Math 40603: Solutions to Spring 2013 Final Exam
1. D
The probability a student is not taking mathematics and is taking English is
(1 0.6)0.75 = 0.3. Thus, 0.4 0.3 = 0.1 are taking neither. If x is the
percentage taking mathematics and not taking English,
Math 40603: Solutions to Exam #2
1. A
We subtract the complementary probability of at most one accident
from 1.
1 0.93 0.952 3 0.1 0.92 0.952 2 0.93 0.05 0.95 0.0535.
2. B
6(x + y + 1)
3
2
4
2(x + y + 1)3
dy dx =
2(x + 3)3 dx = (x + 3)2
=
dx
2
3
= 1/36.
3
Math 40603: Solutions to Exam #3
1. C
P (3X Y ) =
=
0
0
3x
1 x/5 1 y/10
e
e
dy dx
5
10
1
ex/5 ey/10
5
dx =
3x
0
1 x/2
2
e
dx =
5
5
2. D
600 500
5000
= P (z > 1.4142 . . .) 1 0.9192 = 0.0808
P (X > 600) = P
z>
3. D
E [T ] = 8,
1
1 T /8
T 2 eT /8 dT
dT 8
Math 40603: Solutions to Exam #1 Spring 2011
1. D
Let x be the percentage who drink and smoke. Then x/20 = 1.5(60x)/80, hence
x = 180/11. The ratio of the probability a drinker smoke to that a non-drinker
smokes is
x/60
= 3.
(20 x)/40
2. C
A magnitude 5.0
Math 40603
Exam 2 Spring 2011
Name
1. The times T1 and T2 until a rms two key executives retire are random variables with joint
108
distribution estimated by f (t1 , t2 ) =
, t1 > 0, t2 > 0. Find the probability that
(t1 + 2t2 + 3)4
the maximum of T1 and
Math 40603
Exam 3 Spring 2011
Name
24
, where t > 0 is
(t + 2)4
measured in months. What is the least number of machines that must be on hand to be sure
the probability of at least one working after 4 months is 99% or better?
1. The time to failure for a
Math 40603: Solutions to Exam #3 Spring 2011
1. E
The probability of one given machine working after 4 months is
4
24
8
dt =
(t + 2)4
(t + 2)3
=
4
1
.
27
The probability of at least one out of n working after 4 months is
n
n
ln 0.01
26
26
= 122.02 . . .
Math 40603: Solutions to Final Exam Spring 2011
1. D
P (Not Med and Not Lab) = 1 P (Not Med or Not Lab)
= 1 P (Med) P (Lab) + P (Med and Lab),
0.62 = 1 0.22 0.33 + P (Med and Lab).
P (Med and Lab) = 0.17
2. C
Only car Not only car
= 0.25
= 0.15 0.4 (ii)
=
Math 40603
Exam 1
Spring 2013
1. The probability a random client of a young insurance company has held a policy for n years
is proportional to 5 n, n = 1, 2, 3, 4, and is 0 otherwise. What is the probability a random
client has held a policy at least thre
Math 40603: Solutions to Exam #1 Spring 2011
1. B
P (n 3) =
3
(5 3) + (5 4)
=
(5 1) + (5 2) + (5 3) + (5 4) 10
2. C
For an exponentially distributed time with mean , the probability T 15 is
1 t/
e
dt = e15/ . Thus, the overall probability is
15
0.6 e0.5
Math 40603
Exam 2
Spring 2013
1. The annual number of losses M and N for two types of losses have joint probability
f (m, n) =
10 m 2n
,
90
m = 0, 1, 2, . . . ,
n = 0, 1, 2, . . . ,
m + n 4.
Find the probability that N = 3, given that M = 1.
A. 1/30
B. 1/
Math 40603: Solutions to Exam #2 Spring 2013
1. D
The probability is
f (1, 3)
. Dropping the denominators of 90 that
f (1, 0) + + f (1, 3)
cancel, this equals
1
3
=
9+7+5+3 8
2. A
P (X > 1, Y > 1) =
1
=
(4y + 4)e2(x+1)(y+1)+2 dy dx
1
2(x+1)(y +1)+2
dx =
2
Math 40603
Final Exam
Spring 2013
1. In their rst semester at a college, 60% of freshmen are taking mathematics. Of those who
are not taking mathematics, 75% are taking English. Moreover, given that someone is not
taking English, the probability that the
Math 40603: Solutions to Exam #1
1. E
Comprehensive
No Comprehensive
One Vehicle
0.3 0.6 = 0.18 1 0.4 = 0.6
Two+ Vehicles 0.4 0.02 = 0.38 0.2 0.18 = 0.02
0.4
0.8
1 0.8 = 0.2
The percentage of those insuring two or more vehicles that have
comprehensive pro