2.42
Given:
P(H=1) = 0.2, P(H=2) = 0.05, P(H=0) = 0.75
Where H = number of hurricanes in a year
P(J=H=1) = P(D
⏐
H=1) = 0.99
P(J=H=2) = P(D
⏐
H=2) = 0.80
Where J and D denote survival of jacket and deck substructure respectively
Assume J and D are statistically independent
(a)
For the case of one hurricane, i.e. H=1
P(damage) = 1-P(JD
⏐
H=1)
= 1- P(J
⏐
H=1) P(D
⏐
H=1)
= 1 - 0.99x0.99
= 0.0199
(b)
For next year where the number of hurricanes is not known.
P(no damage) = P(JD
⏐
H=0)P(H=0)+ P(JD
⏐
H=1)P(H=1)+ P(JD
⏐
H=2)P(H=2)
= 1x0.75 + 0.99x0.99x0.2 + 0.8x0.8x0.05
= 0.75 + 0.196 + 0.032
= 0.978
(c)
P(H=0
⏐
JD) =
(
0)
(
0)
1
0.75
0.767
(
)
0.978
P JD H
P H
P JH
=
=
×
=
=

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2.54
(a)
T = wait time in queue (in min.)
N = number of trucks in queue
P(T<5
⏐
N=2) = P(loading time for both trucks ahead will take only 2 min. each)
= 0.5x0.5
= 0.25
(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)
Note that if the number of trucks in queue is 3 or more, the waiting time will definitely
exceed 5.
Hence those items will not contribute any probabilities.
Hence P(T<5) = 1x0.175 + 1x0.125 + 0.25x0.3 = 0.375