SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY
EXAM P PROBABILITY
EXAM P SAMPLE SOLUTIONS
Copyright 2005 by the Society of Actuaries and the Casualty Actuarial Society
Some of the questions in this study note are taken from past SOA/CAS examinations.
P-09-05
PRINTED IN U.S.A.
Page 1 of 54
1. Solution: D
Let
G ! event that a viewer watched gymnastics
B ! event that a viewer watched baseball
S ! event that a viewer watched soccer
Then we want to find
c
Pr $" G & B & S # % ! 1 ' Pr " G & B & S #
(
)
! 1 ' $( Pr " G # * Pr " B # * Pr " S # ' Pr " G + B # ' Pr " G + S # ' Pr " B + S # * Pr " G + B + S # %)
! 1 ' " 0.28 * 0.29 * 0.19 ' 0.14 ' 0.10 ' 0.12 * 0.08 # ! 1 ' 0.48 ! 0.52
-------------------------------------------------------------------------------------------------------2.
Solution: A
Let R = event of referral to a specialist
L = event of lab work
We want to find
P[R+L] = P[R] + P[L] – P[R&L] = P[R] + P[L] – 1 + P[~(R&L)]
= P[R] + P[L] – 1 + P[~R+~L] = 0.30 + 0.40 – 1 + 0.35 = 0.05 .
-------------------------------------------------------------------------------------------------------3.
Solution: D
First note
P , A & B - ! P , A- * P , B - ' P , A + B -
P , A & B '- ! P , A- * P , B '- ' P , A + B 'Then add these two equations to get
P , A & B - * P , A & B '- ! 2 P , A- * " P , B - * P , B '- # ' " P , A + B - * P , A + B '- #
0.7 * 0.9 ! 2 P , A- * 1 ' P $(" A + B # & " A + B '# %)
1.6 ! 2 P , A- * 1 ' P , AP , A- ! 0.6
Page 2 of 54
4.
Solution: A
For i ! 1, 2, let
Ri ! event that a red ball is drawn form urn i
Bi ! event that a blue ball is drawn from urn i .
Then if x is the number of blue balls in urn 2,
0.44 ! Pr[" R1 ! R2 # " " B1 ! B2 #] ! Pr[ R1 ! R2 ] * Pr , B1 ! B2 ! Pr , R1 - Pr , R2 - * Pr , B1 - Pr , B2 !
4 . 16 / 6 . x /
0
1* 0
1
10 2 x * 16 3 10 2 x * 16 3
Therefore,
32
3x
3x * 32
*
!
x * 16 x * 16 x * 16
2.2 x * 35.2 ! 3x * 32
0.8 x ! 3.2
x!4
2.2 !
-------------------------------------------------------------------------------------------------------5.
Solution: D
Let N(C) denote the number of policyholders in classification C . Then
N(Young + Female + Single) = N(Young + Female) – N(Young + Female + Married)
= N(Young) – N(Young + Male) – [N(Young + Married) – N(Young + Married +
Male)] = 3000 – 1320 – (1400 – 600) = 880 .
-------------------------------------------------------------------------------------------------------6.
Solution: B
Let
H = event that a death is due to heart disease
F = event that at least one parent suffered from heart disease
Then based on the medical records,
210 ' 102 108
!
P $( H + F c %) !
937
937
937 ' 312 625
P $( F c %) !
!
937
937
c
P $ H + F %) 108 625 108
!
!
! 0.173
and P $( H | F c %) ! (
937 937 625
P $( F c %)
Page 3 of 54
7.
Solution: D
Let
A ! event that a policyholder has an auto policy
H ! event that a policyholder has a homeowners policy
Then based on the information given,
Pr " A + H # ! 0.15
Pr " A + H c # ! Pr " A # ' Pr " A + H # ! 0.65 ' 0.15 ! 0.50
Pr " Ac + H # ! Pr " H # ' Pr " A + H # ! 0.50 ' 0.15 ! 0.35
and the portion of policyholders that will renew at least one policy is given by
0.4 Pr " A + H c # * 0.6 Pr " Ac + H # * 0.8 Pr " A + H #
! " 0.4 #" 0.5 # * " 0.6 #" 0.35 # * " 0.8 #" 0.15 # ! 0.53
" ! 53% #
-------------------------------------------------------------------------------------------------------100292
01B-9
8.
Solution: D
Let
C = event that patient visits a chiropractor
T = event that patient visits a physical therapist
We are given that
Pr ,C - ! Pr ,T - * 0.14
Pr " C ! T # ! 0.22
Pr " C c ! T c # ! 0.12
Therefore,
0.88 ! 1 ' Pr $(C c ! T c %) ! Pr ,C " T - ! Pr ,C - * Pr ,T - ' Pr ,C ! T ! Pr ,T - * 0.14 * Pr ,T - ' 0.22
! 2 Pr ,T - ' 0.08
or
Pr ,T - ! " 0.88 * 0.08 # 2 ! 0.48
Page 4 of 54
9.
Solution: B
Let
M ! event that customer insures more than one car
S ! event that customer insures a sports car
Then applying DeMorgan’s Law, we may compute the desired
probability as follows:
c
Pr " M c + S c # ! Pr $" M & S # % ! 1 ' Pr " M & S # ! 1 ' $( Pr " M # * Pr " S # ' Pr " M + S # %)
(
)
! 1 ' Pr " M # ' Pr " S # * Pr " S M # Pr " M # ! 1 ' 0.70 ' 0.20 * " 0.15 #" 0.70 # ! 0.205
-------------------------------------------------------------------------------------------------------10.
Solution: C
Consider the following events about a randomly selected auto insurance customer:
A = customer insures more than one car
B = customer insures a sports car
We want to find the probability of the complement of A intersecting the complement of B
(exactly one car, non-sports). But P ( Ac + Bc) = 1 – P (A & B)
And, by the Additive Law, P ( A & B ) = P ( A) + P ( B ) – P ( A + B ).
By the Multiplicative Law, P ( A + B ) = P ( B 4 A ) P (A) = 0.15 * 0.64 = 0.096
It follows that P ( A & B ) = 0.64 + 0.20 – 0.096 = 0.744 and P (Ac + Bc ) = 0.744 =
0.256
-------------------------------------------------------------------------------------------------------11.
Solution: B
Let
C = Event that a policyholder buys collision coverage
D = Event that a policyholder buys disability coverage
Then we are given that P[C] = 2P[D] and P[C + D] = 0.15 .
By the independence of C and D, it therefore follows that
0.15 = P[C + D] = P[C] P[D] = 2P[D] P[D] = 2(P[D])2
(P[D])2 = 0.15/2 = 0.075
P[D] = 0.075 and P[C] = 2P[D] = 2 0.075
Now the independence of C and D also implies the independence of CC and DC . As a
result, we see that P[CC + DC] = P[CC] P[DC] = (1 – P[C]) (1 – P[D])
= (1 – 2 0.075 ) (1 – 0.075 ) = 0.33 .
Page 5 of 54
12.
Solution: E
“Boxed” numbers in the table below were computed.
High BP Low BP Norm BP
Total
Regular heartbeat
0.09
0.20
0.56
0.85
Irregular heartbeat
0.05
0.02
0.08
0.15
Total
0.14
0.22
0.64
1.00
From the table, we can see that 20% of patients have a regular heartbeat and low blood
pressure.
-------------------------------------------------------------------------------------------------------13.
Solution: C
The Venn diagram below summarizes the unconditional probabilities described in the
problem.
In addition, we are told that
P, A + B + C1
x
! P , A + B + C | A + B- !
!
P , A + Bx * 0.12
3
It follows that
1
1
x ! " x * 0.12 # ! x * 0.04
3
3
2
x ! 0.04
3
x ! 0.06
Now we want to find
c
P $" A & B & C # %
c
)
P $" A & B & C # | Ac % ! (
c
(
)
P $( A %)
1' P, A & B & C!
1 ' P , A!
1 ' 3 " 0.10 # ' 3 " 0.12 # ' 0.06
1 ' 0.10 ' 2 " 0.12 # ' 0.06
!
0.28
! 0.467
0.60
Page 6 of 54
14.
Solution: A
k
1
11
1 1 1
.1/
pk ' 2 ! 5 5 pk '3 ! ... ! 0 1 p0
pk = pk '1 !
5
55
5 5 5
253
k
7
p0
5
.1/
p
!
! p0
8
8
k
0 1 p0 !
1
4
k !0
k !0 2 5 3
1'
5
p0 = 4/5 .
Therefore, P[N > 1] = 1 – P[N 91] = 1 – (4/5 + 4/5 5 1/5) = 1 – 24/25 = 1/25 = 0.04 .
1=
7
k 60
-------------------------------------------------------------------------------------------------------15.
Solution: C
A Venn diagram for this situation looks like:
We want to find w ! 1 ' " x * y * z #
1
1
5
We have x * y ! , x * z ! , y * z !
4
3
12
Adding these three equations gives
1 1 5
" x * y# * " x * z# * " y * z# ! * *
4 3 12
2" x * y * z# !1
x* y* z !
1
2
1 1
!
2 2
Alternatively the three equations can be solved to give x = 1/12, y = 1/6, z =1/4
. 1 1 1/ 1
again leading to w ! 1 ' 0 * * 1 !
2 12 6 4 3 2
w ! 1' " x * y * z # ! 1'
Page 7 of 54
16.
Solution: D
Let N1 and N 2 denote the number of claims during weeks one and two, respectively.
Then since N1 and N 2 are independent,
Pr , N1 * N 2 ! 7 - ! 8 n !0 Pr , N1 ! n - Pr , N 2 ! 7 ' n 7
7 . 1 /. 1 /
! 8 n !0 0 n *1 10 8' n 1
2 2 32 2 3
1
7
! 8 n !0 9
2
8
1
1
! 9 ! 6 !
2
2
64
-------------------------------------------------------------------------------------------------------17.
Solution: D
Let
O ! Event of operating room charges
E ! Event of emergency room charges
Then
0.85 ! Pr " O & E # ! Pr " O # * Pr " E # ' Pr " O + E #
! Pr " O # * Pr " E # ' Pr " O # Pr " E #
Since
So
" Independence #
Pr " E c # ! 0.25 ! 1 ' Pr " E # , it follows Pr " E # ! 0.75 .
0.85 ! Pr " O # * 0.75 ' Pr " O #" 0.75 #
Pr " O #"1 ' 0.75 # ! 0.10
Pr " O # ! 0.40
-------------------------------------------------------------------------------------------------------18.
Solution: D
Let X1 and X2 denote the measurement errors of the less and more accurate instruments,
respectively. If N(:,;) denotes a normal random variable with mean : and standard
deviation ;, then we are given X1 is N(0, 0.0056h), X2 is N(0, 0.0044h) and X1, X2 are
X1 * X 2
0.00562 h 2 * 0.00442 h 2
is N (0,
) = N(0,
2
4
0.00356h) . Therefore, P['0.005h 9 Y 9 0.005h] = P[Y 9 0.005h] – P[Y 9 '0.005h] =
P[Y 9 0.005h] – P[Y 6 0.005h]
0.005h %
$
= 2P[Y 9 0.005h] – 1 = 2P < Z 9
' 1 = 2P[Z 9 1.4] – 1 = 2(0.9192) – 1 = 0.84.
0.00356h =)
(
independent. It follows that Y =
Page 8 of 54
19.
Solution: B
Apply Bayes’ Formula. Let
A ! Event of an accident
B1 ! Event the driver’s age is in the range 16-20
B2 ! Event the driver’s age is in the range 21-30
B3 ! Event the driver’s age is in the range 30-65
B4 ! Event the driver’s age is in the range 66-99
Then
Pr " A B1 # Pr " B1 #
Pr " B1 A # !
Pr " A B1 # Pr " B1 # * Pr " A B2 # Pr " B2 # * Pr " A B3 # Pr " B3 # * Pr " A B4 # Pr " B4 #
!
" 0.06 #" 0.08#
! 0.1584
" 0.06 #" 0.08# * " 0.03#" 0.15# * " 0.02 #" 0.49 # * " 0.04 #" 0.28#
--------------------------------------------------------------------------------------------------------
20.
Solution: D
Let
S = Event of a standard policy
F = Event of a preferred policy
U = Event of an ultra-preferred policy
D = Event that a policyholder dies
Then
P , D | U - P ,U P ,U | D - !
P , D | S - P , S - * P , D | F - P , F - * P , D | U - P ,U !
" 0.001#" 0.10 #
" 0.01#" 0.50 # * " 0.005 #" 0.40 # * " 0.001#" 0.10 #
! 0.0141
-------------------------------------------------------------------------------------------------------21.
Solution: B
Apply Baye’s Formula:
Pr $(Seri. Surv.%)
!
!
Pr $(Surv. Seri.%) Pr ,Seri.-
Pr $(Surv. Crit.%) Pr , Crit.- * Pr $(Surv. Seri.%) Pr ,Seri.- * Pr $(Surv. Stab.%) Pr ,Stab.-
" 0.9 #" 0.3#
! 0.29
" 0.6 #" 0.1# * " 0.9 #" 0.3# * " 0.99 #" 0.6 #
Page 9 of 54
22.
Solution: D
Let
H ! Event of a heavy smoker
L ! Event of a light smoker
N ! Event of a non-smoker
D ! Event of a death within five-year period
1
Now we are given that Pr $( D L %) ! 2 Pr $( D N %) and Pr $( D L %) ! Pr $( D H %)
2
Therefore, upon applying Bayes’ Formula, we find that
Pr $( D H %) Pr , H Pr $( H D %) !
Pr $( D N %) Pr , N - * Pr $( D L %) Pr , L - * Pr $( D H %) Pr , H 2 Pr $( D L %) " 0.2 #
0.4
!
!
! 0.42
1
0.25
0.3
0.4
*
*
Pr $ D L %) " 0.5 # * Pr $( D L %) " 0.3# * 2 Pr $( D L %) " 0.2 #
2 (
-------------------------------------------------------------------------------------------------------23.
Solution: D
Let
C = Event of a collision
T = Event of a teen driver
Y = Event of a young adult driver
M = Event of a midlife driver
S = Event of a senior driver
Then using Bayes’ Theorem, we see that
P[C Y ]P[Y ]
P[Y>C] =
P[C T ]P[T ] * P[C Y ]P[Y ] * P[C M ]P[ M ] * P[C S ]P[ S ]
=
(0.08)(0.16)
= 0.22 .
(0.15)(0.08) * (0.08)(0.16) * (0.04)(0.45) * (0.05)(0.31)
-------------------------------------------------------------------------------------------------------24.
Solution: B
Observe
Pr ,1 9 N 9 4- $ 1 1 1
1 % $1 1 1 1
1%
Pr $( N 6 1 N 9 4 %) !
!< * * * = < * * * * =
Pr , N 9 4( 6 12 20 30 ) ( 2 6 12 20 30 )
!
10 * 5 * 3 * 2
20 2
!
!
30 * 10 * 5 * 3 * 2 50 5
Page 10 of 54
25.
Solution: B
Let
Y = positive test result
D = disease is present (and ~D = not D)
Using Baye’s theorem:
P[Y | D]P[ D]
(0.95)(0.01)
!
= 0.657 .
P[D|Y] =
P[Y | D]P[ D] * P[Y |~ D]P[~ D] (0.95)(0.01) * (0.005)(0.99)
-------------------------------------------------------------------------------------------------------26.
Solution: C
Let:
S = Event of a smoker
C = Event of a circulation problem
Then we are given that P[C] = 0.25 and P[S>C] = 2 P[S>CC]
Now applying Bayes’ Theorem, we find that P[C>S] =
=
2 P[ S C C ]P[C ]
2 P[ S C ]P[C ] * P[ S C ](1 ' P[C ])
C
C
!
P[ S C ]P[C ]
P[ S C ]P[C ] * P[ S C C ]( P[C C ])
2(0.25)
2
2
!
! .
2(0.25) * 0.75 2 * 3 5
-------------------------------------------------------------------------------------------------------27.
Solution: D
Use Baye’s Theorem with A = the event of an accident in one of the years 1997, 1998 or
1999.
P[ A 1997]P[1997]
P[1997|A] =
P[ A 1997][ P[1997] * P[ A 1998]P[1998] * P[ A 1999]P[1999]
=
(0.05)(0.16)
= 0.45 .
(0.05)(0.16) * (0.02)(0.18) * (0.03)(0.20)
--------------------------------------------------------------------------------------------------------
Page 11 of 54
28.
Solution: A
Let
C = Event that shipment came from Company X
I1 = Event that one of the vaccine vials tested is ineffective
P , I1 | C - P , C Then by Bayes’ Formula, P , C | I1 - !
P , I1 | C - P , C - * P $( I1 | C c %) P $(C c %)
Now
1
P ,C - !
5
1 4
P $(C c %) ! 1 ' P , C - ! 1 ' !
5 5
P , I1 | C - ! " 130 # " 0.10 #" 0.90 # ! 0.141
29
P $( I1 | C c %) ! " 130 # " 0.02 #" 0.98 # ! 0.334
29
Therefore,
P , C | I1 - !
" 0.141#"1/ 5#
! 0.096
" 0.141#"1/ 5# * " 0.334 #" 4 / 5#
-------------------------------------------------------------------------------------------------------29.
Solution: C
Let T denote the number of days that elapse before a high-risk driver is involved in an
accident. Then T is exponentially distributed with unknown parameter ? . Now we are
given that
50
0.3 = P[T 9 50] =
@ ?e
' ?t
dt ! 'e ' ?t
0
50
0
= 1 – e–50?
Therefore, e–50? = 0.7 or ? = ' (1/50) ln(0.7)
80
It follows that P[T 9 80] =
=1–e
(80/50) ln(0.7)
@ ?e
0
' ?t
dt ! 'e ' ?t
80
0
= 1 – e–80?
= 1 – (0.7)80/50 = 0.435 .
-------------------------------------------------------------------------------------------------------30.
Solution: D
e' ? ? 2
e' ? ? 4
!3
= 3 5 P[N
Let N be the number of claims filed. We are given P[N = 2] =
2!
4!
= 4]24 ?2 = 6 ?4
?2 = 4 A ? = 2
Therefore, Var[N] = ? = 2 .
Page 12 of 54
31.
Solution: D
Let X denote the number of employees that achieve the high performance level. Then X
follows a binomial distribution with parameters n ! 20 and p ! 0.02 . Now we want to
determine x such that
Pr , X B x - 9 0.01
or, equivalently,
x
k
20 ' k
0.99 9 Pr , X 9 x - ! 8 k !0 " 20k # " 0.02 # " 0.98 #
The following table summarizes the selection process for x:
x
Pr , X ! x Pr , X 9 x 0
1
2
" 0.98# ! 0.668
19
20 " 0.02 #" 0.98 # ! 0.272
2
18
190 " 0.02 # " 0.98 # ! 0.053
20
0.668
0.940
0.993
Consequently, there is less than a 1% chance that more than two employees will achieve
the high performance level. We conclude that we should choose the payment amount C
such that
2C ! 120, 000
or
C ! 60, 000
-------------------------------------------------------------------------------------------------------32.
Solution: D
Let
X = number of low-risk drivers insured
Y = number of moderate-risk drivers insured
Z = number of high-risk drivers insured
f(x, y, z) = probability function of X, Y, and Z
Then f is a trinomial probability function, so
Pr , z 6 x * 2- ! f " 0, 0, 4 # * f "1, 0,3# * f " 0,1,3# * f " 0, 2, 2 #
! " 0.20 # * 4 " 0.50 #" 0.20 # * 4 " 0.30 #" 0.20 # *
4
3
! 0.0488
Page 13 of 54
3
4!
2
2
" 0.30 # " 0.20 #
2!2!
33.
Solution: B
Note that
1 /
.
0.005 " 20 ' t # dt ! 0.005 0 20t ' t 2 1 20
x
x
2 3
2
1 /
1 /
.
.
! 0.005 0 400 ' 200 ' 20 x * x 2 1 ! 0.005 0 200 ' 20 x * x 2 1
2 3
2 3
2
2
where 0 C x C 20 . Therefore,
2
Pr , X B 16- 200 ' 20 "16 # * 1 2 "16 #
8 1
!
!
!
Pr $( X B 16 X B 8%) !
2
1
Pr , X B 872 9
200 ' 20 " 8 # * " 8 #
2
Pr , X B x - ! @
20
-------------------------------------------------------------------------------------------------------34.
Solution: C
'2
We know the density has the form C "10 * x # for 0 C x C 40 (equals zero otherwise).
First, determine the proportionality constant C from the condition
@
40
0
f ( x)dx !1 :
C C
2
'
! C
0
10 50 25
so C ! 25 2 , or 12.5 . Then, calculate the probability over the interval (0, 6):
6
'2
'1 6
.1 1/
! 0 ' 1 "12.5 # ! 0.47 .
12.5@ "10 * x # dx ! ' "10 * x #
0
0
2 10 16 3
40
1 ! @ C "10 * x # dx ! ' C (10 * x) '1
'2
0
40
!
-------------------------------------------------------------------------------------------------------35.
Solution: C
Let the random variable T be the future lifetime of a 30-year-old. We know that the
density of T has the form f (x) = C(10 + x)'2 for 0 < x < 40 (and it is equal to zero
otherwise). First, determine the proportionality constant C from the condition
@ 040 f ( x)dx !1:
40
2
1 = @ f ( x)dx ! ' C (10 * x) '1 |040 ! C
0
25
25
= 12.5. Then, calculate P(T < 5) by integrating f (x) = 12.5 (10 + x)'2
so that C =
2
over the interval (0.5).
Page 14 of 54
36.
Solution: B
To determine k, note that
1
k
k
4
5
1 = @ k "1 ' y # dy ! ' "1 ' y # 1 !
0
5
5
0
k=5
We next need to find P[V > 10,000] = P[100,000 Y > 10,000] = P[Y > 0.1]
1
=
@ 5 "1 ' y # dy ! ' "1 ' y #
4
5 1
0.1
0.1
= (0.9)5 = 0.59 and P[V > 40,000]
1
= P[100,000 Y > 40,000] = P[Y > 0.4] =
@ 5 "1 ' y # dy ! ' "1 ' y #
4
5 1
0.4
0.4
= (0.6)5 = 0.078 .
It now follows that P[V > 40,000>V > 10,000]
P[V B 40, 000 + V B 10, 000] P[V B 40, 000] 0.078
!
!
= 0.132 .
=
P[V B 10, 000]
P[V B 10, 000] 0.590
-------------------------------------------------------------------------------------------------------37.
Solution: D
Let T denote printer lifetime. Then f(t) = ½ e–t/2, 0 9 t 9 !
Note that
1
1
P[T 9 1] = @ e ' t / 2 dt ! e' t / 2 1 = 1 – e–1/2 = 0.393
0
2
0
2
P[1 9 T 9 2] =
1
@ 2e
't / 2
dt ! e' t / 2
2
1
= e –1/2 ' e –1 = 0.239
1
Next, denote refunds for the 100 printers sold by independent and identically distributed
random variables Y1, . . . , Y100 where
with probability 0.393
D 200
E
Yi ! F100
with probability 0.239
i = 1, . . . , 100
E0
with probability 0.368
G
Now E[Yi] = 200(0.393) + 100(0.239) = 102.56
100
Therefore, Expected Refunds =
8 E ,Y - = 100(102.56) = 10,256 .
i !1
i
Page 15 of 54
38.
Solution: A
Let F denote the distribution function of f. Then
F " x # ! Pr , X 9 x - ! @ 3t '4 dt ! 't '3 ! 1 ' x '3
x
1
Using this result, we see
Pr , X < 2| X 6 1.5- !
x
1
Pr $(" X < 2 # + " X 6 1.5 # %)
Pr , X 6 1.5-
!
Pr , X < 2- ' Pr , X 9 1.5Pr , X 6 1.5-
F " 2 # ' F "1.5 # "1.5 # ' " 2 #
!
!
'3
1 ' F "1.5 #
"1.5 #
'3
'3
.3/
! 1' 0 1
243
3
! 0.578
-------------------------------------------------------------------------------------------------------39.
Solution: E
Let X be the number of hurricanes over the 20-year period. The conditions of the
problem give x is a binomial distribution with n = 20 and p = 0.05 . It follows that
P[X < 2] = (0.95)20(0.05)0 + 20(0.95)19(0.05) + 190(0.95)18(0.05)2
= 0.358 + 0.377 + 0.189 = 0.925 .
-------------------------------------------------------------------------------------------------------40.
Solution: B
Denote the insurance payment by the random variable Y. Then
if 0 C X 9 C
D0
Y !F
G X ' C if C C X C 1
Now we are given that
0.64 ! Pr "Y C 0.5 # ! Pr " 0 C X C 0.5 * C # ! @
Therefore, solving for C, we find C ! H0.8 ' 0.5
Finally, since 0 C C C 1 , we conclude that C ! 0.3
Page 16 of 54
0.5* C
0
2 x dx ! x 2
0.5 * C
0
! " 0.5 * C #
2
41.
Solution: E
Let
X = number of group 1 participants that complete the study.
Y = number of group 2 participants that complete the study.
Now we are given that X and Y are independent.
Therefore,
P $(" X 6 9 # + " Y < 9 # %) & $(" X < 9 # + " Y 6 9 # %)
I
J
! P $(" X 6 9 # + " Y < 9 # %) * P $(" X < 9 # + " Y 6 9 # %)
! 2 P $(" X 6 9 # + " Y < 9 # %)
(due to symmetry)
! 2 P , X 6 9 - P ,Y < 9 -
! 2 P , X 6 9- P , X < 9-
(again due to symmetry)
! 2 P , X 6 9- "1 ' P , X 6 9- #
9
10
9
10
! 2 $" 10
0.2 0.8 * 10 0.8 % $1 ' 10 0.2 0.8 ' 10 0.8 %
( 9 # " #" # " 10 # " # ) ( " 9 # " #" # " 10 # " # )
! 2 , 0.376-,1 ' 0.376- ! 0.469
-------------------------------------------------------------------------------------------------------42.
Solution: D
Let
IA = Event that Company A makes a claim
IB = Event that Company B makes a claim
XA = Expense paid to Company A if claims are made
XB = Expense paid to Company B if claims are made
Then we want to find
Pr ($ I AC + I B )% & $(" I A + I B # + " X A < X B # %)
I
J
! Pr ($ I AC + I B )% * Pr $(" I A + I B # + " X A < X B # %)
! Pr $( I AC %) Pr , I B - * Pr , I A - Pr , I B - Pr , X A < X B -
(independence)
! " 0.60 #" 0.30 # * " 0.40 #" 0.30 # Pr , X B ' X A 6 0! 0.18 * 0.12 Pr , X B ' X A 6 0Now X B ' X A is a linear combination of independent normal random variables.
Therefore, X B ' X A is also a normal random variable with mean
M ! E , X B ' X A - ! E , X B - ' E , X A - ! 9, 000 ' 10, 000 ! '1, 000
and standard deviation ; ! Var " X B # * Var " X A # !
It follows that
Page 17 of 54
" 2000 # * " 2000 #
2
2
! 2000 2
1000 %
$
Pr , X B ' X A 6 0- ! Pr < Z 6
2000 2 =)
(
1 %
$
! Pr < Z 6
2 2 =)
(
(Z is standard normal)
1 %
$
! 1 ' Pr < Z <
2 2 =)
(
! 1 ' Pr , Z C 0.354-
Finally,
! 1 ' 0.638 ! 0.362
I
J
Pr ($ I AC + I B )% & $(" I A + I B # + " X A < X B # %) ! 0.18 * " 0.12 #" 0.362 #
! 0.223
-------------------------------------------------------------------------------------------------------43.
Solution: D
If a month with one or more accidents is regarded as success and k = the number of
failures before the fourth success, then k follows a negative binomial distribution and the
requested probability is
4
k
3
3* k . 3 / . 2 /
Pr , k 6 4- ! 1 ' Pr , k 9 3- ! 1 ' 8 " k # 0 1 0 1
253 2 53
k !0
. 3/
! 1' 0 1
253
4
$ 3 . 2 /0 4 . 2 /1 5 . 2 / 2 6 . 2 /3 %
<" 0 # 0 1 * " 1 # 0 1 * " 2 # 0 1 * " 3 # 0 1 =
253
253
2 5 3 =)
<( 2 5 3
4
. 3 / $ 8 8 32 %
! 1 ' 0 1 <1 * * * =
2 5 3 ( 5 5 25 )
! 0.2898
Alternatively the solution is
4
4
4
2
4
3
. 2/
4 . 2/ 3
5 . 2/ . 3/
6 . 2/ . 3/
0 1 * " 1 # 0 1 * " 2 # 0 1 0 1 * " 3 # 0 1 0 1 ! 0.2898
253
2 53 5
2 53 2 53
2 53 2 53
which can be derived directly or by regarding the problem as a negative binomial
distribution with
i) success taken as a month with no accidents
ii) k = the number of failures before the fourth success, and
iii) calculating Pr , k 9 3-
Page 18 of 54
44.
Solution: C
If k is the number of days of hospitalization, then the insurance payment g(k) is
100k
for k !1, 2, 3
g(k) =
300 * 50 (k ' 3) for k ! 4, 5.
I
Thus, the expected payment is
5
8 g (k ) p
k !1
k
! 100 p1 * 200 p2 * 300 p3 * 350 p4 * 400 p5
= "100 x5 * 200 x 4 * 300 x3 * 350 x 2 * 400 x1# =
1
(100 * 200 K 2 * 300 K 3 * 350 K 4 * 400 K 5) ! 220
15
-------------------------------------------------------------------------------------------------------45.
Solution: D
0
4
2
4 x
x2
x3
x3
8 64 56 28
dx ! '
*
!' *
!
!
Note that E " X # ! @ ' dx * @
' 2 10
0 10
30 '2 30 0
30 30 30 15
0
-------------------------------------------------------------------------------------------------------46.
Solution: D
The density function of T is
1
f " t # ! e't / 3 , 0 < t < 7
3
Therefore,
E , X - ! E $( max " T , 2 # %)
!@
2
0
7 t
2 't / 3
e dt * @ e ' t / 3 dt
2
3
3
7
! '2e ' t / 3 | 02 'te ' t / 3 | 72 * @ e' t / 3 dt
2
! '2e
'2 / 3
* 2 * 2e
'2 / 3
' 3e' t / 3 | 72
! 2 * 3e '2 / 3
Page 19 of 54
47.
Solution: D
Let T be the time from purchase until failure of the equipment. We are given that T is
exponentially distributed with parameter ? = 10 since 10 = E[T] = ? . Next define the
payment
for 0 9 T 9 1
Dx
Ex
E
for 1 C T 9 3
P under the insurance contract by P ! F
2
E
for T B 3
EG0
We want to find x such that
1
x –t/10
1000 = E[P] = @
e
dt +
10
0
–1/10
3
x 1 –t/10
' t /10
@1 2 10 e dt = ' xe
–3/10
–1/10
+ x – (x/2) e
+ (x/2) e
= 'x e
We conclude that x = 5644 .
1
0
–1/10
= x(1 – ½ e
x
' e 't /10
2
3
1
– ½ e–3/10) = 0.1772x .
-------------------------------------------------------------------------------------------------------48.
Solution: E
Let X and Y denote the year the device fails and the benefit amount, respectively. Then
the density function of X is given by
x '1
f " x # ! " 0.6 # " 0.4 # , x ! 1, 2,3...
and
DE1000 " 5 ' x # if x ! 1, 2,3, 4
y!F
if x B 4
EG0
It follows that
2
3
E ,Y - ! 4000 " 0.4 # * 3000 " 0.6 #" 0.4 # * 2000 " 0.6 # " 0.4 # * 1000 " 0.6 # " 0.4 #
! 2694
-------------------------------------------------------------------------------------------------------49.
Solution: D
Define f ( X ) to be hospitalization payments made by the insurance policy. Then
DE100 X
f (X ) ! F
EG300 * 25 " X ' 3#
if X ! 1, 2,3
if X ! 4,5
and
Page 20 of 54
5
E $( f " X # %) ! 8 f " k # Pr , X ! k k !1
.5/
. 4/
. 3/
. 2/
.1/
! 100 0 1 * 200 0 1 * 300 0 1 * 325 0 1 * 350 0 1
2 15 3
2 15 3
2 15 3
2 15 3
2 15 3
1
640
! ,100 * 160 * 180 * 130 * 70- !
! 213.33
3
3
-------------------------------------------------------------------------------------------------------50.
Solution: C
Let N be the number of major snowstorms per year, and let P be the amount paid to
(3 / 2) n e '3/ 2
, n = 0, 1, 2, . . . and
the company under the policy. Then Pr[N = n] =
n!
for N ! 0
D0
.
P!F
G10, 000( N ' 1) for N 6 1
Now observe that E[P] =
7
810, 000(n ' 1)
n !1
(3 / 2) n e '3/ 2
n!
(3 / 2) n e '3/ 2
= 10,000 e–3/2 + E[10,000 (N – 1)]
n!
n !0
–3/2
= 10,000 e + E[10,000N] – E[10,000] = 10,000 e–3/2 + 10,000 (3/2) – 10,000 = 7,231 .
= 10,000 e–3/2 +
7
810, 000(n ' 1)
-------------------------------------------------------------------------------------------------------51.
Solution: C
Let Y denote the manufacturer’s retained annual losses.
for 0.6 C x 9 2
Dx
Then Y ! F
for x B 2
G2
7
2
$ 2.5(0.6) 2.5 %
$ 2.5(0.6) 2.5 %
2.5(0.6) 2.5
2(0.6) 2.5
x
dx
2
dx
dx
*
!
'
@ < x3.5 =) @2 <( x3.5 =) 0.6@ x 2.5
x 2.5
0.6 (
2
and E[Y] =
= '
2.5(0.6) 2.5
1.5 x1.5
2
0.6
*
2(0.6) 2.5
2.5(0.6) 2.5 2.5(0.6) 2.5 (0.6) 2.5
!
'
*
* 1.5 = 0.9343 .
(2) 2.5
1.5(2)1.5
1.5(0.6)1.5
2
Page 21 of 54
7
2
52.
Solution: A
Let us first determine K. Observe that
. 1 1 1 1/
. 60 * 30 * 20 * 15 * 12 /
. 137 /
1 ! K 01 * * * * 1 ! K 0
1!K0
1
60
2 2 3 4 53
2
3
2 60 3
60
K!
137
It then follows that
Pr , N ! n - ! Pr $( N ! n Insured Suffers a Loss %) Pr , Insured Suffers a Loss 60
3
, N ! 1,...,5
" 0.05# !
137 N
137 N
Now because of the deductible of 2, the net annual premium P ! E , X - where
!
D0
X !F
GN ' 2
, if N 9 2
, if N B 2
Then,
P ! E , X - ! 8 N !3 " N ' 2 #
5
$ 3 %
$ 3 %
3
. 1 /
! "1# 0
= * 3<
= ! 0.0314
1* 2<
137 N
2 137 3
(137 " 4 # )
(137 " 5 # )
-------------------------------------------------------------------------------------------------------53.
Solution: D
for 1 C y 9 10
Dy
Let W denote claim payments. Then W ! F
for y 6 10
G10
10
7
2
2
2 10
10 7
It follows that E[W] = @ y 3 dy * @ 10 3 dy ! '
' 2
= 2 – 2/10 + 1/10 = 1.9 .
y
y
y1
y 10
1
10
Page 22 of 54
54.
Solution: B
Let Y denote the claim payment made by the insurance company.
Then
with probability 0.94
D0
E
Y ! FMax " 0, x ' 1# with probability 0.04
E14
with probability 0.02
G
and
E ,Y - ! " 0.94 #" 0 # * " 0.04 #" 0.5003# @
15
1
" x ' 1# e' x / 2 dx * " 0.02 #"14 #
15
15
! " 0.020012 # $ @ xe ' x / 2 dx ' @ e' x / 2 dx % * 0.28
<( 1
=)
1
15
15
! 0.28 * " 0.020012 # $ '2 xe ' x / 2 | 115 *2@ e ' x / 2 dx ' @ e' x / 2 dx %
1
1
(<
)=
15
! 0.28 * " 0.020012 # $ '30e '7.5 * 2e '0.5 * @ e' x / 2 dx %
<(
=)
1
! 0.28 * " 0.020012 # $( '30e '7.5 * 2e '0.5 ' 2e' x / 2 | 115 %)
! 0.28 * " 0.020012 # " '30e '7.5 * 2e '0.5 ' 2e'7.5 * 2e'0.5 #
! 0.28 * " 0.020012 # " '32e '7.5 * 4e '0.5 #
! 0.28 * " 0.020012 #" 2.408 #
(in thousands)
! 0.328
It follows that the expected claim payment is 328 .
-------------------------------------------------------------------------------------------------------55.
Solution: C
k
, 0 < x < 7 . To find k, note
(1 * x) 4
The pdf of x is given by f(x) =
7
k
k 1
@0 (1 * x)4 dx ! ' 3 (1 * x)3
k=3
1=
7
It then follows that E[x] =
7
0
!
k
3
3x
@ (1 * x)
4
dx and substituting u = 1 + x, du = dx, we see
0
7
7
7
$ u '2 u '3 %
3(u ' 1)
$1 1%
'3
'4
u
u
du
du
(
)
3
E[x] = @
=
3
'
!
'
! 3 < ' = = 3/2 – 1 = ½ .
<
=
4
@
u
( 2 3)
( '2 '3 )1
1
1
Page 23 of 54
56.
Solution: C
Let Y represent the payment made to the policyholder for a loss subject to a deductible D.
for 0 9 X 9 D
D0
That is Y ! F
G x ' D for D C X 9 1
Then since E[X] = 500, we want to choose D so that
1000
1
1
1 ( x ' D) 2 1000 (1000 ' D) 2
500 ! @
( x ' D)dx !
!
D
4
1000
1000
2
2000
D
(1000 – D)2 = 2000/4 5 500 = 5002
1000 – D = H 500
D = 500 (or D = 1500 which is extraneous).
-------------------------------------------------------------------------------------------------------57.
Solution: B
1
for the claim size X in a certain class of accidents.
(1 ' 2500t ) 4
('4)('2500)
10, 000
!
First, compute MxL(t) =
5
(1 ' 2500t )
(1 ' 2500t )5
(10, 000)('5)('2500) 125, 000, 000
!
MxM(t) =
(1 ' 2500t )6
(1 ' 2500t )6
Then E[X] = MxL (0) = 10,000
E[X2] = MxM (0) = 125,000,000
Var[X] = E[X2] – {E[X]}2 = 125,000,000 – (10,000)2 = 25,000,000
Var[ X ] = 5,000 .
We are given that Mx(t) =
-------------------------------------------------------------------------------------------------------58.
Solution: E
Let XJ, XK, and XL represent annual losses for cities J, K, and L, respectively. Then
X = XJ + XK + XL and due to independence
x *x *x t
M(t) = E $(e xt %) ! E ($e" J K L # )% ! E ($ e xJ t )% E $(e xK t %) E $(e xLt %)
= MJ(t) MK(t) ML(t) = (1 – 2t)–3 (1 – 2t)–2.5 (1 – 2t)–4.5 = (1 – 2t)–10
Therefore,
ML(t) = 20(1 – 2t)–11
MM(t) = 440(1 – 2t)–12
MML(t) = 10,560(1 – 2t)–13
E[X3] = MML(0) = 10,560
Page 24 of 54
59.
Solution: B
The distribution function of X is given by
2.5 " 200 #
F " x# ! @
200
t 3.5
x
2.5
' " 200 #
dt !
t 2.5
2.5 x
" 200 #
! 1'
200
x 2.5
2.5
,
x B 200
th
Therefore, the p percentile x p of X is given by
" 200 #
p
! F " xp # ! 1'
2.5
100
xp
2.5
" 200 #
1 ' 0.01 p !
xp
"1 ' 0.01 p #
xp !
25
!
2.5
2.5
200
xp
200
"1 ' 0.01 p #
25
It follows that x 70 ' x 30 !
200
" 0.30 #
25
'
200
" 0.70 #
25
! 93.06
-------------------------------------------------------------------------------------------------------60.
Solution: E
Let X and Y denote the annual cost of maintaining and repairing a car before and after
the 20% tax, respectively. Then Y = 1.2X and Var[Y] = Var[1.2X] = (1.2)2 Var[X] =
(1.2)2(260) = 374 .
-------------------------------------------------------------------------------------------------------61.
Solution: A
The first quartile, Q1, is found by ¾ =
z
7
Q1
f(x) dx . That is, ¾ = (200/Q1)2.5 or
Q1 = 200 (4/3)0.4 = 224.4 . Similarly, the third quartile, Q3, is given by Q3 = 200 (4)0.4
= 348.2 . The interquartile range is the difference Q3 – Q1 .
Page 25 of 54
62.
Solution: C
First note that the density function of X is given by
D1
if
x !1
E2
EE
1C x C 2
f " x # ! F x ' 1 if
E
E
otherwise
EG0
Then
2
2
2
1
1
1 .1
1 /
E " X # ! * @ x " x ' 1# dx ! * @ x 2 ' x dx ! * 0 x3 ' x 2 1
1
1
2
2
2 23
2 31
"
!
" #
E X
2
#
1 8 4 1 1 7
4
* ' ' * ! '1 !
2 3 2 3 2 3
3
2
2
2
1
1
1 .1
1 /
! * @ x 2 " x ' 1# dx ! * @ x 3 ' x 2 dx ! * 0 x 4 ' x3 1
2 1
2 1
2 24
3 31
"
!
#
1 16 8 1 1 17 7 23
* ' ' * ! ' !
2 4 3 4 3 4 3 12
" #
Var " X # ! E X
2
2
23 . 4 /
23 16 5
' $( E " X # %) !
'0 1 !
' !
12 2 3 3 12 9 36
2
-------------------------------------------------------------------------------------------------------63.
Solution: C
D X if 0 9 X 9 4
Note Y ! F
G4 if 4 < X 9 5
Therefore,
41
54
1
4
E ,Y - ! @ xdx * @ dx ! x 2 | 04 * x| 54
0 5
4 5
10
5
16 20 16 8 4 12
! * ' ! * !
10 5 5 5 5 5
41
5 16
1
16
E $(Y 2 %) ! @ x 2 dx * @
dx ! x 3 | 04 * x| 54
0 5
4 5
15
5
64 80 64 64 16 64 48 112
!
* '
!
* !
*
!
15 5 5 15 5 15 15 15
2
2
112 . 12 /
' 0 1 ! 1.71
Var ,Y - ! E $(Y 2 %) ' " E ,Y - # !
15 2 5 3
Page 26 of 54
64.
Solution: A
Let X denote claim size. Then E[X] = [20(0.15) + 30(0.10) + 40(0.05) + 50(0.20) +
60(0.10) + 70(0.10) + 80(0.30)] = (3 + 3 + 2 + 10 + 6 + 7 + 24) = 55
E[X2] = 400(0.15) + 900(0.10) + 1600(0.05) + 2500(0.20) + 3600(0.10) + 4900(0.10)
+ 6400(0.30) = 60 + 90 + 80 + 500 + 360 + 490 + 1920 = 3500
Var[X] = E[X2] – (E[X])2 = 3500 – 3025 = 475 and Var[ X ] = 21.79 .
Now the range of claims within one standard deviation of the mean is given by
[55.00 – 21.79, 55.00 + 21.79] = [33.21, 76.79]
Therefore, the proportion of claims within one standard deviation is
0.05 + 0.20 + 0.10 + 0.10 = 0.45 .
-------------------------------------------------------------------------------------------------------65.
Solution: B
Let X and Y denote repair cost and insurance payment, respectively, in the event the auto
is damaged. Then
if x 9 250
D0
Y !F
G x ' 250 if x B 250
and
2
1500
1
1
2 1500 1250
E ,Y - ! @
x
250
dx
x
250
'
!
'
!
! 521
"
#
"
# 250
250 1500
3000
3000
1500
1
1
12503
2
3
E $(Y 2 %) ! @
x ' 250 # dx !
x ' 250 # 1500
!
! 434, 028
"
"
250
250 1500
4500
4500
Var ,Y - ! E $(Y 2 %) ' I E ,Y -J ! 434, 028 ' " 521#
2
2
Var ,Y - ! 403
-------------------------------------------------------------------------------------------------------66.
Solution: E
Let X1, X2, X3, and X4 denote the four independent bids with common distribution
function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is
given by
G " y # ! Pr ,Y 9 y -
! Pr $(" X 1 9 y # + " X 2 9 y # + " X 3 9 y # + " X 4 9 y # %)
! Pr , X 1 9 y - Pr , X 2 9 y - Pr , X 3 9 y - Pr , X 4 9 y ! $( F " y # %)
4
1
3
5
4
9 y9
"1 * sinN y # ,
16
2
2
It then follows that the density function g of Y is given by
!
Page 27 of 54
g " y # ! G '" y #
!
!
1
3
"1 * sinN y # "N cosN y #
4
N
cosN y "1 * sinN y #
4
3
3
5
9 y9
2
2
,
Finally,
E ,Y - ! @
5/ 2
!@
5/ 2
N
3/ 2
4
3/ 2
yg " y # dy
ycosN y "1 * sinN y # dy
3
-------------------------------------------------------------------------------------------------------67.
Solution: B
The amount of money the insurance company will have to pay is defined by the random
variable
D1000 x if x C 2
Y !F
if x 6 2
G2000
where x is a Poisson random variable with mean 0.6 . The probability function for X is
k
e '0.6 " 0.6 #
p " x# !
k ! 0,1, 2,3! and
k!
k
7 0.6
'0.6
'0.6
E ,Y - ! 0 * 1000 " 0.6 # e * 2000e 8 k ! 2
k!
k
.
/
7 0.6
! 1000 " 0.6 # e '0.6 * 2000 0 e '0.6 8 k !0
' e'0.6 ' " 0.6 # e'0.6 1
k!
2
3
! 2000e
'0.6
7
8
" 0.6 #
k!
k !0
k
' 2000e '0.6 ' 1000 " 0.6 #e'0.6 ! 2000 ' 2000e '0.6 ' 600e'0.6
! 573
E $(Y 2 %) ! "1000 # " 0.6 # e '0.6 * " 2000 # e '0.6 8 k ! 2
2
2
7
0.6k
k!
0.6k
2
2
2
' " 2000 # e '0.6 ' $" 2000 # ' "1000 # % " 0.6 # e '0.6
(
)
k!
2
2
2
' " 2000 # e '0.6 ' $" 2000 # ' "1000 # % " 0.6 # e '0.6
(
)
! " 2000 # e '0.6 8 k !0
2
! " 2000 #
2
7
! 816,893
Var ,Y - ! E $(Y 2 %) ' I E ,Y -J ! 816,893 ' " 573# ! 488,564
2
2
Var ,Y - ! 699
Page 28 of 54
68.
Solution: C
Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the
for x C 250
Dx
claim benefits paid. Then Y ! F
and we want to find m such that 0.50
for x 6 250
G250
m
= @ 0.004e '0.004 x dx ! 'e '0.004 x
m
0
= 1 – e–0.004m
0
This condition implies e–0.004m = 0.5 A m = 250 ln 2 = 173.29 .
-------------------------------------------------------------------------------------------------------69.
Solution: D
The distribution function of an exponential random variable
T with parameter O is given by F " t # ! 1 ' e ' t O , t B 0
Since we are told that T has a median of four hours, we may determine O as follows:
1
! F " 4 # ! 1 ' e '4 O
2
1
! e '4 O
2
4
' ln " 2 # ! '
O
O!
4
ln " 2 #
Therefore, Pr "T 6 5 # ! 1 ' F " 5 # ! e '5 O ! e
'
5ln " 2 #
4
! 2'5 4 ! 0.42
-------------------------------------------------------------------------------------------------------70.
Solution: E
Let X denote actual losses incurred. We are given that X follows an exponential
distribution with mean 300, and we are asked to find the 95th percentile of all claims that
exceed 100 . Consequently, we want to find p95 such that
Pr[100 C x C p95 ] F ( p95 ) ' F (100)
0.95 =
!
where F(x) is the distribution function of X .
P[ X B 100]
1 ' F (100)
Now F(x) = 1 – e–x/300 .
1 ' e ' p95 / 300 ' (1 ' e '100 / 300 ) e '1/ 3 ' e ' p95 / 300
!
! 1 ' e1/ 3e ' p95 / 300
Therefore, 0.95 =
1 ' (1 ' e '100 / 300 )
e '1/ 3
e ' p95 / 300 = 0.05 e –1/3
p95 = '300 ln(0.05 e–1/3) = 999
Page 29 of 54
71.
Solution: A
The distribution function of Y is given by
G " y # ! Pr "T 2 9 y # ! Pr T 9 y ! F y ! 1 ' 4 y
"
#
" #
for y B 4 . Differentiate to obtain the density function g " y # ! 4 y '2
Alternate solution:
Differentiate F " t # to obtain f " t # ! 8t '3 and set y ! t 2 . Then t !
g " y # ! f " t " y # # dt dy ! f
" y # dtd " y # ! 8 y
'3 2
y and
. 1 '1 2 /
'2
0 y 1 ! 4y
22
3
-------------------------------------------------------------------------------------------------------72.
Solution: E
We are given that R is uniform on the interval " 0.04, 0.08 # and V ! 10, 000e R
Therefore, the distribution function of V is given by
F " v # ! Pr ,V 9 v - ! Pr $(10, 000e R 9 v %) ! Pr $( R 9 ln " v # ' ln "10, 000 # %)
1 ln " v #'ln "10,000 #
1
!
dr !
r
@
0.04
0.04
0.04
ln " v # ' ln "10,000 #
! 25ln " v # ' 25ln "10, 000 # ' 1
0.04
$ . v /
%
! 25 < ln 0
' 0.04 =
1
( 2 10, 000 3
)
-------------------------------------------------------------------------------------------------------73.
Solution: E
" #
$
F " y # ! Pr ,Y 9 y - ! Pr $(10 X 0.8 9 y %) ! Pr < X 9 Y
10
(
1
1 . Y / 4 ' Y 10 5 4
Therefore, f " y # ! F L " y # ! 0 1 e " #
8 2 10 3
Page 30 of 54
10
' "Y #
%
10
!
'
1
e
=
)
10
8
8
74.
Solution: E
First note R = 10/T . Then
10 %
$10
%
$
. 10 /
FR(r) = P[R 9 r] = P < 9 r = ! P <T 6 = ! 1 ' FT 0 1 . Differentiating with respect to
r)
(T
)
(
2 r 3
.
. 10 / /
.d
/ . '10 /
r fR(r) = FLR(r) = d/dr 01 ' FT 0 1 1 ! ' 0 FT " t # 1 0 2 1
2 r 33
2 dt
32 r 3
2
d
1
FT (t ) ! fT (t ) ! since T is uniformly distributed on [8, 12] .
dt
4
'1 . '10 /
5
Therefore fR(r) =
0 2 1! 2 .
4 2 r 3 2r
-------------------------------------------------------------------------------------------------------75.
Solution: A
Let X and Y be the monthly profits of Company I and Company II, respectively. We are
given that the pdf of X is f . Let us also take g to be the pdf of Y and take F and G to be
the distribution functions corresponding to f and g . Then G(y) = Pr[Y 9 y] = P[2X 9 y]
= P[X 9 y/2] = F(y/2) and g(y) = GL(y) = d/dy F(y/2) = ½ FL(y/2) = ½ f(y/2) .
-------------------------------------------------------------------------------------------------------76.
Solution: A
First, observe that the distribution function of X is given by
x 3
1
1
F " x # ! @ 4 dt ! ' 3 | 1x ! 1 ' 3 , x > 1
1 t
t
x
Next, let X1, X2, and X3 denote the three claims made that have this distribution. Then if
Y denotes the largest of these three claims, it follows that the distribution function of Y is
given by
G " y # ! Pr , X 1 9 y - Pr , X 2 9 y - Pr , X 3 9 y 3
.
1 /
, y>1
! 01 ' 3 1
y 3
2
while the density function of Y is given by
2
.
1 / . 3 / . 9 /.
1 /
g " y # ! G ' " y # ! 3 01 ' 3 1 0 4 1 ! 0 4 10 1 ' 3 1
y 3
2 y 3 2 y 3 2 y 32
Therefore,
Page 31 of 54
2
, y>1
E ,Y - ! @
7
1
2
7 9 .
9 .
1 /
2
1 /
1 ' 3 1 dy ! @ 3 0 1 ' 3 * 6 1 dy
3 0
1 y
y 2 y 3
y 3
2 y
7
. 9 18 9 /
$ 9
18
9 %
! @ 0 3 ' 6 * 9 1 dy ! < ' 2 * 5 ' 8 =
1
y
y 3
2y
( 2 y 5 y 8 y )1
$1 2 1%
! 9 < ' * = ! 2.025 (in thousands)
(2 5 8)
7
-------------------------------------------------------------------------------------------------------77.
Solution: D
Prob. = 1'
Note
2
@@
1
2
1
1
( x * y )dxdy = 0.625
8
I
Pr $(" X 9 1# " "Y 9 1# %) ! Pr $(" X B 1# # "Y B 1# %)
! 1 ' Pr $(" X B 1# # "Y B 1# %)
1 2$
2
2
y * 2 # ' " y * 1# % dy
"
@
)
16 1 (
18 30
! 1'
!
! 0.625
48 48
! 1'
! 1' @
! 1'
2
1
c
J
(De Morgan's Law)
1
@ 8 " x * y # dxdy
2
1
1 $
3
3
y * 2 # ' " y * 1# %
"
)
48 (
1 21
2
x * y # 12 dy
"
@
8 1 2
1
! 1 ' " 64 ' 27 ' 27 * 8 #
48
! 1'
2
1
-------------------------------------------------------------------------------------------------------78.
Solution: B
That the device fails within the first hour means the joint density function must be
integrated over the shaded region shown below.
This evaluation is more easily performed by integrating over the unshaded region and
subtracting from 1.
Page 32 of 54
Pr $(" X C 1# & "Y C 1# %)
! 1' @
3
1
@
3
1
3
2
3 x * 2 xy
1 3
x* y
dx dy ! 1 ' @
dy ! 1 ' @ " 9 * 6 y ' 1 ' 2 y # dy
1
27
54 1
54 1
3
! 1'
1 3
1
1
32 11
8 * 4 y # dy ! 1 ' " 8 y * 2 y 2 # ! 1 ' " 24 * 18 ' 8 ' 2 # ! 1 '
!
! 0.41
"
@
54 1
54
54
54 27
1
-------------------------------------------------------------------------------------------------------79.
Solution: E
The domain of s and t is pictured below.
Note that the shaded region is the portion of the domain of s and t over which the device
fails sometime during the first half hour. Therefore,
1/ 2 1
1 1/ 2
$.
1/ .
1 /%
Pr <0 S 9 1 & 0 T 9 1 = ! @ @ f " s, t # dsdt * @ @ f " s, t # dsdt
0 0
23 2
2 3 ) 0 1/ 2
(2
(where the first integral covers A and the second integral covers B).
-------------------------------------------------------------------------------------------------------80.
Solution: C
By the central limit theorem, the total contributions are approximately normally
distributed with mean n: ! " 2025 #" 3125 # ! 6,328,125 and standard deviation
; n ! 250 2025 ! 11, 250 . From the tables, the 90th percentile for a standard normal
random variable is 1.282 . Letting p be the 90th percentile for total contributions,
p ' n:
! 1.282, and so p ! n: * 1.282 ; n ! 6,328,125 * "1.282 #"11, 250 # ! 6,342,548 .
; n
Page 33 of 54
-------------------------------------------------------------------------------------------------------81.
Solution: C
1
Let X1, . . . , X25 denote the 25 collision claims, and let X !
(X1 + . . . +X25) . We are
25
given that each Xi (i = 1, . . . , 25) follows a normal distribution with mean 19,400 and
standard deviation 5000 . As a result X also follows a normal distribution with mean
1
19,400 and standard deviation
(5000) = 1000 . We conclude that P[ X > 20,000]
25
$ X ' 19, 400 20, 000 ' 19, 400 %
$ X ' 19, 400
%
= P<
B
! P<
B 0.6 = = 1 ' P(0.6) = 1 – 0.7257
=
1000
( 1000
)
( 1000
)
= 0.2743 .
-------------------------------------------------------------------------------------------------------82.
Solution: B
Let X1, . . . , X1250 be the number of claims filed by each of the 1250 policyholders.
We are given that each Xi follows a Poisson distribution with mean 2 . It follows that
E[Xi] = Var[Xi] = 2 . Now we are interested in the random variable S = X1 + . . . + X1250 .
Assuming that the random variables are independent, we may conclude that S has an
approximate normal distribution with E[S] = Var[S] = (2)(1250) = 2500 .
Therefore P[2450 < S < 2600] =
S ' 2500
$ 2450 ' 2500 S ' 2500 2600 ' 2500 %
$
%
C
C
C 2=
P<
= ! P < '1 C
50
2500
2500
2500 )
(
)
(
$ S ' 2500
%
$ S ' 2500
%
! P<
C 2= ' P <
C '1=
( 50
)
( 50
)
S ' 2500
Then using the normal approximation with Z =
, we have P[2450 < S < 2600]
50
! P[Z < 2] – P[Z > 1] = P[Z < 2] + P[Z < 1] – 1 ! 0.9773 + 0.8413 – 1 = 0.8186 .
-------------------------------------------------------------------------------------------------------83.
Solution: B
Let X1,…, Xn denote the life spans of the n light bulbs purchased. Since these random
variables are independent and normally distributed with mean 3 and variance 1, the
random variable S = X1 + … + Xn is also normally distributed with mean
: ! 3n
and standard deviation
;! n
Now we want to choose the smallest value for n such that
$ S ' 3n 40 ' 3n %
0.9772 9 Pr , S > 40- ! Pr <
>
=
n )
( n
This implies that n should satisfy the following inequality:
Page 34 of 54
40 ' 3n
n
To find such an n, let’s solve the corresponding equation for n:
40 ' 3n
'2 !
n
'2 6
'2 n ! 40 ' 3n
3n ' 2 n ' 40 ! 0
"3
n * 10
#"
#
n '4 !0
n !4
n ! 16
-------------------------------------------------------------------------------------------------------84.
Solution: B
Observe that
E , X * Y - ! E , X - * E ,Y - ! 50 * 20 ! 70
Var , X * Y - ! Var , X - * Var ,Y - * 2 Cov , X , Y - ! 50 * 30 * 20 ! 100
for a randomly selected person. It then follows from the Central Limit Theorem that T is
approximately normal with mean
E ,T - ! 100 " 70 # ! 7000
and variance
Var ,T - ! 100 "100 # ! 1002
Therefore,
$ T ' 7000 7100 ' 7000 %
Pr ,T C 7100- ! Pr <
C
=)
100
( 100
! Pr , Z C 1- ! 0.8413
where Z is a standard normal random variable.
Page 35 of 54
-------------------------------------------------------------------------------------------------------85.
Solution: B
Denote the policy premium by P . Since x is exponential with parameter 1000, it follows
from what we are given that E[X] = 1000, Var[X] = 1,000,000, Var[ X ] = 1000 and P =
100 + E[X] = 1,100 . Now if 100 policies are sold, then Total Premium Collected =
100(1,100) = 110,000
Moreover, if we denote total claims by S, and assume the claims of each policy are
independent of the others then E[S] = 100 E[X] = (100)(1000) and Var[S] = 100 Var[X]
= (100)(1,000,000) . It follows from the Central Limit Theorem that S is approximately
normally distributed with mean 100,000 and standard deviation = 10,000 . Therefore,
110, 000 ' 100, 000 %
$
P[S 6 110,000] = 1 – P[S 9 110,000] = 1 – P < Z 9
= = 1 – P[Z 9 1] = 1
10, 000
(
)
– 0.841 Q 0.159 .
-------------------------------------------------------------------------------------------------------86.
Solution: E
Let X 1 ,..., X 100 denote the number of pensions that will be provided to each new recruit.
Now under the assumptions given,
D0 with probability 1 ' 0.4 ! 0.6
E
X i ! F1 with probability " 0.4 #" 0.25 # ! 0.1
E
G2 with probability " 0.4 #" 0.75 # ! 0.3
for i ! 1,...,100 . Therefore,
E , X i - ! " 0 #" 0.6 # * "1#" 0.1# * " 2 #" 0.3# ! 0.7 ,
2
2
2
2
E $( X i %) ! " 0 # " 0.6 # * "1# " 0.1# * " 2 # " 0.3# ! 1.3 , and
2
2
Var , X i - ! E $( X i %) ' I E , X i -J ! 1.3 ' " 0.7 # ! 0.81
Since X 1 ,..., X 100 are assumed by the consulting actuary to be independent, the Central
2
Limit Theorem then implies that S ! X 1 * ... * X 100 is approximately normally distributed
with mean
E , S - ! E , X 1 - * ... * E , X 100 - ! 100 " 0.7 # ! 70
and variance
Var , S - ! Var , X 1 - * ... * Var , X 100 - ! 100 " 0.81# ! 81
Consequently,
$ S ' 70 90.5 ' 70 %
Pr , S 9 90.5- ! Pr <
9
=)
9
( 9
! Pr , Z 9 2.28! 0.99
Page 36 of 54
-------------------------------------------------------------------------------------------------------87.
Solution: D
Let X denote the difference between true and reported age. We are given X is uniformly
distributed on ('2.5,2.5) . That is, X has pdf f(x) = 1/5, '2.5 < x < 2.5 . It follows that
: x = E[X] = 0
;x2 = Var[X] = E[X2] =
2.5
x2
x3
dx
!
@ 5
15
'2.5
2.5
'2.5
!
2(2.5)3
=2.083
15
;x =1.443
Now X 48 , the difference between the means of the true and rounded ages, has a
1.443
distribution that is approximately normal with mean 0 and standard deviation
=
48
0.2083 . Therefore,
1%
0.25 %
$ 1
$ '0.25
= P['1.2 9 Z 9 1.2] = P[Z 9 1.2] – P[Z 9 –
P < ' 9 X 48 9 = ! P <
9Z9
4)
0.2083 =)
( 4
( 0.2083
1.2]
= P[Z 9 1.2] – 1 + P[Z 9 1.2] = 2P[Z 9 1.2] – 1 = 2(0.8849) – 1 = 0.77 .
-------------------------------------------------------------------------------------------------------88.
Solution: C
Let X denote the waiting time for a first claim from a good driver, and let Y denote the
waiting time for a first claim from a bad driver. The problem statement implies that the
respective distribution functions for X and Y are
F " x # ! 1 ' e' x / 6 , x > 0
and
G " y # ! 1 ' e' y / 3 , y B 0
Therefore,
Pr $(" X 9 3# + "Y 9 2 # %) ! Pr , X 9 3- Pr ,Y 9 2-
! F " 3# G " 2 # ! "1 ' e '1/ 2 #"1 ' e '2 / 3 # ! 1 ' e'2 / 3 ' e'1/ 2 * e'7 / 6
Page 37 of 54
89.
Solution: B
D 6
(50 ' x ' y )
E
We are given that f ( x, y ) ! F125, 000
E0
G
for 0 C x C 50 ' y C 50
otherwise
and we want to determine P[X > 20 + Y > 20] . In order to determine integration limits,
consider the following diagram:
y
50
x>20 y>20
(20, 30)
(30, 20)
50
x
30 50 ' x
6
We conclude that P[X > 20 + Y > 20] =
125, 000 20@
@
(50 ' x ' y ) dy dx .
20
-------------------------------------------------------------------------------------------------------90.
Solution: C
Let T1 be the time until the next Basic Policy claim, and let T2 be the time until the next
Deluxe policy claim. Then the joint pdf of T1 and T2 is
.1
/. 1
/ 1
f (t1 , t2 ) ! 0 e ' t1 / 2 1 0 e' t2 / 3 1 ! e' t1 / 2 e' t2 / 3 , 0 < t1 < 7 , 0 < t2 < 7 and we need to find
22
32 3
3 6
7 t1
P[T2 < T1] =
7
1 't / 2 't / 3
$ 1 ' t / 2 ' t / 3 % t1
@0 @0 6e 1 e 2 dt2 dt1 ! @0 <( ' 2 e 1 e 2 =) 0 dt1
7
7
7
3 2
$ 1 ' t1 / 2 1 ' t1 / 2 ' t1 / 3 %
$ 1 't1 / 2 1 '5t1 / 6 %
$ 't1 / 2 3 '5t1 / 6 %
' e e = dt1 ! @ < e
' e
dt1 = < 'e
* e
! 1' !
= @< e
=
=
5
5 5
2
2
2
2
)
)
(
)0
0 (
0 (
= 0.4 .
-------------------------------------------------------------------------------------------------------91.
Solution: D
We want to find P[X + Y > 1] . To this end, note that P[X + Y > 1]
1 2
2
1
1
1 %
$ 2x * 2 ' y %
$1
= @@<
dydx ! @ < xy * y ' y 2 = dx
=
4
2
2
8 ) 1' x
)
0 1' x (
0 (
1
=
1 1
1
1
$
2%
@0 <( x * 1 ' 2 ' 2 x(1 ' x) ' 2 (1 ' x) * 8 (1 ' x) =) dx =
1
=
$5
@ <( 8 x
0
1
2
*
1
$
1
@ <( x * 2 x
0
2
1 1
1 %
* ' x * x 2 = dx
8 4
8 )
3
1 %
3
1%
5 3 1 17
$5
* * !
x * = dx = < x 3 * x 2 * x = =
8
8 ) 0 24 8 8 24
4
8)
( 24
Page 38 of 54
92.
Solution: B
Let X and Y denote the two bids. Then the graph below illustrates the region over which
X and Y differ by less than 20:
Based on the graph and the uniform distribution:
Pr $( X ' Y C 20 %) !
Shaded Region Area
" 2200 ' 2000 #
2
!
1
2
"180 #
2
2002
2002 ' 2 5
1802
2
! 1 ' " 0.9 # ! 0.19
2
200
More formally (still using symmetry)
Pr $( X ' Y C 20 %) ! 1 ' Pr $( X ' Y 6 20 %) ! 1 ' 2 Pr , X ' Y 6 20! 1'
2200
1
1
x ' 20
dydx ! 1 ' 2 @
y 2000
dx
2
2020 2000 200
2020 200 2
2200
2
1
2
! 1'
x ' 20 ' 2000 # dx ! 1 '
x ' 2020 #
2 @ 2020 "
2 "
200
200
! 1 ' 2@
2200
@
x ' 20
2
. 180 /
! 1' 0
1 ! 0.19
2 200 3
Page 39 of 54
2200
2020
-------------------------------------------------------------------------------------------------------93.
Solution: C
Define X and Y to be loss amounts covered by the policies having deductibles of 1 and 2,
respectively. The shaded portion of the graph below shows the region over which the
total benefit paid to the family does not exceed 5:
We can also infer from the graph that the uniform random variables X and Y have joint
1
, 0 C x C 10 , 0 C y C 10
density function f " x, y # !
100
We could integrate f over the shaded region in order to determine the desired probability.
However, since X and Y are uniform random variables, it is simpler to determine the
portion of the 10 x 10 square that is shaded in the graph above. That is,
Pr " Total Benefit Paid Does not Exceed 5 #
! Pr " 0 C X C 6, 0 C Y C 2 # * Pr " 0 C X C 1, 2 C Y C 7 # * Pr "1 C X C 6, 2 C Y C 8 ' X #
!
" 6 #" 2 # * "1#" 5 # * "1 2 #" 5#" 5 # !
100
100
100
12
5 12.5
*
*
! 0.295
100 100 100
-------------------------------------------------------------------------------------------------------94.
Solution: C
Let f " t1 , t2 # denote the joint density function of T1 and T2 . The domain of f is pictured
below:
Now the area of this domain is given by
1
2
A ! 62 ' " 6 ' 4 # ! 36 ' 2 ! 34
2
Page 40 of 54
D1
, 0 C t1 C 6 , 0 C t2 C 6 , t1 * t2 C 10
E
Consequently, f " t1 , t2 # ! F 34
EG 0
elsewhere
and
E ,T1 * T2 - ! E ,T1 - * E ,T2 - ! 2 E ,T1 (due to symmetry)
6 1
6
10 't1 1
6
D 4 $t
%
$t
D 4
R
! 2 F @ t1 @
dt2 dt1 * @ t1 @
dt2 dt1 S ! 2 F @ t1 < 2 06 = dt1 * @ t1 < 2
0 34
4
0
4
34
G 0
T
( 34
G 0 ( 34 )
6 1
D 3t 2
D 4 3t1
R
! 2 F@
10t1 ' t12 # dt1 S ! 2 F 1
dt1 * @
"
4 34
G 0 17
T
G 34
4
0
*
10 ' t1
0
% R
= dt1 S
) T
1 . 2 1 3/ 6R
0 5t1 ' t1 1 4 S
34 2
3 3 T
D 24 1 $
64 % R
! 2 F * <180 ' 72 ' 80 * = S ! 5.7
3 )T
G 17 34 (
-------------------------------------------------------------------------------------------------------95.
Solution: E
t X *Y * t Y ' X
t 't X t *t Y
M " t1 , t2 # ! E $( et1W *t2 Z %) ! E $(e 1 " # 2 " # %) ! E $(e" 1 2 # e" 1 2 # %)
1
t 't X
t *t Y
! E $(e" 1 2 # %) E $( e" 1 2 # %) ! e 2
" t1 't2 #2
1
e2
" t1 * t2 #2
1
! e2
"t
2
2
1 ' 2 t1t2 * t2
# 12 "t
e
2
2
1 * 2 t1t2 * t2
#
2
! et1 *t2
2
-------------------------------------------------------------------------------------------------------96.
Solution: E
Observe that the bus driver collect 21x50 = 1050 for the 21 tickets he sells. However, he
may be required to refund 100 to one passenger if all 21 ticket holders show up. Since
passengers show up or do not show up independently of one another, the probability that
21
21
all 21 passengers will show up is "1 ' 0.02 # ! " 0.98 # ! 0.65 . Therefore, the tour
operator’s expected revenue is 1050 ' "100 #" 0.65 # ! 985 .
Page 41 of 54
97.
Solution: C
We are given f(t1, t2) = 2/L2, 0 9 t1 9 t2 9 L .
L t2
2
2 2
2
2
Therefore, E[T1 + T2 ] = @ @ (t1 * t2 ) 2 dt1dt2 =
L
0 0
t2
3
3
L
L
ED $ t1
ER 2 ED . t2
ER
2 %
3/
F @ < * t2 t1 = dt1 S ! 2 F @ 0 * t2 1 dt2 S
ET L GE 0 2 3
)0
3 TE
GE 0 ( 3
4 L
L
2 4 3
2 $t %
2
= 2 @ t2 dt2 ! 2 < 2 = ! L2
L 03
L ( 3 )0
3
t2
2
L2
(L, L)
t1
-------------------------------------------------------------------------------------------------------98.
Solution: A
Let g(y) be the probability function for Y = X1X2X3 . Note that Y = 1 if and only if
X1 = X2 = X3 = 1 . Otherwise, Y = 0 . Since P[Y = 1] = P[X1 = 1 + X2 = 1 + X3 = 1]
= P[X1 = 1] P[X2 = 1] P[X3 = 1] = (2/3)3 = 8/27 .
D 19
for y ! 0
E 27
E
E8
We conclude that g ( y ) ! F
for y ! 1
27
E
otherwise
E0
E
G
19 8 t
and M(t) = E $(e yt %) !
* e
27 27
Page 42 of 54
99.
Solution: C
We use the relationships Var " aX * b # ! a 2 Var " X # , Cov " aX , bY # ! ab Cov " X , Y # , and
Var " X * Y # ! Var " X # * Var "Y # * 2 Cov " X , Y # . First we observe
17, 000 ! Var " X * Y # ! 5000 * 10, 000 * 2 Cov " X , Y # , and so Cov " X , Y # ! 1000.
We want to find Var $(" X * 100 # * 1.1Y %) ! Var $(" X * 1.1Y # * 100 %)
! Var , X * 1.1Y - ! Var X * Var $("1.1# Y %) * 2 Cov " X ,1.1Y #
! Var X * "1.1# Var Y * 2 "1.1# Cov " X , Y # ! 5000 * 12,100 * 2200 ! 19,300.
2
-------------------------------------------------------------------------------------------------------100.
Solution: B
Note
P(X = 0) = 1/6
P(X = 1) = 1/12 + 1/6 = 3/12
P(X = 2) = 1/12 + 1/3 + 1/6 = 7/12 .
E[X] = (0)(1/6) + (1)(3/12) + (2)(7/12) = 17/12
E[X2] = (0)2(1/6) + (1)2(3/12) + (2)2(7/12) = 31/12
Var[X] = 31/12 – (17/12)2 = 0.58 .
-------------------------------------------------------------------------------------------------------101.
Solution: D
Note that due to the independence of X and Y
Var(Z) = Var(3X ' Y ' 5) = Var(3X) + Var(Y) = 32 Var(X) + Var(Y) = 9(1) + 2 = 11 .
-------------------------------------------------------------------------------------------------------102.
Solution: E
Let X and Y denote the times that the two backup generators can operate. Now the
variance of an exponential random variable with mean U is U 2 . Therefore,
Var , X - ! Var ,Y - ! 102 ! 100
Then assuming that X and Y are independent, we see
Var , X+Y - ! Var , X - * Var , Y - ! 100 * 100 ! 200
Page 43 of 54
103.
Solution: E
Let X 1 , X 2 , and X 3 denote annual loss due to storm, fire, and theft, respectively. In
addition, let Y ! Max " X 1 , X 2 , X 3 # .
Then
Pr ,Y B 3- ! 1 ' Pr ,Y 9 3- ! 1 ' Pr , X 1 9 3- Pr , X 2 9 3- Pr , X 3 9 3-
"
! 1 ' "1 ' e '3 # 1 ' e
#"1 ' e #
# "1 ' e #
'3
1.5
! 1 ' "1 ' e '3 #"1 ' e '2
'3
'5
2.4
*
4
! 0.414
* Uses that if X has an exponential distribution with mean :
7
Pr " X 9 x # ! 1 ' Pr " X 6 x # ! 1 ' @
x
1
:
e ' t : dt ! 1 ' " 'e' t : # 7x ! 1 ' e' x :
-------------------------------------------------------------------------------------------------------104.
Solution: B
Let us first determine k:
1! @
1
0
@
1
0
kxdxdy ! @
1k
1 2 1
k
kx | 0 dy ! @ dy !
0 2
0 2
2
1
k !2
Then
1 1
E , X - ! @ @ 2 x 2 dydx ! @ 2 x 2 dx !
1
0
0 0
1 1
E ,Y - ! @ @ y 2 x dxdy ! @ ydy !
1
0
0 0
E , XY - ! @
!
1
0
@
1
0
2x 2 ydxdy ! @
2 31 2
x |0!
3
3
1 2 1 1
y |0!
2
2
12
2 3 1
x y | 0 dy ! @ ydy
0 3
0 3
1
2 2 1 2 1
y |0 ! !
6
6 3
1 . 2 /. 1 / 1 1
Cov , X , Y - ! E , XY - ' E , X - E ,Y - ! ' 0 10 1 ! ' ! 0
3 2 3 32 2 3 3 3
(Alternative Solution)
Define g(x) = kx and h(y) = 1 . Then
f(x,y) = g(x)h(x)
In other words, f(x,y) can be written as the product of a function of x alone and a function
of y alone. It follows that X and Y are independent. Therefore, Cov[X, Y] = 0 .
Page 44 of 54
105.
Solution: A
The calculation requires integrating over the indicated region.
E"X # ! @
1
0
E "Y # ! @
1
0
@
2x
x
@
18
8 2
xy dy dx ! @ xy 3
0 9
3
2x
x
E " XY # ! @
1
0
14
8 2
x y dy dx ! @ x 2 y 2
0 3
3
@
2x
x
2x
x
2x
x
1
1
4
4
4
dx ! @ x 2 " 4 x 2 ' x 2 # dx ! @ 4 x 4 dx ! x5 !
0 3
0
5 0 5
1
1
1 56
8
56 5
56
dy dx ! @ x " 8 x3 ' x3 # dx ! @
x 4 dx !
x !
0 9
0 9
45 0 45
18
8 2 2
x y dy dx ! @ x 2 y 3
0 9
3
Cov " X , Y # ! E " XY # ' E " X # E "Y # !
1
2x
x
1 56
56 28
8 2
!
x " 8 x3 ' x3 # dx ! @
x5 dx !
0 9
0 9
54 27
dx ! @
1
28 . 56 /. 4 /
' 0 10 1 ! 0.04
27 2 45 32 5 3
-------------------------------------------------------------------------------------------------------106.
Solution: C
The joint pdf of X and Y is f(x,y) = f2(y|x) f1(x)
= (1/x)(1/12), 0 < y < x, 0 < x < 12 .
Therefore,
12 x
12
12
1
y x
x
x 2 12
=6
dydx ! @
dx ! @ dx !
E[X] = @ @ x 5
12 x
12 0
12
24 0
0 0
0
0
x
12
12
$ y2 %
y
x
x 2 12 144
E[Y] = @ @
=3
dydx ! @ <
dx ! @ dx !
!
12 x
24 x =) 0
24
48 0
48
0 0
0 (
0
12 x
x
12
12 2
$ y2 %
y
x
x3 12
(12)3
= 24
dydx
!
dx
!
dx
!
!
@0 @0 12
@0 <( 24 =)
@0 24 72 0
72
0
12 x
E[XY] =
Cov(X,Y) = E[XY] – E[X]E[Y] = 24 ' (3)(6) = 24 – 18 = 6 .
Page 45 of 54
107.
Solution: A
Cov " C1 , C2 # ! Cov " X * Y , X * 1.2Y #
! Cov " X , X # * Cov "Y , X # * Cov " X ,1.2Y # * Cov " Y,1.2Y #
! Var X * Cov " X , Y # * 1.2Cov " X , Y # * 1.2VarY
! Var X * 2.2 Cov " X , Y # * 1.2VarY
Var X ! E " X 2 # ' " E " X # # ! 27.4 ' 52 ! 2.4
2
Var Y ! E "Y 2 # ' " E "Y # # ! 51.4 ' 7 2 ! 2.4
2
Var " X * Y # ! Var X * Var Y * 2 Cov " X , Y #
1
1
Var " X * Y # ' Var X ' Var Y # ! " 8 ' 2.4 ' 2.4 # ! 1.6
"
2
2
Cov " C1 , C2 # ! 2.4 * 2.2 "1.6 # * 1.2 " 2.4 # ! 8.8
Cov " X , Y # !
-------------------------------------------------------------------------------------------------------107.
Alternate solution:
We are given the following information:
C1 ! X * Y
C2 ! X * 1.2Y
E,X - ! 5
E $( X 2 %) ! 27.4
E ,Y - ! 7
E $(Y 2 %) ! 51.4
Var , X * Y - ! 8
Now we want to calculate
Cov " C1 , C2 # ! Cov " X * Y , X * 1.2Y #
! E $(" X * Y #" X * 1.2Y # %) ' E , X * Y -# E , X * 1.2Y -
! E $( X 2 * 2.2 XY * 1.2Y 2 %) ' " E , X - * E ,Y -# " E , X - * 1.2 E ,Y -#
! E $( X 2 %) * 2.2 E , XY - * 1.2 E $(Y 2 %) ' " 5 * 7 # " 5 * "1.2 # 7 #
! 27.4 * 2.2 E , XY - * 1.2 " 51.4 # ' "12 #"13.4 #
! 2.2 E , XY - ' 71.72
Therefore, we need to calculate E , XY - first. To this end, observe
Page 46 of 54
2
2
8 ! Var , X * Y - ! E $" X * Y # % ' " E , X * Y -#
(
)
! E $( X 2 * 2 XY * Y 2 %) ' " E , X - * E ,Y -#
! E $( X 2 %) * 2 E , XY - * E $(Y 2 %) ' " 5 * 7 #
2
2
! 27.4 * 2 E , XY - * 51.4 ' 144
! 2 E , XY - ' 65.2
E , XY - ! " 8 * 65.2 # 2 ! 36.6
Finally, Cov " C1,C2 # ! 2.2 " 36.6 # ' 71.72 ! 8.8
-------------------------------------------------------------------------------------------------------108.
Solution: A
The joint density of T1 and T2 is given by
f " t1 , t2 # ! e ' t1 e' t2 , t1 B 0 , t2 B 0
Therefore,
Pr , X 9 x - ! Pr , 2T1 * T2 9 x !@
x
0
@
1
" x ' t2 #
2
0
$
x
e ' t1 e 't2 dt1dt2 ! @ e ' t2 < 'e' t1
0
<(
1
" x ' t2 #
2
0
%
= dt2
=)
1 1
1
1
' x * t2 %
' x ' t2 /
x
x.
$
! @ e ' t2 <1 ' e 2 2 = dt2 ! @ 0 e ' t2 ' e 2 e 2 1dt2
0
0
(
)
2
3
1
1
' x ' t2 %
$
! < ' e ' t 2 * 2e 2 e 2 =
(
)
'
1
x
0
! ' e ' x * 2e
x
'
1
x
'
1
1
x ' x
2
2
e
* 1 ' 2e
! 1 ' e ' x * 2e ' x ' 2 e 2 ! 1 ' 2 e 2 * e ' x , x B 0
It follows that the density of X is given by
1
1
' x
' x
%
d $
'x
2
2
'
*
!
' e' x , x B 0
1
2
g " x# !
e
e
e
<
=
dx (
)
Page 47 of 54
'
1
x
2
109.
Solution: B
Let
u be annual claims,
v be annual premiums,
g(u, v) be the joint density function of U and V,
f(x) be the density function of X, and
F(x) be the distribution function of X.
Then since U and V are independent,
.1
/ 1
g " u, v # ! " e'u # 0 e' v / 2 1 ! e'u e' v / 2 , 0 < u < 7 , 0 < v < 7
22
3 2
and
$u
%
F " x # ! Pr , X 9 x - ! Pr < 9 x = ! Pr ,U 9 Vx (v
)
7 vx
7 vx 1
e 'u e' v / 2 dudv
! @ @ g " u, v #dudv ! @ @
0 0
0 0 2
7
7.
1
1
1
/
! @ ' e ' u e ' v / 2 | 0vx dv ! @ 0 ' e' vx e' v / 2 * e' v / 2 1 dv
0
0
2
2
2 2
3
7.
1 ' v x *1/ 2# 1 ' v / 2 /
! @ 0' e "
* e
1 dv
0
2
2 2
3
7
1
$ 1
%
' v x *1/ 2 #
*1
e "
!<
' e' v / 2 = ! '
2x * 1
( 2x * 1
)0
2
Finally, f " x # ! F ' " x # !
2
" 2 x * 1#
-------------------------------------------------------------------------------------------------------110.
Solution: C
Note that the conditional density function
.
1 / f "1 3, y #
2
f 0y x! 1!
, 0C yC ,
f x "1 3#
33
3
2
23
23
2 3
16
.1/
f x 0 1 ! @ 24 "1 3# y dy ! @ 8 y dy ! 4 y 2 !
0
9
0
2 33 0
.
1/ 9
9
2
It follows that f 0 y x ! 1 !
f "1 3, y # ! y , 0 C y C
3 3 16
2
3
2
139
9
Consequently, Pr $(Y C X X ! 1 3%) ! @
y dy ! y 2
0 2
4
Page 48 of 54
13
!
0
1
4
111.
Solution: E
3 f " 2, y #
Pr $(1 C Y C 3 X ! 2 %) ! @
dy
1
f x " 2#
f " 2, y # !
2
1
' 4 '1 2 '1
y " # ! y '3
4 " 2 ' 1#
2
7
7
1
1
1
f x " 2 # ! @ y '3 dy ! ' y '2 !
2
4
4
1
1
Finally, Pr $(1 C Y C 3 X ! 2 %) !
@
3
1
1 '3
y dy
3
1 8
2
! ' y '2 ! 1 ' !
1
1
9 9
4
-------------------------------------------------------------------------------------------------------112.
Solution: D
We are given that the joint pdf of X and Y is f(x,y) = 2(x+y), 0 < y < x < 1 .
x
Now fx(x) = @ (2 x * 2 y )dy ! $( 2 xy * y 2 %)
0
x
0
= 2x2 + x2 = 3x2, 0 < x < 1
f ( x, y ) 2( x * y ) 2 . 1 y /
!
! 0 * 2 1, 0 < y < x
f x ( x)
3x 2
32 x x 3
2$ 1
y % 2
f(y|x = 0.10) = <
*
! ,10 * 100 y - , 0 < y < 0.10
3 ( 0.1 0.01 =) 3
so f(y|x) =
0.05
P[Y < 0.05|X = 0.10] =
2
100
$ 20
@ 3 ,10 * 100 y - dy ! <( 3 y * 3 y
0
2
5
% 0.05 1 1
! * !
= 0.4167 .
=)
3 12 12
0
-------------------------------------------------------------------------------------------------------113.
Solution: E
Let
W = event that wife survives at least 10 years
H = event that husband survives at least 10 years
B = benefit paid
P = profit from selling policies
Then Pr , H - ! P , H + W - * Pr $( H + W c %) ! 0.96 * 0.01 ! 0.97
and
Pr ,W | H - !
Pr ,W + H - 0.96
!
! 0.9897
Pr , H 0.97
Pr $(W c | H %) !
Pr $( H + W c %)
Pr , H -
!
0.01
! 0.0103
0.97
Page 49 of 54
It follows that
E , P - ! E ,1000 ' B - ! 1000 ' E , B - ! 1000 ' " 0 # Pr ,W | H - * "10, 000 # Pr $(W c | H %)
! 1000 ' 10, 000 " 0.0103# ! 1000 ' 103 ! 897
I
J
-------------------------------------------------------------------------------------------------------114.
Solution: C
Note that
P(Y = 0>X = 1) =
P( X ! 1, Y ! 0)
P( X ! 1, Y ! 0)
0.05
!
!
P( X ! 1)
P ( X ! 1, Y ! 0) * P( X ! 1, Y ! 1) 0.05 * 0.125
= 0.286
P(Y = 1>X=1) = 1 – P(Y = 0 > X = 1) = 1 – 0.286 = 0.714
Therefore, E(Y>X = 1) = (0) P(Y = 0>X = 1) + (1) P(Y = 1>X = 1) = (1)(0.714) = 0.714
E(Y2>X = 1) = (0)2 P(Y = 0>X = 1) + (1)2 P(Y = 1>X = 1) = 0.714
Var(Y>X = 1) = E(Y2>X = 1) – [E(Y>X = 1)]2 = 0.714 – (0.714)2 = 0.20
-------------------------------------------------------------------------------------------------------115.
Solution: A
Let f1(x) denote the marginal density function of X. Then
f1 " x # ! @
x *1
x
2 xdy ! 2 xy | xx *1 ! 2 x " x * 1 ' x # ! 2 x
,
0< x<1
Consequently,
f " x, y # D1 if: x < y < x * 1
!F
f1 " x #
G0 otherwise
x *1
1
1
1
1
1 1
1
2
E ,Y | X - ! @ ydy ! y 2 | xx *1 ! " x * 1# ' x 2 ! x 2 * x * ' x 2 ! x *
x
2
2
2
2
2 2
2
x *1
1
1
1
3
E $(Y 2 | X %) ! @ y 2 dy ! y 3 | xx *1 ! " x * 1# ' x3
x
3
3
3
1
1 1
1
! x3 * x 2 * x * ' x3 ! x 2 * x *
3
3 3
3
f " y| x # !
1 .
1/
Var ,Y | X - ! E $(Y | X %) ' I E ,Y | X -J ! x * x * ' 0 x * 1
3 2
23
1
1 1
! x2 * x * ' x2 ' x ' !
3
4 12
2
2
Page 50 of 54
2
2
116.
Solution: D
Denote the number of tornadoes in counties P and Q by NP and NQ, respectively. Then
E[NQ|NP = 0]
= [(0)(0.12) + (1)(0.06) + (2)(0.05) + 3(0.02)] / [0.12 + 0.06 + 0.05 + 0.02] = 0.88
E[NQ2|NP = 0]
= [(0)2(0.12) + (1)2(0.06) + (2)2(0.05) + (3)2(0.02)] / [0.12 + 0.06 + 0.05 + 0.02]
= 1.76 and Var[NQ|NP = 0] = E[NQ2|NP = 0] – {E[NQ|NP = 0]}2 = 1.76 – (0.88)2
= 0.9856 .
-------------------------------------------------------------------------------------------------------117.
Solution: C
The domain of X and Y is pictured below. The shaded region is the portion of the domain
over which X<0.2 .
Now observe
Pr , X < 0.2- ! @
0.2
0
@
1' x
0
0.2
6 $(1 ' " x * y # %)dydx ! 6 @
0
1' x
1 2%
$
<( y ' xy ' 2 y =) dx
0
0.2 $
0.2 $
1
1
2%
2
2%
! 6@ <1 ' x ' x "1 ' x # ' "1 ' x # = dx ! 6 @ <"1 ' x # ' "1 ' x # = dx
0
0
2
2
(
)
(
)
0.2 1
2
3
3
! 6@
"1 ' x # dx ! ' "1 ' x # | 0.2
0 ! ' " 0.8 # * 1 ! 0.488
0 2
-------------------------------------------------------------------------------------------------------118.
Solution: E
The shaded portion of the graph below shows the region over which f " x, y # is nonzero:
We can infer from the graph that the marginal density function of Y is given by
y
y
g " y # ! @ 15 y dx ! 15 xy
y
! 15 y
y
"
#
y ' y ! 15 y 3 2 "1 ' y1 2 # , 0 C y C 1
Page 51 of 54
12
32
ED15 y "1 ' y # , 0 C y C 1
or more precisely, g " y # ! F
otherwise
EG0
-------------------------------------------------------------------------------------------------------119.
Solution: D
The diagram below illustrates the domain of the joint density f " x, y # of X and Y .
We are told that the marginal density function of X is f x " x # ! 1 , 0 C x C 1
while f y x " y x # ! 1 , x C y C x * 1
if 0 C x C 1 , x C y C x * 1
otherwise
D1
It follows that f " x, y # ! f x " x # f y x " y x # ! F
G0
Therefore,
Pr ,Y B 0.5- ! 1 ' Pr ,Y 9 0.5- ! 1 ' @
1
0
2
@
1
x
2
dydx
1 % 1
1 1 7
.1
/
$1
' x 1 dx ! 1 ' < x ' x 2 = 0 2 ! 1 ' * !
0
0
0
2 )
4 8 8
22
3
(2
[Note since the density is constant over the shaded parallelogram in the figure the
solution is also obtained as the ratio of the area of the portion of the parallelogram above
y ! 0.5 to the entire shaded area.]
! 1' @
1
2
y
1
2
x
dx ! 1 ' @
1
2
Page 52 of 54
120.
Solution: A
We are given that X denotes loss. In addition, denote the time required to process a claim
by T.
D3 2 1 3
E x 5 ! x , x C t C 2 x, 0 9 x 9 2
Then the joint pdf of X and T is f ( x, t ) ! F 8
x 8
EG0,
otherwise.
Now we can find P[T 6 3] =
4
4
2
4
3
12
$ 3 2%
. 12 3 2 /
$12 1 3 %
. 36 27 /
xdxdt
!
x
dt
!
'
t
dt
!
'
t
@3 t@/ 2 8
@3 <(16 =) t / 2 @3 20 16 64 31 (<16 64 )= 3 ! 4 ' 1 ' 20 16 ' 64 31
= 11/64 = 0.17 .
t
t = 2x
4 2
4
3
2
1
t=x
x
1 2
-------------------------------------------------------------------------------------------------------121.
Solution: C
The marginal density of X is given by
1
1
1 .
xy 3 /
1 .
x/
2
fx " x# ! @
10
'
xy
dy
!
10
y
'
! 010 ' 1
"
#
0
1
0 64
64 2
3 3 0 64 2
33
1
10
10
10
2
2
Then E ( X ) ! @ x f x ( x)dx ! @
=
1 . 2 x3 /
1 .
x2 /
0 5x ' 1
010 x ' 1 dx =
64 2
9 32
64 2
33
1 $.
1000 / .
8 /%
0 500 '
1 ' 0 20 ' 1 = = 5.778
<
64 (2
9 3 2
9 3)
Page 53 of 54
122.
Solution: D
y
The marginal distribution of Y is given by f2(y) = @ 6 e e
–x
y
–2y
dx = 6 e
–2y
0
@e
'x
dx
0
= '6 e–2y e–y + 6e–2y = 6 e–2y – 6 e–3y, 0 < y < 7
7
Therefore, E(Y) =
@y
0
7
7
7
f2(y) dy = @ (6 ye
'2 y
' 6 ye
'3 y
7
) dy = 6
0
@ ye
'2 y
7
dy – 6
0
@y
e–3y dy =
0
6
6
2 ye–2y dy ' @ 3 y e–3y dy
@
20
30
7
7
0
0
But @ 2 y e–2y dy and @ 3 y e–3y dy are equivalent to the means of exponential random
7
variables with parameters 1/2 and 1/3, respectively. In other words, @ 2 y e–2y dy = 1/2
0
7
and @ 3 y e–3y dy = 1/3 . We conclude that E(Y) = (6/2) (1/2) – (6/3) (1/3) = 3/2 – 2/3 =
0
9/6 ' 4/6 = 5/6 = 0.83 .
-------------------------------------------------------------------------------------------------------123.
Solution: C
Observe
Pr , 4 C S C 8- ! Pr $( 4 C S C 8 N ! 1%) Pr , N ! 1- * Pr $( 4 C S C 8 N B 1%) Pr , N B 1'8
1 '4
1 '1
! e 5 ' e 5 * e 2 ' e '1 *
3
6
! 0.122
*Uses that if X has an exponential distribution with mean :
"
# "
#
7
Pr " a 9 X 9 b # ! Pr " X 6 a # ' Pr " X 6 b # ! @
a
1
:
e
Page 54 of 54
't :
7
dt ' @
b
1
:
e
't :
dt ! e
'
a
:
'e
'
b
: