3.18
(a)
Let E and T denote the number of earthquakes and tornadoes in one year, respectively.
They are both Poisson random variables with respective means
1
1 year = 0.1; T = 0.3
E = Et =
10 years
Also, the (yearly) probability of flooding, P(F) = 1/5 = 0.2, hence, due to statistical
independence among E, T, F
-0.4
P(good) = P(E = 0)P(T = 0)P(F’) = e-0.1 e-0.3 (1 – 0.2) = e
0.8 0.536
Note: alternatively, we can let D be the combined number of earthquakes or/and
tornadoes, with mean rate D = E + T = 0.1 + 0.3 = 0.4 (disasters per year), and compute
-0.4
P(D = 0)P(F’) = e
0.8 instead
p
0.536 (from (a)). Hence P(2 out of 5 years are good)
5 2
=
p (1 p) 3 0.287
2
(b)
In each year, P(good year)
(c)
Let’s work with D as defined in (a).
P(only one incidence of natural hazard) = P(D = 0)P(F) + P(D = 1)P(F’)
–0.4
–0.4
=e
0.2 + (e
0.4) (1 – 0.2)
0.349
3.23
= mean flaw rate = 1 (flaw) / 50m2;
t = area of a panel = 3m x 5m = 15m2
Let X be the number of flaws found on area t. X is Poisson distributed, i.e.
Given:
x
f(x) = e
x!
with
= t = 15 / 50 = 0.3 (flaws)
(a)
P(replacement) = 1 - P(0 or 1 flaw)
= 1 - f(0) - f(1)
= 1 - e-0.3 - 0.3 e-0.3
0.037
(b) Since the probability of replacement is 0.037 and there are 100 panels, we would expect
0.037 100 = 3.7 replacements on average,
which gives the expected replacement cost
3.7 $5000 $18500
(c) For the higher-grade glass,
calculate the new probability
= 1 (flaw) / 80m2
= 15/80 = 0.1875, with which we can
P(replacement) = 1 - e-0.1875 - 0.1875 e-0.1875
0.0155,
which gives rise to an expected replacement cost of
100 0.0155 $5100 $7920
(assuming each higher grade panel is also $100 more expensive to replace)
We can now compare the total costs:
Let C = initial cost of old type panels
Old type: Expected total cost = C + Expected replacement cost
= C + $18500
New type: Expected total cost = C + Extra initial cost + Expected replacement cost
= C + $100 100 + $7920
= C + $17920,
which is less than that of the old type, the higher grade panels are recommended.
3.27
(a)
Let X be the total number of excavations along the pipeline over the next year; X has a
Poisson distribution with mean = (1/50 miles)(100 miles) = 2, hence
P(at least two excavations)
= 1 – P(X = 0) – P(X = 1)
= 1 – e- (1 + ) = 1 – e-2(3)
0.594
(b) For each excavation that takes place, the pipeline has 0.4 probability of getting damaged, and
hence (1 – 0.4) = 0.6 probability of having no damage. Hence
P(any damage to pipeline | X = 2)
= 1 – P(no damage | X = 2)
= 1 – 0.62 = 1 – 0.36
= 0.64
Alternative method: Let Di denote “damage to pipeline in i-th excavation”; the desired
probability is
P(D1
D2) = P(D1) + P(D2) – P(D1D2)
= P(D1) + P(D2) – P(D1 | D2) P(D2)
= P(D1) + P(D2) – P(D1 ) P(D2)
= 0.4 + 0.4 – 0.42 = 0.8 – 0.16
= 0.64
(c) Any number (x) of excavations could take place, but there must be no damage no matter what
x value, hence we have the total probability
P(no damage | x excavations) P(x excavations)
x 0
x
0.6
=
x
x!
(0.6 ) x
0.6
=e e
x!
x 0
= ex 0
-0.4
e
-0.4(2)
=e
=e
0.449
-0.8
=e
Alternative method: recall the meaning of in a Poisson process—it is the mean rate, i.e. the
true proportion of occurrence over a large period of observation. Experimentally, it would be
determined by
nE
N
where nE is the number of excavations observed over a very large number (N) of miles of
pipeline. Since 40% of all excavations are damaging ones, damaging excavations must also
occur as a Poisson process, but with the “diluted” mean rate of
0.4n E
0.4 , hence
D
N
D = (0.4)(1/50) = (1/125) (damaging excavations per mile)
Hence
P(no damaging excavation over a 100 mile pipeline)
–(1/125 mi.)(100 mi.)
= e–100/125 = e–0.8
=e
0.449
3.51
The capacity C is gamma distributed with a mean of 2,500 tone and a c.o.v. of 35%
k/
(a)
(b)
= 2500,
1
k
= 0.35
Hence, k = 8.16, = 0.00327
P(C >3000 C > 1500)
P (C 3000) 1 I u (0.00327 3000, 8.16)
=
P (C 1500) 1 I u (0.00327 1500, 8.16)
1 I u (9.81, 8.16)
0.745
I u (4.905, 8.16)
0.889
P(C<2000) = Iu (0.00327x2000, 8.16) = 0.312
Mean rate of damaging quake = 0.312 x 1/20 = 0.0156
P(no damage over 50 years) = e-0.0156x50 = 0.458
(c)
P(at least 4 buildings damaged)
= 1 – P(none of the the 4 will be damaged)
= 1 – (0.688)4
= 0.776
0.838
3.52
Ti = time for loading/unloading operation of ith ship
T = waiting time of the merchant ship
(a)
Given that there are already 2 ships in the queue,
P(T>5) = P(T>5 T1 = 2)P(T1 = 2) + P(T>5 T1 = 3)P(T1 = 3)
= P(T2>3)P(T1 = 2) + P(T2>2)P(T1 = 2)
= 0x1/4 + 1/4x3/4 = 0.1875
(b)
P(T>5) = P(T>5 N=0)P(N=0) + P(T>5 N=1)P(N=1) + P(T>5 N=2)P(N=2) +
P(T>5 N=3)P(N=3)
Given 0 ship in the queue, P(T>5) = 0
Given 1 ship in the queue, P(T>5) = P(T1>5) = 0
Given 2 ships in the queue, P(T>5) = 0.1875 from above
Given 3 ships in the queue, the waiting time will be at least 6 days. Hence it is certain that
P(T>5) = 1
In summary,
P(T>5) = 0x0.1 + 0x0.3 + 0.1875x0.4 + 1x0.2 = 0.275
3.33
Let J1 and J2 denote the events that John’s scheduled connection time is 1 and 2 hours,
respectively, where P(J1) = 0.3 and P(J2) = 0.7. Also, let X be the delay time of the flight in hours.
Note that
P(X > x) = 1 – P(X
-x/0.5
x) = 1 – F(x) = 1 – (1 – e
-x/0.5
=e
)
(a) Let M denote the event that John misses his connection, i.e. the flight delay time exceeded his
scheduled time for connection. Using the theorem of total probability,
P(M) = P(M | J1)P(J1) + P(M | J2)P(J2)
= P(X > 1) 0.3 + P(X > 2) 0.7
= e– 1/0.5 0.3 + e– 2/0.5 0.7
= 0.135 0.3 + 0.0183 0.7 0.053
(b) Regardless of whether John has a connection time of 1 hour and Mike has 2, or the opposite,
for them to both miss their connections the flight must experience a delay of more than two
hours, and the probability of such an event is
P(X > 2) = e
– 2/0.5
0.018
(c) Since Mary has already waited for 30 minutes, the flight will have a delay time of at least 0.5
hours when it arrives. Hence the desired probability is
P(X > 1 | X > 0.5)
= P(X > 1 and X > 0.5) / P(X > 0.5)
= P(X > 1) / P(X > 0.5)
–1/0.5
–0.5/0.5
–2
–1
=e
/e
= e / e = 1/e 0.368
3.42
H = annual maximum wave height = N(4,3.2)
(
6 4
(a)
P(H>6) = 1
(b)
Design requirement is
P(no exceedance over 3 years) = 0.8 = (1-p)3
Where p = p(no exceedance in a given year)
3.2
) 1
(0.625)
0.266
Hence, p = 1-(0.8)1/3 = 0.0717
That is,
(c)
h 4
where h is the designed wave height
) 0.0717
3.2
h = 3.2 -1(0.0717) + 4
= 3.2x0.374 + 4 = 5.2 m
(
Probability of damaging wave height in a year
= 0.266x0.4 = 0.1064
Hence mean rate of damaging wave height = 0.1064 per year
P(no damage in 3 years)
= P(no damaging wave in 3 years)
= e-0.1064x3
=0.727
3.50
T = time until breakdown = gamma with mean of 40 days and standard deviation of 10
days
(a)
k/
= 40,
c.o.v. = 0.25 =
Hence, k = 16,
1
k
= 0.4
Let t be the required maintenance schedule interval
Hence, P(T<t)
0.95
or
Iu(0.4 t, 16)
0.95
I u (26,16)
I u (23.2,16)
1
By trial and error,
t
40
50
60
58
57
Hence, t
(b)
58 days
P(T < 58+7 T > 58)
=1
(c)
Probability
0.533
0.8435
0.9656
0.9520
0.944
P(T
P(T
65)
58)
1
I u (0.4 65,16)
I u (0.4 58,16)
1
P(at least one out of three will breakdown)
= 1 – P(none of the three will breakdown)
= 1 – (0.95)3
= 0.143
0.01417
0.048
0 .7
3.53
Traffic volume= V = Beta between 600 and 1100 vph with mean of 750 vph and c.o.v. of
0.20
A = accidenct
(a)
P(Jamming occurs on the bridge) = P(J)
= P(J A)P(A)+P(J A )P( A )
= 1x0.02 + P(V>1000)x0.98
For the parameters of Beta distribution for V,
750 = 600+
q
q
(0.2x750)2 =
Hence,
r
(q
(1100-600)
qr
r ) (q
2
r 1)
(1100-600)2
r 0.95,q 0.41
P(V>1000) = 1- u (0.41, 0.95)
But u = (1000-600)/(1100-600) = 0.8
P(V>1000) = 1- 0.8 (0.41, 0.95) = 0.097
P(J) = 0.02 + 0.097x0.98 = 0.971
3.56
Seismic capacity C is lognormal with median of 6.5 and standard deviation of 1.5
= ln 6.5
xm
1.5 / 6.5 0.23
For lognormal distribution,
/
hence
/ xm
and
(1.5 / ) 2
6.7; and
Since =ln 6.5 = ln
By trial and error,
(a)
P(Damage) = P(C<5.5) =
(b)
P(C>5.5 C>4) =
1
=
1
(c)
(
P(C
(
ln 5.5 ln 6.5
)
0.198
4 C 5.5)
P(C 4)
ln 5.5 ln 6.5
)
0.198
ln 4 ln 6.5
(
)
0.198
1.5 / 6.7 0.198
0.907
0.9929
( 0.84) 0.2
P(C 5.5)
P (C 4)
0.915
Mean rate of damaging earthquake = 0.2x1/500 = 0.0004
P(building survives 100 years)
= P(no damaging earthquake in 100 years)
= e-0.0004x100 = e-0.04 = 0.96
(d)
P(survival of at least 4 structures during the earthquake)
=
5
0.8
4
4
0.2
1
5
0.8
5
5
0.2
0
= 0.4096+0.3277 = 0.737
mean rate earthquake that causes damage to at most 3 structures
= 0.737x1/500 = 0.00147
P(at least four building will survive 100 years)
= e-0.00147x100 = e-0.147 = 0.863