = 0) = 1 − e−1.75(1.75)0
0! = 0.826.
d) Applying the formula for the PMF of the P(1.75) distribution, presented at the beginning of the solution to this exercise, we get the following table:
x 0 1 2 3 4 5 6 7
pX(x ) 0.174 0.304 0.266 0.155 0.068 0.024 0.007 0.002
e) A partial probability histogram for the random variable X is as follows. Note that, in the graph, we have used p(x) in place of pX(x).
f) From part (e), the probability histogram of X is right skewed. This property holds for all Poisson random variables, as we learned in Proposition 5.8 on page 225.
5.157
a) We first recall that the number of possible five-card draw poker hands is%52
5
&
. A hand of four of a kind is of the form {x, x, x, x, y}, where x and y are distinct denominations. There are%13
1
&
choices for x and then%12
1
&
choices for y. Then there are%4
4
&
ways to choose the four xs from the four xs, and there are%4
1
&
ways to choose the one y from the four ys. Hence, the probability of getting four of a kind is '13
1 ('12
1 ('4
4 ('4
1 ( '52
5
( = 624
2,598,960 =0.00024.
b) Let X denote the number of times that we get four of a kind in 10,000 hands of five-card draw.
Then X ∼ B(10,000, 0.00024). Noting that 10,000 · 0.00024 = 2.4, the Poisson approximation to this binomial distribution yields
pX(x)≈ e−2.4 (2.4)x
x! , x= 0, 1, . . . , 10,000.
Hence,
P (X = 2) = pX(2) ≈ e−2.4 (2.4)2
2! = 0.261.
Also, from the complementation rule and the FPF, P (X ≥ 2) = 1 − P (X < 2) = 1 −/
x<2
pX(x)= 1 − pX(0) − pX(1)
≈ 1 − e−2.4 (2.4)0
0! − e−2.4(2.4)1
1! = 1 − 3.4e−2.4= 0.692.
5.158 Suppose that X ∼ B(n, p), where n is large and p is large (i.e., close to 1). The random vari-able Y = n − X gives the number of failures in the n Bernoulli trials and, hence, by interchanging the roles of success and failure, we see that Y ∼ B(n, 1 − p). As n is large and 1 − p is small, we can apply Proposition 5.7 on page 220 to conclude that
pY(y)≈ e−n(1−p)
%n(1 − p)&y
y! , y = 0, 1, . . . , n.
Consequently, from Equation (5.54) on page 248,
pX(x)= pn−Y(x)= pY(n− x) ≈ e−n(1−p)
%n(1 − p)&n−x
(n− x)! , x = 0, 1, . . . , n.
Thus, in the sense of the preceding display, we can use the Poisson distribution with parameter n(1 − p) to approximate probabilities for X.
5.159
a) Let X denote weekly demand to the nearest gallon. Then X ∼ P(λ). Letting Y denote weekly gas sales, we see that Y = min{X, m}. The range of Y is {0, 1, . . . , m}. For y = 0, 1, . . . , m − 1,
pY(y)= P (Y = y) = P%
min{X, m} = y&
= P (X = y) = pX(y)= e−λλy y!. Also, from the complementation rule and the FPF,
pY(m)= P (Y = m) = P%
min{X, m} = m&
= P (X ≥ m) = 1 − P (X < m)
= 1 −/
x<m
pX(x)= 1 −
m/−1 k=0
e−λλk
k! =1 − e−λ
m/−1 k=0
λk k!. Hence,
pY(y)=
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ e−λ λy
y!, if y = 0, 1, . . . , m − 1;
1 − e−λ
m/−1 k=0
λk
k!, if y = m;
0, otherwise.
Note: We could also apply Proposition 5.14 on page 247 to solve this problem.
b) The gas station runs out of gas before the end of the week if and only if the demand equals or exceeds m, which is
P (X ≥ m) = /
x≥m
pX(x)= /∞ k=m
e−λλk k! =e−λ
/∞ k=m
λk k!.
5.160 For convenience, set p = e−6.9(6.9)7/7!. For each n ∈ N , let Endenote the event that exactly seven patients arrive during the nth hour after 12:00P.M.. By assumption, E1, E2, . . . are independent events and, moreover, P (En)= p for all n ∈ N . Referring to Proposition 4.5 on page 152 and applying the complementation rule, we conclude that
P (X = k) = P%
E1c∩ · · · ∩ Ekc−1∩ Ek&
= P% E1c&
· · · P% Ekc−1&
P (Ek)= p(1 − p)k−1, k∈ N . Therefore, X ∼ G(p).
a) We have
P (X = 4) = p(1 − p)3 = 0.0918.
b) From Proposition 5.10 on page 231,
P (X >6) = (1 − p)6= 0.380.
c) From the lack-of-memory property of geometric random variables, Exercise 5.106, and part (b), we deduce that
P (X >15 | X > 9) = P (X > 9 + 6 | X > 9) = P (X > 6) = 0.380.
5.161
a) Referring to Proposition 5.10 on page 231, we get
P (X >6) = (1 − p)6 ='6 0 (
p0(1 − p)6 = P (Y = 0).
b) Consider Bernoulli trials with success probability p. Let X denote the number of trials up to and including the first success, and let Y denote the number of successes in the first six trials. Then X ∼ G(p) and Y ∼ B(6, p). Now, X > 6 if and only if the first success occurs after the sixth trial, which happens if and only if no successes occur during the first six trials, which means that Y = 0. Consequently, we have {X > 6} = {Y = 0} and, hence, in particular, P (X > 6) = P (Y = 0).
5.162
a) From Proposition 5.13 on page 241, we know that X ∼ N B(3, p) and that pX(x)=
'x− 1 2
(
p3(1 − p)x−3, x = 3, 4, . . . , and pX(x)= 0 otherwise.
b) From the FPF,
P (X≤ 40) = /
x≤40
pX(x)= /40 k=3
pX(k).
Hence, we see that 38 terms of the PMF of X must be added to obtain P (X ≤ 40). We have P (X≤ 40) =
/40 k=3
'k− 1 2
(
p3(1 − p)k−3.
c) Let Y denote the number of successes in the first 40 trials. Then, we have Y ∼ B(40, p). We note that {X > 40} = {Y < 3}. Hence, from the complementation rule and the FPF,
P (X ≤ 40) = 1 − P (X > 40) = 1 − P (Y < 3) = 1 −/
y<3
pY(y)= 1 − /2 k=0
pY(k).
Using the preceding display and the PMF of a B(40, p) random variable, we get P (X≤ 40) = 1 −
/2 k=0
'40 k
(
pk(1 − p)40−k.
d) To evaluate P (X ≤ 40), it is far easier to use the method of part (c) than the direct method of part (b).
5.163 Consider a “success” a year in which at least one serious accident occurs. Let X denote the number of years until the third success (i.e., until the third year in which at least one serious accident occurs).
From the assumptions, we know that X ∼ N B(3, 0.4). The problem is to determine the probability that there will be at least 3 years in which no serious accidents occur before the third year in which at least one serious accident occurs, that is, the probability of at least three failures before the third success.
Equivalently, we want P (X ≥ 6). From the complementation rule and the FPF, P (X ≥ 6) = 1 − P (X < 6) = 1 −/
x<6
pX(x)= 1 − /5 k=3
'k− 1 2
(
(0.4)3(0.6)k−3 = 0.683.
Advanced Exercises
5.164 Let Xj, 1 ≤ j ≤ n, denote the number of the jth member sampled. Because the sampling is with replacement, X1, . . . , Xn are independent random variables. And, because the sampling is random, each Xj has the discrete uniform distribution on the set S = {1, 2, . . . , N}.
a) Let X = max{Xj : 1 ≤ j ≤ n}. From the FPF, for k ∈ S,
P (Xj ≤ k) =/
x≤k
pXj(x)= /k i=1
1 N = k
N. Hence,
P (X ≤ k) = P%
max{Xj : 1 ≤ j ≤ n} ≤ k&
= P (X1 ≤ k, . . . , Xn≤ k)
= P (X1 ≤ k) · · · P (Xn≤ k) = 'k
N (n
. Consequently,
P (X= k) = P (X ≤ k) − P (X ≤ k − 1) = 'k
N (n
−
'k− 1 N
(n
= kn− (k − 1)n
Nn .
Thus,
pX(x)= xn− (x − 1)n
Nn , x = 1, 2, . . . , N, and pX(x)= 0 otherwise.
b) Let Y = min{Xj : 1 ≤ j ≤ n}. From the FPF, for k ∈ S, P (Xj > k)=/
x>k
pXj(x)= /N i=k+1
1
N = N − k
N .
Hence,
P (Y > k)= P%
min{Xj : 1 ≤ j ≤ n} > k&
= P (X1 > k, . . . , Xn> k)
= P (X1> k)· · · P (Xn> k)=
'N − k N
(n
.
Consequently,
a) Let X denote the largest numbered member sampled. Note that event {X = k} occurs if and only if exactly n − 1 of the members sampled have numbers at most k − 1 and one member sampled has number k. Hence,
b) Let Y denote the smallest numbered member sampled. Note that event {Y = k} occurs if and only if exactly n − 1 of the members sampled have numbers at least k + 1 and one member sampled has number k. Hence,
a) For z = 0, 1, . . . , n + m, we have where the last equality follows from Vandermonde’s identity. Thus, X + Y ∼ B(n + m, p).
b) For x = 0, 1, . . . , z, we have, in view of the conditional probability rule, the independence condition, and part (a), that
P (X= x | X + Y = z) = P (X= x, X + Y = z)
5.167 If the sample is drawn with replacement, then the number of nondefective items of the six sampled has the B(6, 2/3) distribution. Hence, its PMF is
'6 k (
(2/3)k(1/3)6−k, k= 0, 1, . . . 6.
However, if the sample is drawn without replacement., then the number of nondefective items of the six sampled has the H(12, 6, 2/3) distribution. Hence, its PMF is
'8
The following table provides both PMFs.
k Binomial Hypergeometric 0 0.00137 0.00000 1 0.01646 0.00000 2 0.08230 0.03030 3 0.21948 0.24242 4 0.32922 0.45455 5 0.26337 0.24242 6 0.08779 0.03030
The engineer will probably estimate the number of nondefectives in the lot of 12 to be roughly twice the number of nondefectives in the sample of six:
M≈ 12 · k 6 =2k.
From the table, we see that it’s more likely for the number of nondefectives in the sample to be close to four—that is, for the engineer’s estimate to be close to eight—if the sampling is without replacement (hypergeometric). For instance, the probability that the number of nondefectives in the sample will be within one of four is 0.812 for the binomial and 0.939 for the hypergeometric. Hence, a better estimate of the number of nondefectives in the lot is obtained by sampling without replacement.
5.168
a) We would expect the binomial approximation to be worse when n = 50 than when n = 8 because, in the former case, the sample size is larger relative to the size of the population than in the latter case. In fact, here, when n = 50, the rule of thumb for use of a binomial distribution to approximate a hypergeometric distribution is just barely met—the sample size of 50 is exactly 5% of the population size of 1000.
b) We have the following table, where the probabilities are rounded to eight decimal places and where we displayed probabilities until they are zero to that number of decimal places:
Successes x
Hypergeometric probability
Binomial probability
Successes x
Hypergeometric probability
Binomial probability 0 0.00001038 0.00001427 15 0.02854165 0.02991866 1 0.00013826 0.00017841 16 0.01507892 0.01636177 2 0.00089640 0.00109274 17 0.00723473 0.00818088 3 0.00377132 0.00437095 18 0.00316049 0.00374957 4 0.01157781 0.01283965 19 0.00125978 0.00157877 5 0.02765180 0.02953120 20 0.00045900 0.00061177 6 0.05349306 0.05537101 21 0.00015309 0.00021849 7 0.08617037 0.08701158 22 0.00004679 0.00007200 8 0.11793007 0.11692182 23 0.00001312 0.00002191 9 0.13921652 0.13640879 24 0.00000337 0.00000616 10 0.14344797 0.13981901 25 0.00000080 0.00000160 11 0.13023588 0.12710819 26 0.00000017 0.00000039 12 0.10498345 0.10327540 27 0.00000003 0.00000009 13 0.07561264 0.07547049 28 0.00000001 0.00000002 14 0.04891210 0.04986443
c) As we see by comparing the table in part (b) to Table 5.15, the accuracy of the binomial approximation to the hypergeometric is superior when n = 8 than when n = 50—just as expected.
5.169
a) The likelihood ratio for M based on an observed value k of X is
Lk(M)
where the last equality is obtained by subtracting and adding (M − k)(N − M + 1) from the numerator of the penultimate expression, and then simplifying algebraically. It follow that Lk(M)takes its maximum value at M =B
(N+ 1)k/nC
. Hence, the maximum likelihood estimate of M is given by
MD= estimate of the student’s vocabulary is
MD = E
(80,000 + 1) · 38 100
F
= ⌊30,400.38⌋ = 30,400.
5.170
a) The random variable X has PMF given by
pX(x)= p(1 − p)x, x = 0, 1, 2, . . . ,
and pX(x)= 0 otherwise. Given that X = n, the random variable Y has the binomial distribution with parameters n and p. Let k be a nonnegative integer. Applying the law of total probability yields
P (Y = k) =/
The terms in the last sum are from the PDF of a negative binomial random variable with parameters r and p0, summed over all possible values of such a random variable. Hence, the sum is 1. Thus,
P (Y = k) =
' pr
(1 − p)r−1
( '(1 − p0)r−1 p0r
(
= 'p
p0
( 'p(1 − p0) p0(1 − p)
(r−1
= ' p
p0
( 'p(1 − p0) p0(1 − p)
(k
. Simple algebra shows that
p
p0 = 1
2 − p and p(1 − p0)
p0(1 − p) = 1 − 1 2 − p. Consequently,
pY(y)=' 1 2 − p
( '
1 − 1 2 − p
(y
, y = 0, 1, 2, . . . , and pY(y)= 0 otherwise.
b) Let Z = Y + 1. We note that the function g(y) = y + 1 is one-to-one. Solving for y in the equa-tion z = y + 1, we see that g−1(z)= z − 1. Applying Equation (5.54) on page 248 and the result of part (a) yields, for z ∈ N ,
pZ(z)= pg(Y )(z)= pY%
g−1(z)&
= pY(z− 1) =' 1 2 − p
( '
1 − 1 2 − p
(z−1
. Thus, Y + 1 ∼ G%
1/(2 − p)&
. 5.171
a) We consider a “success” to be a defective item. From independence, we know that repeated observa-tions of success (defective) or failure (nondefective) constitute Bernoulli trials with success probability p.
Hence, the time (item number) of the kth success (defective item) has the negative binomial distribution with parameters k and p. Consequently, the probability that the nth item is the kth defective one is
'n− 1 k− 1 (
pk(1 − p)n−k, n= k, k + 1, . . . .
b) Here we consider a “success” to be a rejected item (i.e., defective and inspected). From independence, we know that repeated observations of success or failure constitute Bernoulli trials with success proba-bility pq. Hence, the time (item number) of the kth success (rejected item) has the negative binomial distribution with parameters k and pq. Consequently, the probability that the nth item is the kth rejected
one is '
n− 1 k− 1 (
(pq)k(1 − pq)n−k, n= k, k + 1, . . . . c) We give two solutions.
Method 1: Let Z denote the number of items until the first rejection. From the law of partitions, we have, for each nonnegative integer k, that
P (W = k) =/
n
P (W = k, Z = n) = /∞ n=k+1
P (W = k, Z = n).
Event {W = k, Z = n} occurs if and only if among the first n − 1 items exactly k are defective, none of these k defective items are inspected, and the nth item is inspected and defective. The probability of this happening is
P (W = k, Z = n) =
'n− 1 k
(
pk(1 − p)n−1−k· (1 − q)k· pq
= q(1 − q)k 'n− 1
k (
pk+1(1 − p)n−1−k. Therefore, letting r = k + 1, we get
P (W = k) = q(1 − q)k /∞ n=k+1
'n− 1 k
(
pk+1(1 − p)n−1−k
= q(1 − q)k /∞ n=r
'n− 1 r − 1 (
pr(1 − p)n−r = q(1 − q)k,
where, in the last equation, we used the fact that the values of the PMF of a negative binomial random variable with parameters r and p must sum to 1. Consequently,
pW(w)= q(1 − q)w, w= 0, 1, 2, . . . , and pW(w)= 0 otherwise.
Method 2: Let k be a nonnegative integer. The number of defective items until the first rejection equals k if and only if the first k defective items are not inspected and the next defective item is inspected, which has probability (1 − q)kq. Consequently,
pW(w)= q(1 − q)w, w= 0, 1, 2, . . . , and pW(w)= 0 otherwise.
d) Let Z = W + 1. We note that the function g(w) = w + 1 is one-to-one. Solving for w in the equation z = w + 1, we see that g−1(z)= z − 1. Applying Equation (5.54) on page 248 and the result of part (c) yields, for z ∈ N ,
pZ(z)= pg(W )(z)= pW%
g−1(z)&
= pW(z− 1) = q(1 − q)z−1.
Thus, W + 1 ∼ G(q) and represent the number of defective items up to and including the first rejection.
e) In the preceding (first) scenario, a defective item is rejected if and only if it is inspected, which has probability q. Here, in the second scenario, a defective item is rejected if and only if it is inspected and detected as defective, which has probability qr. Consequently, in the second scenario, the answers to parts (b)–(d) are obtained from those in the first scenario by replacing q by qr.
5.172 Let X denote the number of 6s in four throws of one balanced die and let Y denote the number of pairs of 6s in 24 throws of two balanced dice. Then X ∼ B(4, 1/6) and Y ∼ B(24, 1/36).
a) From the complementation rule,
P (X ≥ 1) = 1 − P (X = 0) = 1 −'4 0 (
(1/6)0(5/6)4−0= 1 − (5/6)4= 0.518 and
P (Y ≥ 1) = 1 − P (Y = 0) = 1 −'24 0
(
(1/36)0(35/36)24−0= 1 − (35/36)24= 0.491.
Hence, the gambler is not right. The chance of getting at least one 6 when one balanced die is thrown four times exceeds that of getting at least one pair of 6s when two balanced dice are thrown 24 times.
b) From the complementation rule,
P (X≥ 2) = 1 − P (X < 2) = 1 − /1 k=0
'4 k (
(1/6)k(5/6)4−k = 0.132 and
P (Y ≥ 2) = 1 − P (Y < 2) = 1 − /1 k=0
'24 k
(
(1/36)k(35/36)24−k = 0.143.
5.173 Let C and D denote the events that the coin and die experiments are chosen, respectively. Given that D occurs, X ∼ G(1/6), whereas, given that C occurs, X ∼ G(1/2). Also, P (C) = P (D) = 1/2.
a) From the lack-of-memory property of the geometric distribution,
P (X = 8 | {X > 5} ∩ D) = (1/6)(5/6)3−1 = 25 216. b) From the lack-of-memory property of the geometric distribution,
P (X = 8 | {X > 5} ∩ C) = (1/2)(1/2)3−1 = 1 8. c) From Bayes’s rule
P (D| X > 5) = P (D)P (X >5 | D)
P (C)P (X >5 | C) + P (D)P (X > 5 | D)
= (1/2)P (X > 5 | D)
(1/2)P (X > 5 | C) + (1/2)P (X > 5 | D) = 1
1 + P (X >5 | C) P (X >5 | D) .
Now, by Proposition 5.10 on page 231,
P (X >5 | D) = (5/6)5 and P (X >5 | C) = (1/2)5. Hence,
P (D| X > 5) = 1 1 +(1/2)5
(5/6)5
= 3125 3368. d) From the conditional form of the complementation rule and part (c),
P (C| X > 5) = 1 − P (D | X > 5) = 1 −3125 3368 =
243 3368.
e) Applying the conditional form of the law of total probability, Exercise 4.20, and the results of parts (a)–(d), we get
P (X= 8 | X > 5)
= P{X>5}(X= 8) = P{X>5}(C)P{X>5}(X= 8 | C) + P{X>5}(D)P{X>5}(X= 8 | D)
= P (C | X > 5)P (X = 8 | {X > 5} ∩ C) + P (D | X > 5)P (X = 8 | {X > 5} ∩ D)
= 243 3368 ·
1 8 +
3125 3368 ·
25
216 =0.116.
f) From the law of total probability,
Therefore, X does not have the lack-of-memory property.
h) If five failures have occurred (i.e., the first five trials resulted in failure), then it is more likely that the die experiment was performed than the coin experiment, because the success probability in the former is much smaller than that in the latter (1/6 vs. 1/2). Therefore, you should feel less sure of a success on the sixth trial, given these five failures, than you were of a success on the first trial, when both experiments were equally likely to be performed.
i) Knowing that the die experiment has been chosen, X ∼ G(1/6). In particular, then, X has the
i) Knowing that the die experiment has been chosen, X ∼ G(1/6). In particular, then, X has the