#1 HW Solutions (Fall 2011)
Assigned problems:
2.5, 2.10, 2.18, 2.19, 2.23
2.5 (Confirmed)
(a)
Locations Locations
Load at B Load at C
of W1
of W2
MA
Probability
E1
0
0
0
B
500
0
5,000
0.045
C
0
500
10,000
0.075
B
200
0
2,000
0.05
B
B
700
0
7,000
0.075
X
B
C
200
500
12,000
0.125
X
C
0
200
4,000
0.12
C
B
500
200
9,000
0.18
X
C
C
0
700
14,000
0.30
X
MA
Probability
E1
E3
0.150.2=0.03
E2
X
X
X
X
X
X
X
X
X
or, sorted by MA:
Locations Locations
Load at B Load at C
of W1
of W2
E2
E3
X
0
0
0
0.03
B
200
0
2,000
0.05
X
C
0
200
4,000
0.12
X
B
500
0
5,000
0.045
X
B
B
700
0
7,000
0.075
X
X
C
B
500
200
9,000
0.18
X
X
C
0
500
10,000
0.075
X
X
B
C
200
500
12,000
0.125
X
C
C
0
700
14,000
0.30
X
X
(b) E1 and E2 are not mutually exclusive because these two events can occur together, for example, when
the weight W2 is applied at C, MA is 10,000 ft-lb; hence both E1 and E2 will occur.
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(c) The probability of each possible value of MA is tabulated in the last column of the table above.
(d) P(E1) = P(MA > 5,000) = 0.075 + 0.075 + 0.125 + 0.18 + 0.30 = 0.755
P(E2) = P(1,000 MA 12,000) = 0.045 + 0.075 + 0.05 + 0.075 + 0.12 + 0.18 = 0.545
P(E3) = 0.05 + 0.075 = 0.125
P(E1 E2) = P(5,000 < MA 12,000) = 0.075 + 0.075 + 0.18 = 0.33
P(E1 E2) = 1 0.03 = 0.97
P( E 2 ) = 0.03 + 0.125 + 0.3 = 0.455
or
P( E 2 ) = 1 P(E2) = 1 0.545 = 0.455
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2.10
(a) The event both subcontractors will be available = AB, hence
since P(AB) = P(A) + P(B) - P(AB)
P(AB) = P(A) + P(B) - P(AB)
= 0.6 + 0.8 - 0.9 = 0.5
(b) P(B is availableA is not available) = P(B A )
=
P( BA )
P( A )
while it is clear from the following Venn diagram that P( BA ) = P(B) - P(AB).
A
AB
B
A
Hence
P( BA ) P( B) P( AB)
=
P( A )
1 P( A)
= (0.8 - 0.5)/(1 - 0.6)
= 0.3/0.4 = 0.75
(c)
(i) If A and are B s.i., we must have P(BA) = P(B) = 0.8. However, using Bayes rule,
P(BA) = P(AB)/P(A)
= 0.5/0.6 = 0.8333
So A and B are not s.i. (As being available boosts the chances that B will be available)
(ii) From (a), P(AB) is nonzero, hence AB , i.e. A and B are not m.e.
(iii) Given: P(AB) = 0.9 AB does not generate the whole sample space (otherwise the
probability would be 1), i.e. A and B are not collectively exhaustive.
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2.18
C = Collapse of superstructure
E = Excessive settlement
Given:
P(E) = 0.1;
P(C) = 0.05
P(CE) = 0.2
(a)
P(Failure) = P(C E)
= P(C) + P(E) P(CE)P(E)
= 0.05 + 0.1 0.2x0.1
= 0.13
(b)
P( ECE C)
=
P(( EC ) ( E C ))
P( E C )
= P(EC) / 0.13
= 0.2x0.1 / 0.13
= 0.154
(c)
P(E C E C) = P(E C ) + P( E C)
= P(EC) P(EC)
= 0.13 0.02
=0.11
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2.19
M = failure of master cylinder
W = failure of wheel cylinders
B = failure of brake pads
Given:
P(M) = 0.02; P(W) = 0.05; P(B) = 0.5
P(MW) = 0.01
B is statistically independent of M or W
(a)
P(W M B ) = P(W M )P( B )
= [P(W) P(MW)] [1 P(B)]
= (0.05 0.01)(1 0.5)
= 0.02
(b)
P(system failure) = 1 P(no component failure)
= 1- P( M W B )
= 1- P( M W ) P( B )
= 1- {1 - P(M W)}P( B )
= 1- {1-[P(M)+P(W)-P(MW)]}P( B )
= {1 - [0.02+0.05-0.01]}(0.5)
= 0.53
(c)
P(only one component failed) = P( W M B) + P( W M B ) + P(W M B )
P(W M B ) = 0.02 from part (a)
also P( W M B ) = P( W M) x P( B )
= [P(M)-P(MW)]P( B )
= (0.02 0.01)x0.5
= 0.005
also P( W M B) = P( W M ) x P( B ) = [1-P(M)-P(W)+P(MW)]P(B)
= (1-0.02-0.005+0.01)x0.5 = 0.47
Since the three events W M B, W M B and W M B are all within the event of system failure
P(only one component failuresystem failure) = (0.47+0.005+0.02) / 0.53 = 0.934
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2.23
P(Damage) = P(D) = 0.02
Assume damages between earth quakes are statistically independent
(a)
P(no damage in all three earthquakes)
= P3 ( D )
= 0.983
= 0.9412
(b)
P( D1 D2) = P( D1 )P(D2) = 0.98x0.02 = 0.0196
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