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(CIVL181)99m1_soln.pdf
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CIVL 281 Midterm 1 Solutions

Problem 1

Given probabilities:
P(D) = 0.001 (
);
P(S) = 0.02 (
);
P(W) = 0.01 (
);

= 0.998;
P(D | S) = P(D | W) = P(D);

(a) Let F denote the slope failure event. Then, non-failure is the event


Hence
P(F) = 1 C P(
)
= 1 C P(
)
= 1 C P(


D (and hence
) is s.i. of all other events (or their compliments),
= 1 C P(

= 1 C

= 1 C (0.999)(0.998)(0.98)
= 0.02293804
0.0229

(b) Expected damage cost =

= 50*0.001 + 20*0.02 + 10*0.01
= 0.55
(all costs are in million dollars)
(c)
(i) If option I did indeed work, the expected cost would be lowered to
50*0.001 + 20*0.02/2 + 10*0.01/2
= 0.3
but this is achieved only 80% of the time. For the other 20% of the time, the expected cost remains the same at 0.55. Hence option I results in an expected cost of
0.3*0.8 + 0.55*0.2
= 0.35,
i.e. option I reduces expected damages by the amount
(0.55 C 0.35) = 0.2 million dollars

On the other hand, option II always ensures P(S) = P(W) =0, hence the expected cost associated with this option is simply
50*0.001 = 0.05,
hence option II brings down the expected damages by
(0.55 C 0.05) = 0.5 million dollars
(ii) Its more worthwhile to spend $0.4M on something thats worth $0.5M (in terms of cost reduction), than paying $0.3M to buy something of $0.2M in value. So option II is better.

(d) Let
I denote device Indicates presence of weak soil that may cause a deep-seated slide
E denote there Exists weak soil that may cause a deep-seated slide

Given probabilities:


Initially,

Now, given the experimental fact
(i.e. device reveals no weak soil layer), Bayes rule gives us the updated probabilities
P(E) = P(E |
)
=

=

= 0.3*0.01 / (0.3*0.01 + 0.8*0.99)
= 0.003773585
0.00377
Hence
P(
= 1 - 0.003773585
= 0.99622642
0.996

Problem 2

Let
R = police sets up a Radar station
T = getting Ticketed by the police
S = Speeding

Given probabilities:
P(R) = 0.01
P(S) = 0.60
Also, when a radar is on,
P(T | S) = 0.95
P(T |
) = 0.02

al probability,


= P(T | S)P(S) + P(T | S)P(S)
= 0.950
= 0.578 But this is the probability relative to the reduced sample spa
0.578P(R)
= 0.5780.0





) = P(T | S) P(S) / P
= 0.020.40 / 0.578
erator an

ets per month
= Monthly traffic flo
= 0.00578100

Now, let
P(B | ) = 0.7;

P(B |R) = 0.05;
P(S | B = 0.05




= P(B | R)P(R) + P(B | R
= 0.70.01



P(S | B)P(B) + P(S |B)P(B)
= 0.050
P



0
.
(2)
Note: if you carried out the integration and solved (1) and (2), the answer would be
x0 = 0.3606737602t and c = 2.957427971 / t

Hence, when t = 25 (mm), we have f(x) = 0.1182971188 e-0.1109035489x
(b) P(X > 16) = 0.114

(c) P(X <5 | X < 15) = P(X < 5) / P(X < 15) 0.525