Homework2EnineeringRiskAnalysis
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School
University of California, Berkeley *
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Course
193
Subject
Industrial Engineering
Date
Apr 3, 2024
Type
Pages
5
Uploaded by JudgeThunder13189 on coursehero.com
Problem 1: A logistics company attempts to deliver ten packages to the Davis Hall, in which five are MacBook Air, three are MacBook Pro, and two are iPad. One package is lost during the transportation process, but which one is not clear. Now we randomly open (with no preference) two packages from the remaining nine. (a) Determine the probability that the opened two packages are both MacBook Air. Let A be the probability that opened two packages that are both MacBook Air. P(A) = !
"
×
#
$
×
%
&
+
#
"
×
’
$
×
%
&
=
’&
%##
=
&
"
Thus, the probability that the opened two packages are both MacBook Air is equal to &
"
. (b) Given that the opened two packages are both MacBook Air, what is the probability that the lost package is also a MacBook Air? Let B be the probability that lost package is a Macbook Air. We want to find ࠵?
(
(࠵?) =
)((∩,)
)(()
We know the formula: ࠵?
(
(࠵?) =
)((∩,)
)(()
We know P(A) =
&
"
Let’s find P
(࠵? ∩ ࠵?)
P
(࠵? ∩ ࠵?) =
!
%.
×
#
"
×
’
$
=
%
%&
So ࠵?
(
(࠵?) =
)((∩,)
)(()
=
!
!"
"
#
= ’
$
Problem 2
Cities C1 and C2 are located in the vicinity of an active fault, which is expected to generate an earthquake having a 30km rupture. Roads connecting the two cities include bridges B1 and B2 and tunnel T. Either of the bridges will fail if the respective shortest distance to the fault rupture is less than 20km, while the tunnel will fail if the rupture intersects it. Assuming the rupture is equally likely to occur anywhere within the fault, determine: (a) The probability of failure of each bridge and the tunnel •
Let B1 be the event that bridge B1 collapses. This can be translated as B1 < 20km from the fault rupture. Let's calculate the area from which B1 can be located within 20km of the fault rupture using the Pythagorean theorem. Area = √20
&
− 15
&
= 13,23 The area is 26,4 km long. Then we can calculate P(B1) P(B1) = %’,&’×&1’.
&..2’.
=
!3,!
%4.
= 0,33 •
Let B2 be the event that bridge B2 collapses. This can be translated as B2 < 20km from the fault rupture. Let's calculate the area from which B2 can be located within 20km of the fault rupture using the Pythagorean theorem. Area = √20
&
− 18
&
= 8,7 The area is 26,4 km long. Then we can calculate P(B2) P(B2) = $,4×&1’.
&..2’.
=
#4,#
%4.
= 0,28 •
Let T be the event that bridge T collapses. This can be translated as T intersect the fault rupture P(T) = ’.
%4.
=
= 0,18
(b) The probability that, after the earthquake, travel between the two cities will not be possible. Let C
be the event that travel between the two cites is impossible
.
P(C) = P(B1) + P(B2
∩ ࠵?)
•
We know that P(B1) = 0,33 •
Let’s calculate P(B1
∩ ࠵?)
P(B2
∩ ࠵?)
= %.1$,4
%4.
=
%$,4
%4.
= 0,11 So we finally get P(C) = 0,33 + 0,11 = 0,44 (c) If the tunnel is known to have survived, what is the probability that bridge B2 has survived? In this question we are looking for ࠵?
6
7
( ࠵?
3
2)
According to the formula we have: ࠵?
6
7
( ࠵?
3
)
= 8 (,&
7777
∩ 6
7)
) (6
7
)
Let’s first calculate P (࠵?2
3333
∩ ࠵?
3
):
P (࠵?2
3333
∩ ࠵?
3
)
= %2)(,&)2)(6) 1 8(9&∩6)
%2)(6)
=
%2.,&$2.,%$1.,%%
%2.,%$
=
.,3!
.,$’
So, ࠵?
6
7
( ࠵?
3
)
= 0,8 Problem 3 A stray rocket randomly falls in a 10km×15km rectangular region. Upon impact, it creates a circular crater of 0.1km radius. Every object within the crater is expected to be damaged. There are three assets within the rectangular region: a 1km×1km square residential area, an 8km long pipeline, and an industrial plant having a circular plan with 1km radius. These assets have the relative locations within the rectangular region shown in the figure below.
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