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(JA)P(A)+P(J A )P( A ) = 1x0.02 + P(V>1000)x0.98 For the parameters of Beta distribution for V,
q (1
3.10
(a) The only region where fX(x) is non-zero is between x = 0 and x = 20, where
fX(x) = FX(x) = - 0.005x + 0.1
which is a straight line segment decreasing from y = 0.1 (at x = 0) to y = 0 (at x = 20)
fX(x
)
x
0
xm
2
(b) The median divides the triangul
2.45
(a) Let A, D, I denote the respective events that a driver encountering the amber light will
accelerate, decelerate, or be indecisive. Let R denote the event that s/he will run the red light.
The given probabilities are
P(A) = 0.10, P(D) = 0.85, P(I)
2.46
A, B, C denote branch A, B, C will be profitable respectively
E denotes bonus received
P(A) = 0.7, P(B) = 0.7, P(C) = 0.6
P(AB) = 0.9 = P(BA)
C is statistically independent of A or B
(a)
P(Exactly two branches profitable)
= P(AB C )+P(A B C)+P( A BC)
3.12
x
(a) For 3 x 6, FX(x) =
24
t
3
dt = 4 / 3 12 / x 2 ; elsewhere FX is either 0 (for x < 3) or 1 (for x >
3
6)
FX is a parabola going from x = 3 to x = 6, followed by a horizontal line when plotted
6
(b) E(X) =
24
x( x
3
)dx = 24(1/3 1/6) = 4 (tons)
3.13
(a)
Since the return period, , is 200 years, this means the yearly probability of exceedance (i.e.
encountering waves exceeding the design value) is
p = 1 / = 1 / (200 years) = 0.005 (probability per year)
(b)
For each year, the probability of non-ex
2.48
P(E) = 0.001, P(L) = 0.002, P(T) = 0.0015
P(LE) = 0.1
T is statistically independent of E or L
(a)
P(ELT) = P(LE)P(E)P(T)
= 0.1x0.001x0.0015
= 1.5x10-6
(b)
P(ELT) = 1-P( E L T )
= 1-P( T )P( E L )
= 1 0.9985x[1-P(E)-P(L)+P(LE)P(E)]
= 1 0.0015[1-0.001
2.49
(a) Let E1, E2 denote the respective events of using 100 and 200 units, and S denote shortage of
material.
P(S) = P(S | E1)P(E1) + P(S | E2)P(E2)
= 0.10.6 + 0.30.4
= 0.06 + 0.12 = 0.18
(b) Using Bayes theorem with the result from part (a),
P(E1 | S)
3.16
(a) Let C be the number of microscopic cracks along a 20-feet beam. C has a Poisson
distribution with mean rate C = 1/40 (number per foot), and length of observation t =
20 feet, hence the parameter = (1/40)(20) = 0.5, thus
P(C = 2) = e-0.5 (0.52 / 2
2.50
Let H and S denote Hard and Soft ground, respectively, and let L denote a successful landing.
Given probabilities:
P(L | H) = 0.9; P(L | S) = 0.5;
P(H) = 3P(S) , but since ground is either hard or soft
P(H) = 0.75, P(S) = 0.25
(a) Using theorem of t
3.2
Let X be the profit (in $1000) from the construction job.
(a) P(lose money) = P(X < 0)
= Area under the PDF where x is negative
= 0.0210 = 0.2
(b) Given event is X > 0 (i.e. money was made), hence the conditional probability,
P (X > 4 0 | X > 0)
= P(
2.43
(a) Sample space = cfw_FA, HA, EA, FB, HB, EB
in which FA, FB denote fully loaded truck in lane A, B respectively
HA, HB denote half loaded truck in lane A, B respectively
EA, EB denote empty truck in lane A, B respectively
(b)
Assume the events of F
2.9
E1, E2, E3 denote events tractor no. 1, 2, 3 are in good condition respectively
(a) A = only tractor no. 1 is in good condition
= E1 E 2 E 3
B = exactly one tractor is in good condition
= E1 E 2 E 3 E1 E2 E 3 E1 E 2 E3
C = at least one tractor is in g
3.54
By using the binomial distribution,
P(2 out of 8 students will fail)
8
= (0.2 ) (0.8)
2
2
6
= 28x(0.2)2(0.8)6
= 0.294
By using the hyper geometric distribution, we need
number of passing students in a class of 30 = 24
number of failing students in a
2.37
(a) Let subscripts 1 and 2 denote after first earthquake and after second earthquake. Note
that (i) occurrence of H1 makes H2 a certain event; (ii) H, L and N are mutually exclusive
events and their union gives the whole sample space.
P(heavy damage
3.55 (a) T = trouble-free operational time follows a gamma distribution with mean of 35 days and c.o.v. of 0.25 k/ = 35,
1 k
= 0.25
Hence, k = 16, = 0.457 P(T>40) = 1-Iu(0.457x40, 16) = Iu(18.3, 16) = 0.736 (b) For the total of 50 road graders, number of
3.56
Seismic capacity C is lognormal with median of 6.5 and standard deviation of 1.5
= ln 6.5
For lognormal distribution, > x m
/ < / x m = 1.5 / 6.5 = 0.23
and
Since =ln 6.5 = ln (1.5 / ) 2
6.7; and 1.5 / 6.7 = 0.198
By trial and error,
hence
(a)
(b
2.39 (a) P(shipment accepted on a given day) = P(at most 1 defective panel) = 0.2 + 0.5 = 0.7 P(exactly one shipment will be rejected in 5 days) = 5xP(0.7)4(0.3) = 0.36 P(acceptance of shipment on a given day) = P(A) = P(AD=0)P(D=0)+ P(AD=1)P(D=1)+ P(AD=2
3.57 (a) Let A and B denote the pressure at nodes A and B, respectively. Since A is log-normal with mean = 10 and c.o.v. = 0.2 (small), we have A 0.2, and A = ln(A)
A2
2
ln(10) 0.02 = 2.282585093
P(satisfactory performance at node A) ln14 - 2.283 - ln
2.40
The possible scenario of the working condition and respective probabilities are as follows:
(a)
P(completion on schedule) = P(E)
= P(EG) P(G) + P(EN) P(N) + P(EBC) P(BC) + P(EB C ) P(B C )
= 1x0.2 + 0.9x0.4 + 0.8x0.4x0.5 + 0.2x0.4x0.5
= 0.76
(b)
P(No
3.58
(a) fX(x) is obtained by integrating out the independence on y,
1
y3
6
6
fX(x) =
( x + y 2 )dy = xy +
5
5
3 0
0
2
(0 < x < 1)
= (3x + 1)
5
1
(b) fY|X(y|x) =
f X ,Y ( x, y )
=
f X ( x)
(6 / 5)( x + y 2 )
x + y2
=3
(2 / 5)(3 x + 1)
3x + 1
1
f
Hence P(
2.41
Let L, N, H denote the event of low, normal and high demand respectively; also O and G
denote oil and gas supply is low respectively.
(a)
Given normal energy demand, probability of energy crisis
= P(EN)
= P(OG)
= P(O)+P(G)-P(OG) = P(O)+P(G)-P(GO)P(O)
3.59 (a) Summing over the last row, 2nd & 3rd columns of the given joint PMF table, we obtain P(X 2 and Y > 20) = 0.1 + 0.1 = 0.2 (b) Given that X = 2, the new, reduced sample space corresponds to only the second column, where probabilities sum to (0.15 +
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(DH=1) = 0.99
P(J=H=2) = P(DH=2) = 0.80
Where J and D denote survival of jacket and deck substructure respectively
Assume J and D are statistically