EXAM P SAMPLE SOLUTIONS

SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY 

EXAM P PROBABILITY 

EXAM P SAMPLE SOLUTIONS 

Copyright 2007 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-09-07 PRINTED IN U.S.A. 

Page 1 of 55

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 . 

-------------------------------------------------------------------------------------------------------- 


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' = 2P A + P B + P B' − P A∩ B + P A∩B' [ ] [ ] [ ] ( [ ] [ ] ) ( [ ] [ ] ) 

+ = P[ A] + −P⎡⎣( A∩B) ∪( A∩B ) ⎤⎦ 

= P[ A] + −P[ 

A] 

[ ] = 0.6 

0.7 0.9 2 1 ' 

1.6 

P A 

2 1 

Page 2 of 55


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[ R IR U B IB ] = Pr[ R IR ] + Pr B IB 

= Pr[ R1] Pr[ R2] + Pr[ B1] Pr[ 

B2] 

4 ⎛ 16 ⎞ 6 ⎛ x ⎞ 

= ⎜ ⎟+ ⎜ ⎟ 

10 ⎝ x+ 16 ⎠ 10 ⎝ x+ 

16 ⎠ 

Therefore, 

32 3x 3x+ 32 

2.2 = + = 

x+ 16 x+ 16 x+ 

16 

2.2x+ 35.2 = 3x+ 32 

0.8x = 3.2 

x = 4 

( ) ( ) [ ] 

1 2 1 2 1 2 1 2 

-------------------------------------------------------------------------------------------------------- 


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, 

c 210 −102 

108 

P⎡ ⎣H ∩ F ⎤ 

⎦ = = 

937 937 

c 937 − 312 625 

P⎡ ⎣F ⎤ 

⎦ = = 

937 937 

c 

P⎡H ∩ F ⎤ 

c 

108 625 108 

and P⎡H | F 

⎣ ⎦ 

⎣ 

⎤ 

⎦ = = = = 0.173 

c 

P⎡ ⎣F ⎤ 

⎦ 

937 937 625 

Page 3 of 55


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 

( ) 

( A 

c 

H ) ( A) ( A H) 

c ( A H) ( H) ( A H) 

Pr ∩ = Pr −Pr∩ = 0.65 − 0.15 = 0.50 

Pr ∩ = Pr −Pr∩ = 0.50 − 0.15 = 0.35 

and the portion of policyholders that will renew at least one policy is given by 

0.4 Pr 

c 

A∩ H + 0.6 Pr 

c 

A ∩ H + 0.8 Pr A∩H ( ) ( ) ( ) 

( )( ) ( )( ) ( )( ) ( 

= 0.4 0.5 + 0.6 0.35 + 0.8 0.15 = 0.53 = 53% 

-------------------------------------------------------------------------------------------------------- 

100292 01B-9 


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 CIT = 0.22 

( ) 

c c ( C IT 

) 

Pr 

Therefore, 

= 0.12 

c c 

0.88 = 1− Pr ⎡ 

⎣C IT ⎤ 

⎦ = Pr CUT = Pr[ T] + 0.14 + Pr[ T] 

−0.22 

= 2Pr [ T ] −0.08 

= Pr C + Pr T −PrCIT 

or 

Pr[ T ] = ( 0.88 + 0.08) 2 = 0.48 

[ ] [ ] [ ] [ ] 

Page 4 of 55 

)


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 c 

c 

Pr ( M ∩ S ) = 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 ( A c ∩ B c ) = 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 | 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 (A c ∩ B c ) = 0.744 = 

0.256 

-------------------------------------------------------------------------------------------------------- 


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 

2 

(P[D]) = 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 C C and D C . As a 

result, we see that P[C C ∩ D C ] = P[C C ] P[D C ] = (1 – P[C]) (1 – P[D]) 

= (1 – 2 0.075 ) (1 – 0.075 ) = 0.33 . 

Page 5 of 55

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. 

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The Venn diagram below summarizes the unconditional probabilities described in the 

problem. 

In addition, we are told that 

1 

P[ A∩B∩C] x 

= P[ A∩B∩C| A∩ B] 

= = 

3 P A∩ B x+ 

0.12 

[ ] 

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 c 

P⎡( A∪B∪ C) | A ⎤ = 

⎣ ⎦ 

⎣ ⎦ c 

P⎡ ⎣A ⎤ 

⎦ 

1−P[ 

A∪B∪C] = 

1− 

P A 

[ ] 

( ) ( ) 

( ) 

1−30.10 −30.12 −0.06 

= 

1−0.10 −20.12 −0.06 

0.28 

= = 0.467 

0.60 

Page 6 of 55


1 11 1 1 1 ⎛1⎞ p k = pk−1 = pk−2 = ⋅ ⋅ pk−3 = ... = ⎜ ⎟ p0 k ≥ 0 

5 55 5 5 5 ⎝5⎠ k 

∞ 

1 = ∑ pk k= 0 

p 0 = 4/5 . 

∞ 

⎛1⎞ = ∑⎜ 

⎟ 

k= 

0⎝5⎠ 

p0 

p0 = 

1 

1− 5 

5 

= p0 

4 

Therefore, P[N > 1] = 1 – P[N ≤1] = 1 – (4/5 + 4/5 ⋅ 1/5) = 1 – 24/25 = 1/25 = 0.04 . 

-------------------------------------------------------------------------------------------------------- 


A Venn diagram for this situation looks like: 

We want to find 

( ) 

w= 1− 

x+ y+z 

1 1 

We have x+ y = , x+ z = , y+ z = 

4 3 

5 

12 

Adding these three equations gives 

1 1 5 

( x+ y) + ( x+ z) + ( y+ z) 

= + + 

4 3 12 

2 x+ y+ z = 1 

( ) 

1 

x+ y+ z = 

2 

1 1 

w= 1− ( x+ y+ z) 

= 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− 

⎜ + + ⎟= 

⎝12 6 4 ⎠ 2 

k 

Page 7 of 55


Let 1 denote the number of claims during weeks one and two, respectively. 

Then since are independent, 

and N 2 N 

N N 

1 and 

2 

[ N1+ N2 = 

7 

] = ∑ n= 

0 [ N1 = n] [ N2 = −n] 

7 

= ∑ 

⎛ 

⎜ 

1 ⎞⎛ 

n+ 1⎟⎜ 1 ⎞ 

8−n 

⎟ 

Pr 7 Pr Pr 7 

n= 

0 

⎝2 7 1 

= ∑ n= 

0 9 

2 

⎠⎝2 ⎠ 

= 

8 

= 

1 

= 

1 

9 6 

2 2 64 

-------------------------------------------------------------------------------------------------------- 


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) 

( Independence) 

Since 

c 

Pr ( E ) = 0.25 = 1− Pr ( E) 

, it follows Pr ( E ) = 0.75 . 

So 0.85 = Pr O + 0.75 −PrO 

0.75 

( ) ( )( ) 

Pr ( O)( 1− 0.75) = 0.10 

Pr ( O ) = 0.40 

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Let X 1 and X 2 denote the measurement errors of the less and more accurate instruments, 

respectively. If N(μ,σ) denotes a normal random variable with mean μ and standard 

deviation σ, then we are given X1 is N(0, 0.0056h), X 2 is N(0, 0.0044h) and X 1, X2 

are 

X1+ X2 independent. It follows that Y = 

2 

is N (0, 

2 2 2 2 

0.0056 h + 0.0044 h 

) = N(0, 

4 

0.00356h) . Therefore, P[−0.005h ≤ Y ≤ 0.005h] = P[Y ≤ 0.005h] – P[Y ≤ −0.005h] = 

P[Y ≤ 0.005h] – P[Y ≥ 0.005h] 

⎡ 0.005h 

⎤ 

= 2P[Y ≤ 0.005h] – 1 = 2P 

⎢ 

Z ≤ 

⎣ 0.00356h 

⎦ 

⎥ 

− 1 = 2P[Z ≤ 1.4] – 1 = 2(0.9192) – 1 = 0.84. 

Page 8 of 55


Apply Bayes’ Formula. Let 

A = Event of an accident 

B 1 = Event the driver’s age is in the range 16-20 

B 2 = Event the driver’s age is in the range 21-30 

B = Event the driver’s age is in the range 30-65 

Then 

Pr 

3 

B = 

4 

( B1A) Event the driver’s age is in the range 66-99 

Pr ( AB1) Pr ( B1) 

( A B1) ( B1) + ( A B2) ( B2) + ( A B3) ( B3) + ( A B4) ( B4) 

= 

Pr Pr Pr Pr Pr Pr Pr Pr 

( 0.06)( 0.08) 

( 0.06)( 0.08) + ( 0.03)( 0.15) + ( 0.02)( 0.49) + ( 0.04)( 0.28) 

= = 0.1584 

-------------------------------------------------------------------------------------------------------- 


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] 

PU [ | 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 

-------------------------------------------------------------------------------------------------------- 


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 55


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 ⎡⎣DL⎤ ⎦ = 2 Pr ⎡⎣DN⎤⎦ and Pr ⎡⎣DL⎤ ⎦ = Pr ⎡DH 

2 

⎣ ⎤⎦ Therefore, upon applying Bayes’ Formula, we find that 

Pr ⎡DH⎤Pr[ H] 

Pr ⎡HD ⎣ ⎦ 

⎣ ⎤ ⎦ = 

Pr ⎡⎣DN⎤ ⎦Pr[ N] + Pr ⎡⎣DL⎤ ⎦Pr[ L] + Pr ⎡⎣DH⎤⎦Pr[ H] 

( ) 

2Pr⎡DL⎤ 0.2 0.4 

= 

⎣ ⎦ 

= 

= 0.42 

1 

Pr⎡DL⎤ 0.25 0.3 0.4 

( 0.5) + Pr⎡DL⎤ ( 0.3) + 2Pr⎡DL⎤( 0.2 

+ + 

) 

2 

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 

-------------------------------------------------------------------------------------------------------- 


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 

PCY [ ] PY [ ] 

P[Y⏐C] = 

PCT [ ] PT [ ] + PCY [ ] PY [ ] + PCM [ ] PM [ ] + PCS [ ] PS [ ] 

(0.08)(0.16) 

= = 0.22 . 

(0.15)(0.08) + (0.08)(0.16) + (0.04)(0.45) + (0.05)(0.31) 

-------------------------------------------------------------------------------------------------------- 


Observe 

Pr 1≤ Pr ⎡⎣N ≥1N ≤4⎤ ⎦ = 

Pr 

≤4 ≤ 4 

⎡1 1 1 1 ⎤ 

= 

⎢ 

+ + + 

⎣6 12 20 30⎥ ⎦ 

⎡1 1 1 1 1 ⎤ 

⎢ 

+ + + + 

⎣2 6 12 20 30⎥ 

⎦ 

[ N ] 

[ N ] 

10 + 5 + 3+ 2 20 2 

= = = 

30 + 10 + 5 + 3+ 2 50 5 

Page 10 of 55


Let Y = positive test result 

D = disease is present (and ~D = not D) 

Using Baye’s theorem: 

PY [ | DPD ] [ ] (0.95)(0.01) 

PY [ | DPD ] [ ] PY [ |~ DP ] [~ D ] (0.95)(0.01) (0.005)(0.99) 

= 

P[D|Y] = = 0.657 . 

+ + 

-------------------------------------------------------------------------------------------------------- 


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⏐C C ] 

PSCPC [ ] [ ] 

Now applying Bayes’ Theorem, we find that P[C⏐S] = 

C C 

PSCPC [ ] [ ] + PSC [ ]( PC [ ] ) 

C 

2 PSC [ ] PC [ ] 2(0.25) 2 2 

= 

= = = . 

C C 

2 PSC [ ] PC [ ] + PSC [ ](1 −PC 

[ ]) 2(0.25) + 0.75 2 + 3 5 

-------------------------------------------------------------------------------------------------------- 


Use Baye’s Theorem with A = the event of an accident in one of the years 1997, 1998 or 

1999. 

PA [ 1997] P[1997] 

P[1997|A] = 

PA [ 1997][ P[1997] + PA [ 1998] P[1998] + PA [ 1999] P[1999] 

(0.05)(0.16) 

= = 0.45 . 

(0.05)(0.16) + (0.02)(0.18) + (0.03)(0.20) 

-------------------------------------------------------------------------------------------------------- 

Page 11 of 55


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, PC [ | I1] 

= 

c c 

P[ I1| C] P[ C] + P⎡ ⎣I1| C ⎤ 

⎦P⎡ ⎣C ⎤ 

⎦ 

Now 

1 

PC [ ] = 

5 

c 

1 4 

P⎡ ⎣C ⎤ 

⎦ = 1− P[ C] 

= 1− = 

5 5 

P I | C = 0.10 0.90 = 0.141 

[ 1 ] 30 ( 1 )( )( 

29 

) 

1 

c 30 ( 1 )( )( ) 

P⎡ ⎣I Therefore, 

| C ⎤ 

⎦ = 0.02 0.98 = 0.334 

PC [ | I 1] 

= 

( 0.141)( 1/ 5) 

0.141 1/ 5 + 0.334 4 / 5 

= 0.096 

29 

( )( ) ( )( ) 

-------------------------------------------------------------------------------------------------------- 


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 λ . Now we are 

given that 

50 

∫ 

0.3 = P[T ≤ 50] = λe dt =−e 

= 1 – e 

0 

−λt −λt 

50 

0 

Therefore, e –50λ = 0.7 or λ = − (1/50) ln(0.7) 

80 

−λt −λt 

80 

∫ λe dt =−e 

0 

0 

(80/50) ln(0.7) 80/50 

–50λ 

It follows that P[T ≤ 80] = = 1 – e –80λ 

= 1 – e = 1 – (0.7) = 0.435 . 

-------------------------------------------------------------------------------------------------------- 


−λ 2 −λ 

4 

e λ e λ 

Let N be the number of claims filed. We are given P[N = 2] = = 3 = 3 ⋅ P[N 

2! 4! 

2 4 

= 4]24 λ = 6 λ 

2 

λ = 4 ⇒ λ = 2 

Therefore, Var[N] = λ = 2 . 

Page 12 of 55


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 > x ≤ 0.01 

[ ] 

or, equivalently, 

[ ] ( )( ) ( ) 20 

x 20 

k −k 

0.99 ≤Pr X ≤ x =∑ 0.02 0.98 

k = 0 k 

The following table summarizes the selection process for x: 

x Pr X = x Pr X ≤ x 

[ ] [ ] 

( 

20 

) = 

0 0.98 0.668 0.668 

( )( ) 

1 20 0.02 0.98 = 0.272 0.940 

19 

( ) ( ) 

2 18 

2 190 0.02 0.98 = 0.053 

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 

-------------------------------------------------------------------------------------------------------- 


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 ≥ x+ 2] = f ( 0,0, 4) + f ( 1,0,3) + f ( 0,1,3 ) + f ( 0, 2, 2) 

= 

= 0.0488 

+ + 

4! 

+ 

2!2! 

4 3 3 2 

( 0.20) 4( 0.50)( 0.20) 4( 0.30)( 0.20) ( 0.30) ( 0.20) 

Page 13 of 55 

2


Note that 

20 

⎛ 1 2⎞ 20 

Pr[ X > x] = ∫ 0.005( 20 − t) dt = 0.005 20 

x 

⎜ t − t ⎟ x 

⎝ 2 ⎠ 

⎛ 1 2⎞ ⎛ 

1 2 

= 0.005⎜400 −200 − 20x + x ⎟ = 0.005⎜200 − 20x+ 

x 

⎝ 2 ⎠ ⎝ 

2 

⎞ ⎟ 

⎠ 

where 0 < x < 20 . Therefore, 

2 

Pr[ X > 16] 200 − 20( 16) + 1 ( 16) 

8 1 

Pr ⎡ 2 

⎣X > 16 X > 8⎤ 

⎦ = = = = 

2 

Pr[ X > 8] 200 − 20( 8) + 1 ( 8) 

72 9 

2 

-------------------------------------------------------------------------------------------------------- 


C x − 

We know the density has the form + for 0< x < 40 (equals zero otherwise). 

( ) 2 

10 

First, determine the proportionality constant C from the condition 

∫ 

40 

0 

f ( xdx= ) 1 

40 40 

−2 −1 

C C 2 

1= ∫ C( 10 + x) dx=− C(10 + x) 

= − = C 

0 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 ⎞ 

12.5∫ ( 10 + x) dx=− ( 10 + x) 

= ( 12.5) 0.47 

0 ⎜ − ⎟ = . 

0 ⎝10 16 ⎠ 

-------------------------------------------------------------------------------------------------------- 


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 

40 

∫ 0 f ( x) dx= 

1: 

40 

−1 

40 2 

1 = ∫ f ( xdx ) =− C(10 + x) | 0 = C 

0 

25 

25 

so that C = = 12.5. Then, calculate P(T < 5) by integrating f (x) = 12.5 (10 + x)−2 

2 

over the interval (0.5). 

Page 14 of 55 

:


To determine k, note that 

1 

4 k 5 1 

1 = ∫ k( 1− y) dy =− ( 1− 

y) 

5 0 

k 

= 

5 

0 

k = 5 

We next need to find P[V > 10,000] = P[100,000 Y > 10,000] = P[Y > 0.1] 

1 

∫ 

0.1 

( − y) dy =−( − y) 

4 5 1 

= 51 1 = (0.9) 5 = 0.59 and P[V > 40,000] 

= P[100,000 Y > 40,000] = P[Y > 0.4] = 

0.1 

1 

∫ 

0.4 

( − y) dy =−( − y) 

4 5 1 

51 1 

It now follows that P[V > 40,000⏐V > 10,000] 

PV [ > 40,000 ∩ V > 10,000] PV [ > 40,000] 0.078 

= 

= = = 0.132 . 

PV [ > 10,000] PV [ > 10,000] 0.590 

-------------------------------------------------------------------------------------------------------- 


Let T denote printer lifetime. Then f(t) = ½ e –t/2 , 0 ≤ t ≤ ∞ 

Note that 

1 

1 −t/2 −t/21–1/2 

P[T ≤ 1] = ∫ e dt = e = 1 – e = 0.393 

2 

0 

0 

2 

0.4 

= (0.6) 5 = 0.078 . 

1 −t/2 −t/22 

P[1 ≤ T ≤ 2] = ∫ e dt = e 1 

2 1 

–1/2 –1 

= e − e = 0.239 

Next, denote refunds for the 100 printers sold by independent and identically distributed 

random variables Y 1, . . . , Y100 

where 

⎧200 

⎪ 

Yi 

= ⎨100 ⎪⎩ 0 

with probability 0.393 



i = 1, . . . , 100 

Now E[Y ] = 200(0.393) + 100(0.239) = 102.56 

i 

Therefore, Expected Refunds = 

100 

∑ E [ Yi 

] = 100(102.56) = 10,256 . 

i= 

1 

Page 15 of 55


Let F denote the distribution function of f. Then 

4 3 

( ) [ ] ∫ 1 

x x 

F x Pr X x 3t dt t 1 x 3 

− − − 

= ≤ = =− = − 

Using this result, we see 

Pr 2 ≥ 1.5 

Pr 

= 

2 1.5 Pr 

= 

2 Pr 1.5 

[ X < | X ] 

1 

⎡⎣( X < ) ∩( X ≥ 

Pr[ X ≥1.5] ) ⎤⎦ [ X < ] − [ X ≤ 

Pr[ X ≥1.5] 

] 

( 2) − ( 1.5) 1−F( 1.5) −3 −3 

( 1.5) −( 2) −3 

( 1.5) 

1 

3 

⎛3⎞ ⎜ ⎟ 

⎝4⎠ 0.578 

F F 

= = = − = 

-------------------------------------------------------------------------------------------------------- 


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 

20 0 19 

18 2 

P[X < 2] = (0.95) (0.05) + 20(0.95) (0.05) + 190(0.95) (0.05) 

= 0.358 + 0.377 + 0.189 = 0.925 . 

-------------------------------------------------------------------------------------------------------- 


Denote the insurance payment by the random variable Y. Then 

⎧0 

if 0< 

X ≤ C 

Y = ⎨ 

⎩X 

− C if C < X < 1 

Now we are given that 

0.5 

0.5+ 

C 

2 

∫ C 

0 

0 

+ C 

2 

( Y ) ( X C) x dx x ( ) 

0.64 = Pr < 0.5 = Pr 0 < < 0.5 + = 2 = = 0.5 + 

Therefore, solving for C, we find C = ± 0.8 −0.5 

Finally, since 0 < C < 1, 

we conclude that C = 0.3 

Page 16 of 55


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 ≥9) ∩( Y < 9) ⎤⎦∪⎡⎣( X < 9) ∩( Y ≥9) 

⎤⎦} 

= P⎡⎣( X ≥9) ∩ ( Y < 9) ⎤⎦+ P⎡⎣( X < 9) ∩( Y ≥9) 

⎤⎦ 

= 2P⎡⎣( X ≥9) ∩( 

Y < 9 ) ⎤⎦ 

(due to symmetry) 

= 2P[ X ≥9] 

P[ Y < 9] 

= 2P X ≥9P 

X < 9 (again due to symmetry) 

[ ] [ ] 

[ ] ( [ ] ) 

= 2P X ≥91−P X ≥9 

10 

9 10 10 10 9 10 

( 9 ) ( 2)( 0.8) + ( 10 )( 0.8) 1−( 9 )( 0.2)( 0.8) −( 

10 )( 0.8) 

= 2 ⎡ 

⎣ 

0. 

⎤⎡ 

⎦⎣ 

= 2 0.376 1− 0.376 = 0.469 

[ ][ ] 

-------------------------------------------------------------------------------------------------------- 


Let 

IA = Event that Company A makes a claim 

I B = Event that Company B makes a claim 

XA = Expense paid to Company A if claims are made 

X B = Expense paid to Company B if claims are made 

Then we want to find 

C 

Pr{ 

⎡ 

⎣IA ∩I ⎤ B⎦∪⎡⎣( IA∩IB) ∩( 

X A < XB) 

⎤⎦} 

C 

= Pr ⎣ 

⎡IA ∩ I ⎤ B⎦+ Pr ⎡⎣( IA∩IB) ∩( 

X A < XB) 

⎤⎦ 

C 

= Pr ⎡ 

⎣I ⎤ A⎦Pr[ IB] + Pr[ IA] Pr[ IB] Pr [ XA < XB] 

(independence) 

= ( 0.60)( 0.30) + ( 0.40)( 0.30) Pr[ XB −X A ≥0] 

= 0.18 + 0.12Pr[ XB −X A ≥0] 

Now XB − XA 

is a linear combination of independent normal random variables. 

Therefore, XB − XA 

is also a normal random variable with mean 

M = E X − X = E X − E X = − =− 

[ ] [ ] [ ] 9,000 10,000 1,000 

B A B A 

2 2 

and standard deviation ( X ) ( X ) ( ) ( ) 


σ = Var + Var = 2000 + 2000 = 2000 2 

B A 

Page 17 of 55 

10 ⎤⎦

Finally, 

⎡ 1000 ⎤ 

Pr[ XB −X A ≥ 0] = Pr ⎢Z ≥ 

( Z is standard normal) 

2000 2 

⎥ 

⎣ ⎦ 

⎡ 1 ⎤ 

= Pr ⎢Z≥ 2 2 

⎥ 

⎣ ⎦ 

⎡ 1 ⎤ 

= 1−Pr⎢Z< 2 2 

⎥ 

⎣ ⎦ 

= 1− Pr < 0.354 

[ Z ] 

= 1− 0.638 = 0.362 

C { ⎡IA ∩I ⎤ B ∪⎡( IA∩IB) ∩ ( X A XB) 

⎤} 

= + ( )( 

Pr ⎣ ⎦ ⎣ < ⎦ 0.18 

= 0.223 

0.12 0.362) 

-------------------------------------------------------------------------------------------------------- 


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 

3 

3+ 

k ⎛3⎞ ⎛2⎞ Pr[ k ≥ 4] = 1−Pr[ k ≤ 3] = 1−∑( 

k ) ⎜ ⎟ ⎜ ⎟ 

k = 0 ⎝5⎠ ⎝5⎠ 4 0 1 2 

3 

⎛3⎞ ⎡ 3 ⎛2⎞ 4 ⎛2⎞ 5 ⎛2⎞ 6 ⎛2⎞ ⎤ 

= 1− 

⎜ ⎟ ⎢( 0) ⎜ ⎟ + ( 1) ⎜ ⎟ + ( 2) ⎜ ⎟ + ( 3) 

⎜ ⎟ ⎥ 

⎝5⎠ ⎢⎣ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎥⎦ 

⎛3⎞ ⎡ 8 8 32⎤ 

= 1− ⎜ ⎟ 1+ 

+ + 

⎝5⎠ ⎣ 

⎢ 5 5 25⎦ 

⎥ 

= 0.2898 

Alternatively the solution is 

4 

4 4 4 2 4 3 

⎛2⎞ 4 ⎛2⎞ 3 5 ⎛2⎞ ⎛3⎞ 6 ⎛2⎞ ⎛3⎞ ⎜ ⎟ + ( 1) ⎜ ⎟ + ( 2) ⎜ ⎟ ⎜ ⎟ + ( 3) 

⎜ ⎟ ⎜ ⎟ = 0.2898 

⎝5⎠ ⎝5⎠ 5 ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 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 

Pr k ≤ 3 

iii) calculating [ ] 

Page 18 of 55 

4 

k


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. 

5 

∑ 

k = 1 

1 2 3 4 

5 

Thus, the expected payment is gk ( ) pk= 100 p+ 200 p+ 300 p+ 350 p+ 400 p = 

1 

( 100× 5 + 200× 4 + 300× 3+ 350× 2 + 400× 1) 

=220 

15 

-------------------------------------------------------------------------------------------------------- 


Note that ( ) 

2 2 3 

0 

3 

4 

0 4 

x x x x 8 64 56 28 

E X = ∫ − dx + dx 

−2 

10 ∫ =− + =− + = = 

0 10 30 30 30 30 30 15 

−2 

0 

-------------------------------------------------------------------------------------------------------- 


The density function of T is 

1 −t 

/3 

f () t = e , 0 < t< 

∞ 

3 

Therefore, 

E[ X] = E⎡⎣max ( T, 

2) 

⎤⎦ 

2 2 

∞ 

−t/3 t −t/3 

= ∫ e dt+ e dt 

0 3 ∫2 

3 

=−2e | − te | + 

∞ 

e 

−t/3 2 −t/3 ∞ −t/3 

0 2 

2 

−2/3 −2/3 −t/3 

∞ 

| 2 

=− 2e + 2 + 2e −3e 

−2/3 

= 2 + 3e 

∫ 

dt 

Page 19 of 55


Let T be the time from purchase until failure of the equipment. We are given that T is 

exponentially distributed with parameter λ = 10 since 10 = E[T] = λ . Next define the 

payment 

⎧x 

for 0 ≤ T ≤ 1 

⎪ x 

P under the insurance contract by P= ⎨ for 1 < T ≤3 

⎪ 2 

⎪⎩ 0 for T > 3 

We want to find x such that 

1 

3 

1 

x –t/10 x 1 –t/10 

−t/10 x −t/10 

3 

1000 = E[P] = ∫ e dt + ∫ e dt = −xe − e 

1 

210 

2 

0 10 

= −x e –1/10 + x – (x/2) e –3/10 + (x/2) e 

We conclude that x = 5644 . 

1 

–1/10 

0 

–1/10 –3/10 

= x(1 – ½ e – ½ e ) = 0.1772x . 

-------------------------------------------------------------------------------------------------------- 


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 

⎧⎪ 1000( 5 − x) if x = 1,2,3,4 

y = ⎨ 

⎪⎩ 0 if x > 4 


2 3 

[ ] ( ) ( )( ) ( ) ( ) ( ) ( 

EY = 4000 0.4 + 3000 0.6 0.4 + 2000 0.6 0.4 + 1000 0.6 0.4 

= 2694 

-------------------------------------------------------------------------------------------------------- 


Define f ( X ) to be hospitalization payments made by the insurance policy. Then 

and 

⎧⎪ 100 X if X = 1, 2,3 

f( X) 

= ⎨ 

⎪⎩ 300 + 25( X − 3 ) if X = 4,5 

Page 20 of 55 

)

5 

( ) = ∑ ( ) Pr[ 

= ] 

E⎡⎣f X ⎤⎦ 

f k X k 

k = 1 

⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ 

= 100⎜ ⎟+ 200⎜ ⎟+ 300⎜ ⎟+ 325⎜ ⎟+ 350⎜ 

⎟ 

⎝15 ⎠ ⎝15 ⎠ ⎝15 ⎠ ⎝15 ⎠ ⎝15 ⎠ 

1 640 

= [ 100 + 160 + 180 + 130 + 70] = = 213.33 

3 3 

-------------------------------------------------------------------------------------------------------- 


Let N be the number of major snowstorms per year, and let P be the amount paid to 

n −3/2 

(3/ 2) e 

the company under the policy. Then Pr[N = n] = 

, n = 0, 1, 2, . . . and 

n! 

⎧0 

for N = 0 

P = ⎨ 

. 

⎩10,000( 

N −1) for N ≥1 

∞ 

n −3/2 

(3/ 2) e 

Now observe that E[P] = ∑10,000( 

n −1) 

n! 

n= 

1 

∞ 

n −3/2 

–3/2 (3/ 2) e 

–3/2 

= 10,000 e + ∑10,000( 

n −1) 

= 10,000 e + E[10,000 (N – 1)] 

n= 

0 

n! 

–3/2 –3/2 

= 10,000 e + E[10,000N] – E[10,000] = 10,000 e + 10,000 (3/2) – 10,000 = 7,231 . 

-------------------------------------------------------------------------------------------------------- 


Let Y denote the manufacturer’s retained annual losses. 

⎧x 

Then Y = ⎨ 

2 

for 0.6 < x≤2 

for x > 2 

and E[Y] = 

⎩ 

2 2.5 ∞ 

2.5 2 

2.5 2.5 

⎡2.5(0.6) ⎤ ⎡2.5(0.6) ⎤ 2.5(0.6) 2(0.6) 

x⎢ dx 2 

dx dx 

x 

⎥ + ⎢ = − 

x 

⎥ 

⎣ ⎦ ⎣ ⎦ x 

x 

∫ ∫ ∫ 

3.5 3.5 2.5 2.5 2 

0.6 2 0.6 

2.5 2.5 2.5 2.5 2.5 

2.5(0.6) 2 2(0.6) 2.5(0.6) 2.5(0.6) (0.6) 

= − + =− + + = 0.9343 . 

1.5 0.6 

2.5 1.5 1.5 1.5 

1.5 x (2) 1.5(2) 1.5(0.6) 2 

Page 21 of 55 

∞


Let us first determine K. Observe that 

⎛ 1 1 1 1 ⎞ ⎛60 + 30 + 20 + 15 + 12 ⎞ ⎛137 ⎞ 

1= K⎜1+ + + + ⎟= K⎜ 

⎟= 

K⎜ ⎟ 

⎝ 2 3 4 5 ⎠ ⎝ 60 ⎠ ⎝ 60 ⎠ 

60 

K = 

137 

It then follows that 

Pr N = n = Pr ⎡ ⎣N = n Insured Suffers a Loss⎤⎦Pr Insured Suffers a Loss 

60 3 

= ( 0.05 ) = 

137N 137N 

, N = 1,...,5 

Now because of the deductible of 2, the net annual premium P= E X where 

Then, 

[ ] [ ] 

⎧0 

, if N ≤ 2 

X = ⎨ 

⎩N 

− 2 , if N > 2 

5 

3 ⎛ 1 ⎞ ⎡ 3 ⎤ ⎡ 3 ⎤ 

P = E[ X] = ∑ ( N − 2) = () 1 2 3 0.0314 

N = 3 

⎜ ⎟+ 

⎢ ⎥+ ⎢ ⎥ = 

137N ⎝137 ⎠ ⎣137( 4) ⎦ ⎣137( 5) 

⎦ 

-------------------------------------------------------------------------------------------------------- 


⎧y 

Let W denote claim payments. Then W = ⎨ 

⎩10 

for 1 < y≤10 

for y ≥10 

10 

∞ 

2 2 2 10 

It follows that E[W] = ∫ y dy+ 10 dy 

3 3 

y ∫ =− 

y y 1 

10 ∞ 

− 2 

y 10 

= 2 – 2/10 + 1/10 = 1.9 . 

1 10 

Page 22 of 55 

[ ]


Let Y denote the claim payment made by the insurance company. 

Then 

⎧0 


⎪ 

Y = ⎨Max 

( 0, x−1 

) with probability 0.04 

⎪ 

⎩14 


and 

EY = 0.94 0 + 0.04 0.5003 

15 

− x /2 

x− 1 e dx+ 

0.02 14 

[ ] ( )( ) ( )( ) ( ) ( )( 

( ) 

15 15 

⎡ −x/2 −x/2 

xe dx e dx⎤ 

∫1 ∫1 

= 0.020012 

⎢⎣ − + 0.28 

⎥⎦ 

15 15 

−x/2 15 −x/2 −x/2 

= 0.28 + ( 0.020012) ⎡− 2xe | 1 + 2 e dx− e dx⎤ 

⎢⎣ ∫1 ∫1 

⎥⎦ 

15 

−7.5 −0.5 −x/ 

2 

= 0.28 + ( 0.020012) ⎡− 30e + 2e 

+ e dx⎤ 

⎢⎣ ∫1 

⎥⎦ 

−7.5 −0.5 −x/ 

2 15 

= 0.28 + ( 0.020012) ⎡ 

⎣− 30e + 2e −2e 

| ⎤ 1 ⎦ 

= 0.28 + 0.020012 

−7.5 −0.5 −7.5 −0.5 

− 30e + 2e − 2e + 2e 

∫ 

( )( ) 

( )( −7.5 e 

−0.5 

e ) 

= 0.28 + 0.020012 − 32 + 4 

( )( ) 

= 0.28 + 0.020012 2.408 

= 0.328 (in thousands) 

It follows that the expected claim payment is 328 . 

-------------------------------------------------------------------------------------------------------- 


k 

The pdf of x is given by f(x) = 4 

(1 + x) 

, 0 < x < ∞ . To find k, note 

∞ 

k k 1 

1 = ∫ dx =− 4 3 

(1 + x) 3 (1 + x) 

0 

k = 3 

∞ 

0 

k 

= 

3 

∞ 

3x 

It then follows that E[x] = ∫ dx and substituting u = 1 + x, du = dx, we see 

4 

(1 + x) 

∞ 

∞ 

0 

3 4 

−2 u 

−3 

u 

3( u −1) 

− − ⎡ ⎤ ⎡1 1⎤ 

E[x] = ∫ du = 3 ( u − u ) du = 3⎢ − = 3 − 

4 

u ∫ 

−2 −3 ⎥ ⎢ 

⎣2 3⎥ 

= 3/2 – 1 = ½ . 

1 

⎣ ⎦ ⎦ 

1 1 

1 

Page 23 of 55 

∞ 

)


Let Y represent the payment made to the policyholder for a loss subject to a deductible D. 

⎧0 

for 0≤ 

X ≤ D 

That is Y = ⎨ 

⎩x− D for D< X ≤ 1 

Then since E[X] = 500, we want to choose D so that 

1000 2 2 

1 1 1 ( x −D) 1000 (1000 −D) 

500 = ( ) 

4 ∫ x− D dx= 

= 

1000 1000 2 D 2000 

D 

2 2 

(1000 – D) = 2000/4 ⋅ 500 = 500 

1000 – D = ± 500 

D = 500 (or D = 1500 which is extraneous). 

-------------------------------------------------------------------------------------------------------- 


1 

We are given that Mx(t) = for the claim size X in a certain class of accidents. 

4 

(1 − 2500 t) 

( −4)( −2500) 

10,000 

First, compute Mx′(t) = 

= 5 5 

(1−2500 t) (1−2500 t) 

(10,000)( −5)( −2500) 

125,000,000 

M x″(t) 

= 

= 

6 6 

(1−2500 t) (1−2500 t) 

Then E[X] = M x′ 

(0) = 10,000 

E[X 2 ] = M x″ 

(0) = 125,000,000 

2 2 2 

Var[X] = E[X ] – {E[X]} = 125,000,000 – (10,000) = 25,000,000 

Var[ X ] = 5,000 . 

-------------------------------------------------------------------------------------------------------- 


Let X J, XK, 

and XL represent annual losses for cities J, K, and L, respectively. Then 

X = X J + X K + XL 

and due to independence 

xt 

( xJ + xK + xL) t xt J xKt xLt 

M(t) = E⎡e E⎡e ⎤ 

⎣ 

⎤ 

⎦ = = E⎣ ⎡e ⎦ 

⎤E⎡ ⎣e ⎤ 

⎦E⎡ ⎣e ⎤ 

⎣ ⎦ 

⎦ 

= MJ(t) MK(t) ML(t) = (1 – 2t) –3 (1 – 2t) –2.5 (1 – 2t) –4.5 = (1 – 2t) –10 

Therefore, 

M′(t) = 20(1 – 2t) –11 

M″(t) = 440(1 – 2t) –12 

M″′(t) = 10,560(1 – 2t) –13 

E[X 3 ] = M″′(0) = 10,560 

Page 24 of 55


The distribution function of X is given by 

Therefore, the 

2.5 2.5 

x 

2.5 

( ) −( 

) ( ) 

x 2.5 200 200 200 

F( x) = ∫ 

dt= = 1 − , x> 

200 

200 3.5 2.5 2.5 

t t x 

( 1 0.01p) 

200 

th 

p of X is given by 

percentile x p 

p 

= F( xp) 

= 1− 

100 

1− 0.01p= 

x 

( 200) 

2.5 

p 

( 1−0.01p) 25 

( 200) 

2.5 

25 200 

− = 

x 

p 

= 

200 

x 

p 

x 

2.5 

p 

200 200 

It follows that x − x = − = 93.06 

2.5 

( 0.30) ( 0.70) 

70 30 25 25 

-------------------------------------------------------------------------------------------------------- 


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 . 

-------------------------------------------------------------------------------------------------------- 


∞z Q1 

The first quartile, Q1, is found by ¾ = 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 55


First note that the density function of X is given by 

⎧1 

⎪ 

if x = 1 

2 

⎪ 

f ( x) 

= ⎨x− 1 if 1< x< 

2 

⎪⎪⎪⎩ 0 otherwise 

Then 

1 1 1 ⎛1 1 ⎞ 

⎜ ⎟ 

2 2 2 ⎝3 2 ⎠ 

2 2 

( ) ∫ ( 1) 

∫ ( ) 

2 3 2 

E X = + x x − dx = + x − x dx = + x − x 

1 1 

1 8 4 1 1 7 4 

= + − − + = − 1 = 

2 3 2 3 2 3 3 

1 1 1 ⎛1 1 ⎞ 

⎜ ⎟ 

2 2 2 ⎝4 3 ⎠ 

2 2 

( ) ∫ ( 1) 

∫ ( ) 

2 2 3 2 4 3 

E X = + x x − dx = + x − x dx = + x − x 

1 1 

1 16 8 1 1 17 7 23 

= + − − + = − = 

2 4 3 4 3 4 3 12 

2 

2 23 ⎛4⎞ 23 16 5 

Var( X) = E( X ) − ⎡⎣E( X) 

⎤⎦ = − ⎜ ⎟ = − = 

12 ⎝3⎠ 12 9 36 

-------------------------------------------------------------------------------------------------------- 


⎧X 

if 0 ≤ X ≤4 

Note Y = ⎨ 

⎩ 4 if 4 < X ≤ 5 

Therefore, 

41 54 

1 2 4 4 5 

EY [ ] = ∫ xdx+ 0 4 

0 5 ∫ dx= x| + x| 

4 5 10 5 

16 20 16 8 4 12 

= + − = + = 

10 5 5 5 5 5 

4 5 

2 1 2 16 1 3 4 16 

E⎡ ⎣Y ⎤ 

⎦ = ∫ x dx+ dx x 0 x 4 

0 5 ∫ = | + | 

4 5 15 5 

64 80 64 64 16 64 48 112 

= + − = + = + = 

15 5 5 15 5 15 15 15 

5 

2 

2 112 ⎛12 ⎞ 

Var[ Y] = E⎡ ⎣Y ⎤ 

⎦− 

( E[ Y] 

) = − ⎜ ⎟ = 1.71 

15 ⎝ 5 ⎠ 

2 

2 

Page 26 of 55 

2 

1 

2 

1


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[X 2 ] = 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 

2 2 

Var[X] = E[X ] – (E[X]) = 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 . 

-------------------------------------------------------------------------------------------------------- 


Let X and Y denote repair cost and insurance payment, respectively, in the event the auto 

is damaged. Then 

⎧0 

if x ≤ 250 

Y = ⎨ 

⎩x− 

250 if x> 

250 

and 

2 

1500 1 1 2 1500 1250 

EY [ ] = ∫ ( x− 250) dx= ( x− 

250) 250 = = 521 

250 1500 3000 3000 

3 

1500 

2 1 2 1 3 1500 1250 

E⎡ ⎣Y ⎤ 

⎦ = ∫ ( x− 250) dx= ( x− 

250) 250 = = 434,028 

250 1500 4500 4500 

2 [ Y] = E⎡Y ⎤− 

{ E[ Y] 

} = −( 

) 

[ Y ] 

2 2 

Var ⎣ ⎦ 

434,028 521 

Var = 403 

-------------------------------------------------------------------------------------------------------- 


Let X1, X 2, X 3, and X4 

denote the four independent bids with common distribution 

function F. Then if we define Y = max (X1, X 2, X 3, X4), 

the distribution function G of Y is 

given by 

G( y) = Pr[ 

Y ≤ y] 

= Pr ⎡⎣( X1 ≤ y) ∩( X2 ≤ y) ∩( X3 ≤ y) ∩( X4 ≤ y) 

⎤⎦ 

= Pr X ≤ y Pr X ≤ y Pr X ≤ y Pr X ≤ y 

[ ] [ ] [ ] [ ] 

( ) 

= ⎡⎣F y ⎤⎦ 

1 2 3 4 

4 

1 4 3 5 

( π ) 

= 1+ sin 

16 

y , ≤ y≤ 

2 2 

It then follows that the density function g of Y is given by 

Page 27 of 55

( ) = '( 

) 

g y G y 

1 

3 

= ( 1+ sinπy) ( πcosπy) 4 

π 

3 3 5 

= cosπy( 1+ sin πy) 

, ≤ y≤ 

4 2 2 

Finally, 

5/2 

EY [ ] = ∫ yg( ydy ) 

3/2 

5/2π 3 

= ∫ ycosπy( 1+ sinπy) 

dy 

3/2 4 

-------------------------------------------------------------------------------------------------------- 


The amount of money the insurance company will have to pay is defined by the random 

variable 

⎧1000 

x if x< 

2 

Y = ⎨ 

⎩2000 

if x ≥ 2 

where x is a Poisson random variable with mean 0.6 . The probability function for X is 

−0.6 

k 

e ( 0.6) 

p( x) = k = 0,1, 2,3 L and 

k! 

k 

−0.6 −0.6 

∞ 0.6 

EY [ ] = 0 + 1000( 0.6) e + 2000e 

∑ k = 2 k! 

k 

−0.6 ⎛ −0.6∞0.6 −0.6 −0.6 

⎞ 

= 1000( 0.6) e + 2000⎜e ∑ −e −( 

0.6) 

e 

k = 0 

⎟ 

⎝ k! 

⎠ 

k 

∞ 

0.6 ( 0.6 

− ) 

−0.6 −0.6 −0.6 

0.6 

= 2000e ∑ −2000e − 1000( 0.6) e = 2000 −2000e −600e 

k = 0 k! 

= 573 

− 

k 

2 2 −0.6 2 −0.6 

∞ 0.6 

E⎡ ⎣Y ⎤ 

⎦ = ( 1000) ( 0.6) e + ( 2000) 

e ∑ k = 2 k! 

k 

2 −0.6 ∞ 0.6 

2 −0.6 

2 2 

0.6 

= ( 2000) e ∑ −( 2000) e −⎡( 2000) −( 

1000) ⎤( 

0.6) 

e 

k = 0 k! 

⎣ ⎦ 

2 2 −0.6 2 2 

−0.6 

= ( 2000) −( 2000) e −⎡( 2000) −( 

1000) ⎤( 

0.6) 

e 

⎣ ⎦ 

= 816,893 

− 

2 [ Y] = E⎡Y ⎤− 

{ E[ Y] 

} = −( 

) 

2 2 

Var ⎣ ⎦ 

816,893 573 

[ Y ] 

Var = 699 

Page 28 of 55 

= 488,564


Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the 

⎧x 

claim benefits paid. Then Y = ⎨ 

⎩250 

for x< 

250 

and we want to find m such that 0.50 

for x ≥ 250 

m 

−0.004x −0.004x 

= 0.004e 

dx =−e 

m 

= 1 – e –0.004m 

∫ 

0 

0 

This condition implies e –0.004m = 0.5 ⇒ m = 250 ln 2 = 173.29 . 

-------------------------------------------------------------------------------------------------------- 


The distribution function of an exponential random variable 

T with parameter θ is given by F t 

−t 

θ 

= − e t > 0 

( ) 1 , 

Since we are told that T has a median of four hours, we may determine θ as follows: 

1 

−4 

θ 

= F( 4) = 1−e 

2 

1 −4 

θ 

= e 

2 

4 

− ln ( 2) 

=− 

θ 

4 

θ = 

ln 2 

( ) 

Therefore, ( ) ( ) 

( ) 

5ln 2 

− 

−5θ4−5 4 

Pr T ≥ 5 = 1− F 5 = e = e = 2 = 0.42 

-------------------------------------------------------------------------------------------------------- 


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 95 th percentile of all claims that 

exceed 100 . Consequently, we want to find p95 such that 

Pr[100 < x< p95] F( p95) −F(100) 

0.95 = = 

where F(x) is the distribution function of X . 

PX [ > 100] 1 −F(100) 

Now F(x) = 1 – e –x/300 . 

−p95 /300 −100/300 −1/3−p95 

/300 

1 −e −(1 −e ) e −e 

1/3 − p95 

/300 

Therefore, 0.95 = 

= = 1−e 

e 

−100/300 −1/3 

1 −(1 −e 

) e 

p95 

/300 

e − 

= 0.05 e –1/3 

p95 = −300 ln(0.05 e –1/3 ) = 999 

Page 29 of 55


The distribution function of Y is given by 

2 

( ) ( ) ( ) ( ) 

G y = Pr T ≤ y = Pr T ≤ y = F y = 1− 4 y 

for y > 4 . Differentiate to obtain the density function 

Alternate solution: 

Differentiate F( t ) f ( t) 3 

8t 

− 

( ) 

g y 4y 

− 

= 

2 

to obtain = and set y = t . Then t = y and 

d 

⎛1⎞ g( y) = f t( y) dt dy = f y y = y ⎜ y ⎟ y 

dt 

⎝2⎠ −32 −12−2 ( ) ( ) ( ) 8 = 4 

-------------------------------------------------------------------------------------------------------- 


We are given that R is uniform on the interval ( 0.04,0.08 ) and 

Therefore, the distribution function of V is given by 

R 

F( v) = Pr[ V ≤ v] = Pr ⎣ 

⎡10,000e ≤ v⎦ ⎤ = Pr ⎡⎣ R≤ln( v) 

−ln( 

10,000) 

⎤⎦ 

1 ln( v) 

−ln( 

10,000 ) 1 

∫ 

0.04 0.04 

0.04 

( v) 

− ( ) 

ln ln 10,000 

0.04 

( ) ( ) 

2 

10,000 R 

V = e 

= dr = r = 25ln v −25ln 

10,000 − 1 

⎡ ⎛ v ⎞ ⎤ 

= 25⎢ln⎜ −0.04⎥ 

10,000 

⎟ 

⎣ ⎝ ⎠ ⎦ 

-------------------------------------------------------------------------------------------------------- 


10 

⎡ 8 ⎤ 

( ) 

( ) [ ] ( ) 10 8 

0.8 10 

= Pr ≤ = Pr ⎡10 ≤ ⎤ = Pr ≤ Y = 1− 

F y Y y X y X 

10 

e − 

⎣ ⎦ ⎢ 

⎣ 

⎥ 

⎦ 

( ) 

( ) ( ) 

54 

1 

4 1 Y Y 10 

f y F y e 

8 10 

− 

′ 

⎛ ⎞ 

Therefore, = = ⎜ ⎟ 

⎝ ⎠ 

Page 30 of 55 

Y


First note R = 10/T . Then 

⎡10 ⎤ ⎡ 10⎤ ⎛10 

⎞ 

F R (r) = P[R ≤ r] = P 

⎢ 

≤ r = P T ≥ = 1−FT⎜ ⎣T ⎥ 

⎦ 

⎢ 

⎣ r ⎥ 

⎟ . Differentiating with respect to 

⎦ ⎝ r ⎠ 

⎛ ⎛10 ⎞⎞ ⎛ d ⎞⎛−10 

⎞ 

r f R(r) = F′ R(r) 

= d/dr ⎜1− FT ⎜ ⎟ =−⎜ 

FT () t ⎟⎜ 2 

r 

⎟ 

⎟⎠ 

⎝ ⎝ ⎠⎠ ⎝dt ⎠⎝ r 

d 1 

FT() t = fT() t = since T is uniformly distributed on [8, 12] . 

dt 

4 

−1⎛−10⎞ 5 

Therefore f R(r) 

= ⎜ = 2 ⎟ . 2 

4 ⎝ r ⎠ 2r 

-------------------------------------------------------------------------------------------------------- 


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 ≤ y] = P[2X ≤ y] 

= P[X ≤ y/2] = F(y/2) and g(y) = G′(y) = d/dy F(y/2) = ½ F′(y/2) = ½ f(y/2) . 

-------------------------------------------------------------------------------------------------------- 


First, observe that the distribution function of X is given by 

x 3 1 x 1 

F ( x) = ∫ dt =− | 4 3 1 = 1 − , x> 

1 

1 

3 

t t x 

Next, let X1, X 2, 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 ≤ y Pr X ≤ y Pr X ≤ y 

( ) [ ] [ ] [ ] 

1 2 3 

⎛ 1 ⎞ 

= ⎜1 − , y 1 

3 ⎟ > 

⎝ y ⎠ 

while the density function of Y is given by 

3 

2 2 

⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 9 ⎞⎛ 1 ⎞ 

g( y) = G'( y) = 3⎜1− 1 , y 1 

3 ⎟ ⎜ 4 ⎟= ⎜ − 4 ⎟⎜ 3 ⎟ > 

⎝ y ⎠ ⎝ y ⎠ ⎝ y ⎠⎝ y ⎠ 

Therefore, 

Page 31 of 55

2 

∞ ∞ 

9 ⎛ 1 ⎞ 9 ⎛ 2 1 ⎞ 

E [ Y ] = ∫ 1 dy 1 dy 

1 3 ⎜ − 3 ⎟ = − + 

1 3 3 6 

y y ∫ ⎜ ⎟ 

⎝ ⎠ y ⎝ y y ⎠ 

∞ ⎛ 9 18 9 ⎞ ⎡ 9 18 9 ⎤ 

= ∫1 

⎜ − + dy 

3 6 9 ⎟ = 2 5 8 

y y y 

⎢− + − 

2y 5y 8y 

⎥ 

⎝ ⎠ ⎣ 

⎦ 

⎡1 2 1⎤ 

= 9 

⎢ 

− + = 2.025 (in thousands) 

⎣2 5 8⎥ 

⎦ 

-------------------------------------------------------------------------------------------------------- 


2 21 

( x ydxdy ) 

1 1 8 + 

Prob. = 1− ∫∫ 

= 0.625 

Note 

c 

{ } 

( X ≤ ) U( Y ≤ ) = ( X > ) I ( Y > ) 

Pr ⎡⎣ 1 1 ⎤⎦ Pr ⎡⎣ 1 1 ⎤⎦ 

(De Morgan's Law) 

= 1− Pr⎡⎣( X > 1) I ( Y > 1) ⎤⎦ 

2 21 = 1− ∫ ( ) 1∫ x+ y dxdy 

1 8 

1 21 2 2 

= 1− 

( ) 1 

8∫ x+ y dy 

1 2 

1 2 

= 1− ⎡ 

16 ∫ 1 ⎣ 

2 

+ 2 − 

2 

+ 1 ⎤ 

⎦ 

1 

= 1− ⎡ 

48 ⎣ 

3 

+ 2 − 

3 

+ 1 ⎤ 

⎦ 

2 1 

1 = 1− 48 

64−27− 2 + 

18 30 

= 1− = = 0.625 

48 48 

( y ) ( y ) dy ( y ) ( y ) ( 7 8) 

-------------------------------------------------------------------------------------------------------- 


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 55 

∞ 

1

( X < ) ∪ ( Y < ) 

Pr ⎡⎣ 1 1 ⎤⎦ 

2 

3 

3 3 3 3 

x+ y x + 2xy 1 

= 1− ∫∫ dx dy = 1− dy 1 ( 9 6y1 2y) 

dy 

27 ∫ = − + − − 

54 54 ∫ 

1 1 1 1 

1 

3 

1 3 

1 2 1 32 11 

( ) ( ) ( ) 

54 ∫ y dy y y 

1 

54 1 54 54 27 

= 1− 8+ 4 = 1− 8 + 2 = 1− 24+ 18−8− 2 = 1− = = 0.41 

-------------------------------------------------------------------------------------------------------- 


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⎞ ⎛ 1⎞⎤ 

1/2 1 1 1/2 

Pr ⎢⎜S ≤ ⎟∪⎜T ≤ ⎟ = f ( s, t) dsdt+ f ( s, t) d 

2 2 

⎥ ∫0 ∫1/2 ∫0∫ sdt 

0 

⎣⎝ ⎠ ⎝ ⎠⎦ 

(where the first integral covers A and the second integral covers B). 

-------------------------------------------------------------------------------------------------------- 


By the central limit theorem, the total contributions are approximately normally 

nμ = 2025 3125 = 6,328,125 and standard deviation 

distributed with mean ( )( ) 

σ n = 250 2025 = 11,250 . From the tables, the 90 th percentile for a standard normal 

random variable is 1.282 . Letting p be the 90 th percentile for total contributions, 

p−nμ = 1.282, and so p= nμ + 1.282σ n = 6,328,125 + ( 1.282)( 11, 250) = 6,342,548 . 

σ n 

Page 33 of 55

-------------------------------------------------------------------------------------------------------- 


1 

Let X 1, . . . , X25 

denote the 25 collision claims, and let X = (X 1 + . . . +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 

19,400 and standard deviation 1 

(5000) = 1000 . We conclude that P[ X > 20,000] 

25 

⎡X −19,400 20,000 −19,400⎤ ⎡X 

−19,400 

⎤ 

= P⎢ > = P 

0.6 

1000 1000 

⎥ ⎢ > 

⎣ ⎦ ⎣ 1000 

⎥ = 1 − Φ(0.6) = 1 – 0.7257 

⎦ 

= 0.2743 . 

-------------------------------------------------------------------------------------------------------- 


Let X 1, . . . , 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[X i] = Var[X i] = 2 . Now we are interested in the random variable S = X 1 + . . . + 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] = 

⎡2450 −2500 S−2500 2600 −2500⎤ ⎡ S−2500 

⎤ 

P⎢ < < ⎥ = P 1 2 

2500 2500 2500 

⎢ 

− < < 

50 ⎥ 

⎣ ⎦ ⎣ ⎦ 

⎡S−2500 ⎤ ⎡S−2500 ⎤ 

= P 

⎢ 

< 2 − P 1] = P[Z < 2] + P[Z < 1] – 1 ≈ 0.9773 + 0.8413 – 1 = 0.8186 . 

-------------------------------------------------------------------------------------------------------- 


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 = X 1 + … + Xn is also normally distributed with mean 

μ = 3n 

and standard deviation 

σ = n 

Now we want to choose the smallest value for n such that 

⎡ S−3n 40−3n⎤ 0.9772 ≤ Pr[ S > 40] = Pr ⎢ > ⎥ 

⎣ n n ⎦ 

This implies that n should satisfy the following inequality: 

Page 34 of 55

40 − 3n 

−2≥ n 

To find such an n, let’s solve the corresponding equation for n: 

40 − 3n 

− 2 = 

n 

− 2 n = 40−3n 3n−2 n − 40= 0 

3 n+ 10 n −4 

= 0 

( )( ) 

n = 4 

n = 16 

-------------------------------------------------------------------------------------------------------- 


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 

ET [ ] = 100( 70) = 7000 

and variance 

2 

Var [ T ] = 100( 100) = 100 

Therefore, 

⎡T−7000 7100 −7000⎤ 

Pr[ T < 7100] = Pr 

⎢ 

< 

⎣ 100 100 ⎥ 

⎦ 

= Pr[ Z < 1] = 0.8413 

where Z is a standard normal random variable. 

Page 35 of 55

-------------------------------------------------------------------------------------------------------- 


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 ≥ 110,000] = 1 – P[S ≤ 110,000] = 1 – P⎢Z ≤ 

10,000 

⎥ = 1 – P[Z ≤ 1] = 1 

⎣ ⎦ 

– 0.841 ≈ 0.159 . 

-------------------------------------------------------------------------------------------------------- 


Let X1,..., X 100 denote the number of pensions that will be provided to each new recruit. 

Now under the assumptions given, 

⎧0 with probability 1− 0.4 = 0.6 

⎪ 

Xi 

= ⎨1 

with probability ( 0.4)( 0.25) = 0.1 

⎪ 

⎩2 

with probability ( 0.4)( 0.75) = 0.3 

for i = 1,...,100 . Therefore, 

E X = 0 0.6 + 1 0.1 + 2 0.3 = 0.7 , 

[ ] ( )( ) ( )( ) ( )( ) 

i 

2 2 2 

( ) ( ) ( ) ( ) ( ) ( ) 

2 

⎡ ⎤ = + + = 

E X 

⎣ i ⎦ 

0 0.6 1 0.1 2 0.3 1.3 , and 

2 

[ Xi] E⎡X ⎤ i { E[ Xi] 

} ( ) 

2 2 

Var = 

⎣ ⎦ 

− = 1.3− 0.7 = 0.81 

Since X1,..., X100 

are assumed by the consulting actuary to be independent, the Central 

Limit Theorem then implies that S = X1 + ... + X100 is approximately normally distributed 

with mean 

E[ S] = E[ X1] + ... + E[ X100] 

= 100( 0.7) = 70 

and variance 

Var[ S] = Var [ X1] + ... + Var[ X100 

] = 100( 0.81) = 81 

Consequently, 

⎡S−70 90.5 −70⎤ 

Pr[ S ≤ 90.5] = Pr 

⎢ 

≤ 

⎣ 9 9 ⎥ 

⎦ 

= Pr[ Z ≤2.28] 

= 0.99 

Page 36 of 55

-------------------------------------------------------------------------------------------------------- 


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 

μ = E[X] = 0 

x 

2.5 2 3 

3 

x x 2.5 2(2.5) 

σx ∫ dx = −2.5 

= 

5 15 15 

−2.5 

2 = Var[X] = E[X 2 ] = =2.083 

σx =1.443 

Now X 48 , the difference between the means of the true and rounded ages, has a 

distribution that is approximately normal with mean 0 and standard deviation 1.443 

48 = 

0.2083 . Therefore, 

⎡ 1 1 ⎤ ⎡ −0.25 

0.25 ⎤ 

P 

⎢ 

− ≤ X48 ≤ = P ≤ Z ≤ 

⎣ 4 4⎥ ⎦ 

⎢ 

⎣0.2083 

0.2083⎥ 

= P[−1.2 ≤ Z ≤ 1.2] = P[Z ≤ 1.2] – P[Z ≤ – 

⎦ 

1.2] 

= P[Z ≤ 1.2] – 1 + P[Z ≤ 1.2] = 2P[Z ≤ 1.2] – 1 = 2(0.8849) – 1 = 0.77 . 

-------------------------------------------------------------------------------------------------------- 


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 

−x 

/6 

= 1 − e , x> 

0 and 

( ) 

( ) − y /3 

G y = 1 − e , y > 0 

Therefore, 

Pr ⎡⎣ X ≤3∩ Y ≤ 2 ⎤⎦ 

= Pr X ≤3PrY ≤ 2 

( ) ( ) [ ] [ ] 

( ) ( ) ( )( ) 

= F 3 G 2 = 1−e 1− e = 1−e 

− e + e 

− 

−1/2 −2/3−2/3−1/2 7/6 

Page 37 of 55


We are given that 

⎧ 6 

⎪ (50 − x− y) for 0 < x< 50 − y< 

50 

f( x, y) 

= ⎨125,000 ⎪⎩ 0 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) 

x 

50 

30 50− 

x 

6 

We conclude that P[X > 20 ∩ Y > 20] = 

(50 − x − y) 

125,000 ∫∫ dy dx . 

-------------------------------------------------------------------------------------------------------- 


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 T 1 and T 2 is 

⎛1 ⎞⎛1 ⎞ 1 

f( t1, t2) = ⎜ e ⎟⎜ e ⎟= 

e e 

⎝2 ⎠⎝3 ⎠ 6 

P[T < T ] = 

2 1 

−t1/2 −t2/3 −t1/2 

−t2/3 

20 20 

∞ t1 

∞ 

1 −t1/2 −t2/3 ⎡ 1 −t1/2 −t2/3⎤ 

t1 

∫∫ e e dt2dt1 = e e dt1 

6 ∫⎢ 

− 

2 ⎥ 

0 0 0⎣ 

⎦0 

, 0 < t 1 < ∞ , 0 < t 2 < ∞ and we need to find 

∞ ∞ 

∞ 

⎡1 −t1/2 1 −t1/2 −t1/3 ⎤ ⎡1 

−t1/2 1 −5 

t1/6⎤ 

⎡ −t1/2 3 −5 

t1/6⎤ 

3 2 

= ∫⎢ e − e e dt1 = e − e d 1 

2 2 ⎥ ∫⎢ t 

2 2 

0⎣ ⎦ 0⎣ 

⎥ 

= e e 1 

⎦ 

⎢ 

− + = − = 

⎣ 5 ⎥ 

⎦05 

5 

= 0.4 . 

-------------------------------------------------------------------------------------------------------- 


We want to find P[X + Y > 1] . To this end, note that P[X + Y > 1] 

= 

1 2 1 

2 

⎡2x+ 2− y⎤ ⎡1 1 1 2 ⎤ 

∫∫⎢ dydx xy y y dx 

4 ⎥ 

= ∫ + − 

2 2 8 01−x ⎣ ⎦ 

⎢ ⎥ 

0⎣ 

⎦ 1−x 

1 

⎡ 1 1 1 1 2 ⎤ 

= x + − − x −x − − x + −x 

dx 

∫ 

0 

⎢ 

1 (1 ) (1 ) (1 ) 

⎣ 2 2 2 8 ⎥ 

⎦ 

1 

⎡5 2 3 1⎤ 

= x + x+ dx 

∫ 

0 

⎢ 

⎣8 4 8⎥ 

⎦ 

1 

3 2 

x + x + x 

0 

⎡ 5 3 1 ⎤ 

= 

⎢ 

⎣24 8 8 ⎥ 

= 

⎦ 

Page 38 of 55 

= 

1 

∫ 

0 

⎡ 1 1 1 1 ⎤ 

⎢ 

+ + − + 

⎣ 2 8 4 8 ⎥ 

⎦ 

2 2 

x x x x dx 

5 3 1 17 

+ + = 

24 8 8 24


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: 

1 

− ⋅ 

Shaded Region Area 

Pr 20 

2 

⎣⎡ X − Y < ⎤ ⎦ = = 2 2 

200 

( 2200 − 2000) 

( ) 

2 

200 2 180 

2 

180 

2 

= 1− = 1− 2 ( 0.9) = 0.19 

200 

More formally (still using symmetry) 

Pr ⎡⎣ X − Y < 20⎤ ⎦ = 1−Pr⎡⎣ X −Y ≥20⎤ ⎦ = 1−2Pr[ X −Y ≥20] 

2200 x−20 

1 2200 1 

= 1− 2∫ 1 2 

2020 ∫ dydx = − 

2000 2 2020 2 

200 ∫ 200 

x−20 

y 2000 dx 

2 

= 1− 2 

200 ∫ 2020 

x−20 − 2000 

1 

dx= 1− 2 

200 

x−2020 

2 

⎛180 ⎞ 

= 1− ⎜ ⎟ = 0.19 

⎝200 ⎠ 

( ) ( ) 

2200 2 2200 

2020 

Page 39 of 55 

2

-------------------------------------------------------------------------------------------------------- 


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 

density function f ( x, y) = , 0< x< 10 , 0< 

y< 

10 

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 < X < 6, 0 < Y < 2 + Pr 0 < X < 1, 2 < Y < 7 + Pr 1< X < 6, 2 < Y < 8− 

X 

( ) ( ) ( 

( )( ) ()( ) ( )( )( ) 

6 2 1 5 1 2 5 5 12 5 12.5 

= + + = + + = 0.295 

100 100 100 100 100 100 

-------------------------------------------------------------------------------------------------------- 


Let f ( t1, t 2) 

denote the joint density function of 1 . The domain of f is pictured 

below: 

and T 2 T 

Now the area of this domain is given by 

( ) 2 

2 1 

A = 6 − 6− 4 = 36− 2=34 

2 

Page 40 of 55 

)

Consequently, 

1 2 1 2 

( , ) 34 

f t t 

1 2 

⎧ 1 

⎪ , 0< t < 6 , 0< t < 6 , t + t < 10 

= ⎨ 

⎪⎩ 0 elsewhere 

and 

ET+ T = ET + ET = 2 ET (due to symmetry) 

[ ] [ ] [ ] [ ] 

1 2 1 2 1 

⎧ 4 6 1 6 10−t11 ⎫ ⎧ 4 ⎡t 6 

2 6 ⎤ ⎡t2 10−t 

⎤ ⎫ 

1 

= 2⎨∫ t1 2 1 1 2 1 1 0 1 1 0 

0 ∫ dt dt + 2 

0 34 ∫ t 

4 ∫ dt dt ⎬= ⎨ 0 34 ∫ t 

0 ⎢ 

34 

⎥dt + ∫ t 

4 ⎢ 

34 

⎥dt1⎬ 

⎩ ⎭ ⎩ ⎣ ⎦ ⎣ ⎦ ⎭ 

2 

⎧ 43t 6 

1 1 2 ⎫ ⎧3t1 4 1 ⎛ 2 1 3⎞6⎫ 

= 2⎨∫ dt1 + ( 10t1 − t1 ) dt 

0 4 

1⎬= 2 0+ ⎜5t1 − t1 

⎩ 17 ∫ 

⎨ ⎟ 4 ⎬ 

34 ⎭ ⎩34 34 ⎝ 3 ⎠ ⎭ 

⎧24 1 ⎡ 64⎤⎫ 

= 2⎨ + 180 −72 − 80 + = 5.7 

17 34 ⎢ ⎬ 

⎣ 3 ⎥ 

⎩ ⎦⎭ 

-------------------------------------------------------------------------------------------------------- 


tW 1 + t2Z t1( X+ Y) + t2( Y−X) ( t1− t2) X ( t1+ t2) Y 

M ( t1, t2) = E⎡e E⎡e ⎤ E⎡e e ⎤ 

⎣ 

⎤ 

⎦ = 

⎣ ⎦ 

= 

⎣ ⎦ 

( 

= E⎡ ⎣ 

e 

) ⎤ ( 

E⎡ ⎦ ⎣ 

e 

) 

1 2 1 2 1 2 2 1 2 2 

( t1− t2) ( t1+ t2) ( t1 − 2t1t2+ t2) ( t1 + 2t1t2+ 

t2) 

⎤ 2 2 2 2 

⎦ 

= e e = e e = e 

2 2 

t1− t2 X t1+ t2 Y t1 + t2 

-------------------------------------------------------------------------------------------------------- 


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 55


We are given f(t1, t ) = 2/L 2 2 , 0 ≤ t1 

≤ t2 ≤ L . 

L t2 

2 2 

2 2 2 

Therefore, E[T1 + T 2 ] = ∫∫( 

t1 + t2 ) dt 2 1dt2 

= 

L 

3 t2 

2 ⎧L ⎪ ⎡t1 2 ⎤ 

2 ⎨ t2 t1 L ∫⎢ + ⎥ 

⎪⎩ 3 0⎣ ⎦0 ⎫ L 3 

⎪ 2 ⎪⎧⎛t 

⎫ 

2 3⎞ 

⎪ 

dt1⎬= + t 2 ⎨ ⎜ 2 ⎟dt2⎬ 

⎪ L ∫ 

⎭ ⎩⎪3 0⎝ 

⎠ ⎭⎪ 

L 

2 4 3 

= t 2 2 dt2 L ∫ 3 

4 L 

2 ⎡t⎤ 2 = 2 ⎢ ⎥ 

L ⎣ 3 ⎦ 

2 2 

= L 

3 

t 2 

0 0 

( L, L) 

t 1 

0 0 

-------------------------------------------------------------------------------------------------------- 


Let g(y) be the probability function for Y = X X X . Note that Y = 1 if and only if 

1 2 3 

X1 = X 2 = X 3 = 1 . Otherwise, Y = 0 . Since P[Y = 1] = P[X 1 = 1 ∩ X2 

= 1 ∩ X 3 = 1] 

= P[X1 = 1] P[X2 = 1] P[X = 1] = (2/3) 3 = 8/27 . 

3 

⎧19 

⎪ 

for y = 0 

27 

⎪ 8 

We conclude that g( y) = ⎨ for y = 1 

⎪27 

⎪0 

otherwise 

⎪ 

⎩ 

y 19 8 

t 

t 

and M(t) = E ⎡ 

⎣e ⎤ 

⎦ = + e 

27 27 

Page 42 of 55


We use the relationships Var aX + b 

2 

= a Var X , Cov aX , bY = ab Cov X , Y , and 

( ) ( ) ( ) ( ) 

( X + Y) = ( X) + ( Y) + ( X Y) 

= ( X+ Y) = + + ( XY) ( XY) 

= 

⎡⎣( X + ) + Y⎤⎦ 

= ar ⎡⎣( X + 1.1Y) + 100⎤⎦ 

[ X Y] X ⎡( ) Y⎤ ( X Y) 

Var Var Var 2 Cov , . First we observe 

17,000 Var 5000 10,000 2 Cov , , and so Cov , 1000. 

We want to find Var 100 1.1 V 

= Var + 1.1 = Var + Var ⎣ 1.1 ⎦+ 

2 Cov ,1.1 

= Var X + 1.1 VarY + 2 1.1 Cov X, Y = 5000 + 12,100 + 2200 = 19,300. 

2 

( ) ( ) ( ) 

-------------------------------------------------------------------------------------------------------- 


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 

2 2 

E[X ] = (0) (1/6) + (1) 2 (3/12) + (2) 2 (7/12) = 31/12 

2 

Var[X] = 31/12 – (17/12) = 0.58 . 

-------------------------------------------------------------------------------------------------------- 


Note that due to the independence of X and Y 

2 

Var(Z) = Var(3X − Y − 5) = Var(3X) + Var(Y) = 3 Var(X) + Var(Y) = 9(1) + 2 = 11 . 

-------------------------------------------------------------------------------------------------------- 


Let X and Y denote the times that the two backup generators can operate. Now the 

2 

variance of an exponential random variable with mean β is β . Therefore, 

2 

[ X] [ Y] 

Var = Var = 10 = 100 

Then assuming that X and Y are independent, we see 

Var X+Y = Var X + Var Y = 100 + 100 = 200 

[ ] [ ] [ ] 

Page 43 of 55


Let X1, X2, and X 3 denote annual loss due to storm, fire, and theft, respectively. In 

( ) 

addition, let Y = Max X , X , X . 

Then 

1 2 3 

[ Y > ] = − [ Y ≤ ] = − [ X ≤ ] [ X ≤ ] [ X ≤ ] 

Pr 3 1 Pr 3 1 Pr 3 Pr 3 Pr 3 

( −3 

e )( 

−3 1.5 e )( 

−3 

2.4 e ) 

( −3 e )( −2 

e )( 

−5 

4 e ) 

= 1− 1− 1− 1− 

= 1− 1− 1− 1− 

1 2 3 

= 0.414 

* Uses that if X has an exponential distribution with mean μ 

1 t t 

Pr ( X x) 1 Pr ( X x) 1 e dt 1 ( e ) x 1 e x 

∞ 

− − ∞ − 

≤ = − ≥ = − ∫ = − − = − 

μ 

x 

* 

μ μ μ 

-------------------------------------------------------------------------------------------------------- 


Let us first determine k: 

1 1 11 1 

2 1 k k 

1 = ∫∫ kxdxdy = 0 

0 0 ∫ kx | dy = dy 

0 2 ∫ = 

0 2 2 

k = 2 

Then 

1 1 

1 

2 2 2 3 1 2 

E[ X] = ∫∫2x dydx= ∫ 2x 

dx= x | 0 = 

0 3 3 

0 0 

1 1 

1 1 2 1 1 

E [ Y ] = ∫∫y 2 x dxdy = ∫ ydy = y | 0 = 

0 2 2 

0 0 

1 1 1 1 

2 2 3 1 2 

E[ XY] = ∫∫ 2x 

ydxdy = 0 

0 0 ∫ x y| dy = 

0 3 ∫ ydy 

0 3 

2 2 1 2 1 

= y | 0 = = 

6 6 3 

1 ⎛2⎞⎛1⎞ 1 1 

Cov [ XY , ] = EXY [ ] − EX [ ] EY [ ] = − ⎜ ⎟⎜ ⎟= 

− = 0 

3 ⎝3⎠⎝2⎠ 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 55


The calculation requires integrating over the indicated region. 

2x1 1 2 8 1 1 1 

2 4 2 2 4 2 2 2 4 4 5 

x 

( ) ∫∫ ∫ ∫ ( ) ∫ 

E X = x y dy dx = x y dx = x 4x− x dx = 4xdx 

= x 

3 3 3 5 

0 x 

0 0 0 

x 

2x 1 

1 2 8 1 1 1 

2 8 3 8 3 3 56 4 56 5 

x 

( ) ∫∫ ( 8 ) 

x ∫ ∫ ∫ 

E Y = xy dy dx = xy dy dx = x x − x dx = x dx = x 

3 9 9 9 45 

0 0 0 0 

x 

2x 

1 1 

2 2 2 3 2 3 3 5 

1 2x8 18 

8 56 56 28 

E( XY) = ∫∫ x y dydx= x y dx x ( 8x 

x ) dx x d 

0 x 3 ∫ 

= − = 

09 

∫09 ∫ x = = 

0 9 54 27 

28 ⎛56 ⎞⎛4⎞ Cov ( XY , ) = E( XY) − E( X) EY ( ) = − ⎜ ⎟⎜ ⎟= 

0.04 

27 ⎝45 ⎠⎝5⎠ -------------------------------------------------------------------------------------------------------- 


The joint pdf of X and Y is f(x,y) = f 2(y|x) f 1(x) 

= (1/x)(1/12), 0 < y < x, 0 < x < 12 . 

Therefore, 

12 x 

12 

1 y x 

E[X] = ∫∫x⋅ dydx= 12x ∫12 0 

12 2 

x x 12 

dx= ∫ dx= 

12 24 0 

= 6 

0 0 0 0 

12 12 2 x 12 

2 

12 

x 

y ⎡ y ⎤ x x 144 

E[Y] = dydx = ⎢ ⎥ dx = dx = = = 3 

E[XY] = 

∫∫ ∫ ∫ 

12x 0 0 0⎣24x⎦024 0 48 0 48 

12 12 2 x 12 2 3 

12 

3 

x 

y ⎡y ⎤ x x (12) 

dydx = ⎢ ⎥ dx = dx = = 

∫∫ ∫ ∫ 

12 0 

0 0 0⎣24⎦ 24 72 72 

0 0 

Cov(X,Y) = E[XY] – E[X]E[Y] = 24 − (3)(6) = 24 – 18 = 6 . 

x 

Page 45 of 55 

= 24 

0 

= 

0 

4 

5 

= 

56 

45


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 + 2.2Cov X, Y + 1.2VarY 

( ) ( ) ( ) ( ) 

( ) ( ( ) ) 

( 2 ) ( E Y ) 

( ) 

2 2 

2 

Var X = E X − E X = 27.4 − 5 = 2.4 

= − ( 

2 

) 

2 

= 51.4 − 7 = 2.4 

( X + Y) = X + Y + ( X Y) 

VarY 

E Y 

Var Var Var 2Cov , 

Cov XY , 

1 

= ( Var 

2 

X+ Y 

1 

−VarX − VarY) = ( 8 −2.4 −2.4) 

= 1.6 

2 

Cov , 2.4 2.2 1.6 1.2 2.4 8.8 

( ) ( ) 

( C C ) = + ( ) + ( ) = 

1 2 

-------------------------------------------------------------------------------------------------------- 

107. Alternate solution: 

We are given the following information: 

C1= X + Y 

C2= X + 1.2Y 

E[ X] 

= 5 

2 

E⎡ ⎣X ⎤ 

⎦ = 27.4 

EY [ ] = 7 

2 

E⎡ ⎣Y ⎤ 

⎦ = 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] 

2 2 

= E⎡ ⎣X + 2.2XY + 1.2Y ⎤ 

⎦− 

E[ X] + E[ Y] E X + 1.2E 

Y 

2 2 

= E⎡ ⎣X ⎤ 

⎦+ 2.2E[ XY] + 1.2E⎡ ⎣Y ⎤ 

⎦− 

( 5 + 7) ( 5 + ( 1.2) 7) 

= 27. 4 + 2.2E[ XY] 

+ 1.2( 51.4) −( 

12)( 13.4) 

= 2.2E[ XY] 

−71.72 

Therefore, we need to calculate 

( ) ( [ ] [ ] ) 

E[ XY ] first. To this end, observe 

Page 46 of 55

2 

[ X Y] E⎡( X Y) ⎤ ( E[ X Y] 

) 

8= Var + = 

⎣ 

+ − 

⎦ 

+ 

= ⎡ 

⎣ + + ⎤ 

⎦− 

+ 

( [ ] [ ] ) 

2 2 

E X 2XY 

Y E X E Y 

2 2 

2 

= E⎡ ⎣X ⎤ 

⎦+ 2E[ XY] + E⎡ ⎣Y ⎤ 

⎦− 

( 5 + 7) 

= 27.4 + 2E[ XY] 

+ 51.4 −144 

= 2E[ XY] 

−65.2 

[ ] = ( 8 + 65.2) 2 = 36.6 

( CC 1, 2 ) = ( ) − = 

E XY 

Finally, Cov 2.2 36.6 71.72 8.8 

-------------------------------------------------------------------------------------------------------- 


The joint density of 1 T2 

is given by 

> 0 

and T 

−t1 −t2 

f ( t1, t2) = e e 

Therefore, 

, t1 > 0 , t2 

Pr X ≤ x = Pr 2T 

+ T ≤ x 

[ ] [ ] 

1 2 

1 

x ( x−t2) x ⎡ 

2 −t1 −t2 −t2 −t1 

= ∫∫ e e dt1dt2 = 

0 0 ∫ e ⎢−e 0 

⎢⎣ 1 

( x−t2) ⎤ 

2 

⎥dt2 

0 ⎥⎦ 

1 1 1 1 

x − x+ t2 x 

− x − t 

−t ⎡ ⎤ ⎛ 2 

2 2 2 −t 

⎞ 

2 2 2 

= ∫ e ⎢1− e ⎥dt2 

= e e e dt2 

0 ∫ ⎜ − ⎟ 

0 

⎣ ⎦ ⎝ ⎠ 

⎡ 

= ⎢− e 

⎣ 

+ e e 

⎤ 

⎥ 

⎦ 

=− e + e e + − e 

1 1 1 1 1 

− x − t2− x − x 

− x 

−t2 2 2 x −x 

2 2 2 

2 0 2 1 2 

1 1 

− x − x 

−x 

−x = 1− e + 2e 

− 2e 2 = 1− 2 e 2 −x 

+ e , x> 

0 

It follows that the density of X is given by 

1 1 

d ⎡ − x ⎤ − x 

2 −x 2 −x 

g( x) = ⎢1− 2 e + e ⎥ = e − e 

dx ⎣ ⎦ 

, x > 0 

Page 47 of 55 

2 

2


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, 

−u ⎛1 −v/2 ⎞ 1 −u −v/2 

g( u, v) = ( e ) ⎜ e ⎟= 

e e , 0 < u< ∞ , 0 < v< 

∞ 

⎝2 ⎠ 2 

and 

⎡u⎤ F( x) = Pr[ X ≤ x] = Pr 

⎢ 

≤ x = Pr[ 

U ≤Vx] 

⎣v⎥ ⎦ 

∞ vx ∞ vx 1 −u −v/2 

= ∫∫ g ( u, v) dudv = 

0 0 ∫∫ e e dudv 

0 0 2 

∞ 1 ∞ 

−u −v/2 vx ⎛ 1 −vx −v/2 1 −v/2⎞ 

= ∫ − e e | 

0 

0 dv = e e e dv 

2 ∫0 

⎜− + ⎟ 

⎝ 2 2 ⎠ 

∞⎛ 

1 − vx ( + 1/2 ) 1 −v 

/2⎞ 

= ∫ e e dv 

0 ⎜− + ⎟ 

⎝ 2 2 ⎠ 

∞ 

⎡ 1 − vx ( + 1/2) 

−v 

/2⎤ 

1 

= 

⎢ 

e −e 

=− + 1 

⎣2x+ 1 

⎥ 

⎦02x+ 

1 

2 

Finally, f ( x) = F'( x) 

= 

2x+ 1 

( ) 2 

-------------------------------------------------------------------------------------------------------- 


Note that the conditional density function 

⎛ 1⎞ f ⎜y x= 3 

⎟= 

⎝ ⎠ 

f ( 13, y) 

2 

, 0 < y< 

, 

fx 

( 1 3) 

3 

⎛1⎞ fx⎜ ⎟= 

3 

23 

24( 1 3) ydy = 

0 

23 

2 

8ydy = 4y 

0 

23 

16 

= 

0 9 


Consequently, 

⎝ ⎠ ∫ ∫ 

⎛ 1⎞ 9 9 

2 

f ⎜y x= f ( 13, y) y , 0 y 

3 

⎟= 

= < < 

⎝ ⎠ 16 2 3 

13 

2 

139 

9 1 

Pr ⎡ ⎣Y < X X = 1 3⎤ 

⎦ = ∫ ydy= y = 

0 2 4 4 

Page 48 of 55 

0


3 f 

Pr ⎡ ⎣1< Y < 3 X = 2⎤ 

⎦ = ∫ 1 f 

y 

dy 

1 

( − ) 

( 2, ) 

( 2) 

2 ( ) 1 

f ( 2, y) = y = y 

4 2 1 2 

∞ 

− 4−1 2−1 −3 

∞ 

−3 −2 

1 1 1 

fx( 2) 

= ∫ y dy =− y = 

2 4 4 

x 

1 

1 

3 

−3 

y dy 

1 2 

−2 

3 1 8 

1 1 9 9 

Finally, Pr ⎡ ⎣1< Y < 3 X = 2⎤ ⎦ = =− y = 1− 

= 

∫ 

-------------------------------------------------------------------------------------------------------- 


We are given that the joint pdf of X and Y is f(x,y) = 2(x+y), 0 < y < x < 1 . 

x 

∫ 

0 

4 

2 

x 

⎡ 

⎣ 

⎤ 

⎦0 

Now fx(x) = (2x+ 2 y) dy = 2xy+ 

y = 2x 2 + x 2 = 3x 2 , 0 < x < 1 

f ( xy , ) 2( x+ y) 2⎛1 y⎞ 

so f(y|x) = = = 2 ⎜ + 2 ⎟ , 0 < y < x 

fx ( x) 3x 3⎝ 

x x ⎠ 

2⎡ 1 y ⎤ 2 

f(y|x = 0.10) = + = [ 10 + 100y 

3 ⎢ 

⎣0.1 0.01⎥ ] , 0 < y < 0.10 

⎦ 3 

0.05 

2 ⎡20 100 2 ⎤ 0.05 1 1 5 

P[Y < 0.05|X = 0.10] = ∫ [ 10 + 100ydy 

] = y+ y = + = 

3 ⎢ 

⎣ 3 3 ⎥ 

= 0.4167 . 

⎦ 3 12 12 

0 

-------------------------------------------------------------------------------------------------------- 


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 

c 

Then Pr[ H] = P[ H ∩ W] + Pr ⎡ 

⎣H ∩ W ⎤ 

⎦ = 0.96 + 0.01 = 0.97 

and 

Pr[ W ∩ H] 

0.96 

Pr[ W| H] 

= = = 0.9897 

Pr H 0.97 

[ ] 

c 

Pr ⎡H ∩W 

⎤ 

c 

0.01 

Pr ⎡W H 

⎣ ⎦ 

⎣ | ⎤ 

⎦ = = = 0.0103 

Pr 0.97 

[ H ] 

Page 49 of 55 

0


c 

{ | ⎡ | ⎤} 

[ ] [ ] [ ] ( ) [ ] ( ) 

E P = E 1000 − B = 1000 − E B = 1000 − 0 Pr W H + 10,000 Pr ⎣W H⎦ 

( ) 

= 1000 − 10,000 0.0103 = 1000 − 103 = 897 

-------------------------------------------------------------------------------------------------------- 


Note that 

PX ( = 1, Y= 0) PX ( = 1, Y= 

0) 0.05 

P(Y = 0⏐X = 1) = 

= = 

PX ( = 1) PX ( = 1, Y= 0) + PX ( = 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(Y 2 ⏐X = 1) = (0) 2 P(Y = 0⏐X = 1) + (1) 2 P(Y = 1⏐X = 1) = 0.714 

Var(Y⏐X = 1) = E(Y 2 2 2 

⏐X = 1) – [E(Y⏐X = 1)] = 0.714 – (0.714) = 0.20 

-------------------------------------------------------------------------------------------------------- 


Let f1(x) 

denote the marginal density function of X. Then 

x+ 

1 

x+ 

1 

f1 ( x) = ∫ 2xdy = 2xy| x = 2x( x+ 1− x) = 2 x , 0 < x< 

1 

x 

Consequently, 

f ( x, y) 

⎧1 

if: x< y< x+ 

1 

f ( y| x) 

= =⎨ 

f1( x) 

⎩0 

otherwise 

x+ 

1 1 2 x+ 

1 1 2 1 2 1 2 1 1 2 1 

EY [ | X] = ∫ ydy= y| x = ( x+ 1) 

− x = x + x+ − x = x+ 

x 2 2 2 2 2 2 2 

x+ 

1 

2 2 1 3 x+ 

1 1 3 1 3 

E⎡ ⎣Y | X⎤ ⎦ = ∫ y dy = y | x = ( x+ 1) 

− x 

x 3 3 3 

1 3 2 1 1 3 2 1 

= x + x + x+ − x = x + x+ 

3 3 3 3 

Var 

[ | ] ⎡ | ⎤ { [ | ] } 

2 

2 

2 

Y X = E⎣Y X⎦− E Y X = x 

1 1 1 

3 4 12 

2 2 

= x + x+ −x −x− = 

Page 50 of 55 

1 ⎛ 1⎞ 

+ x+ − ⎜x+ ⎟ 

3 ⎝ 2⎠ 

2


Denote the number of tornadoes in counties P and Q by NP and NQ, 

respectively. Then 

E[NQ|N P = 0] 

= [(0)(0.12) + (1)(0.06) + (2)(0.05) + 3(0.02)] / [0.12 + 0.06 + 0.05 + 0.02] = 0.88 

E[NQ 2 |N P = 0] 

2 

= [(0) (0.12) + (1) 2 (0.06) + (2) 2 (0.05) + (3) 2 (0.02)] / [0.12 + 0.06 + 0.05 + 0.02] 

2 

= 1.76 and Var[NQ|NP = 0] = E[NQ |NP = 0] – {E[NQ|NP = 0]} 2 = 1.76 – (0.88) 2 

= 0.9856 . 

-------------------------------------------------------------------------------------------------------- 


The domain of X and Y is pictured below. The shaded region is the portion of the domain 

over which X

or more precisely, 

( ) 

g y 

( ) 12 

⎧ 32 

⎪15y 

1 − y , 0 < y< 

1 

= ⎨ 

⎪⎩ 0 otherwise 

-------------------------------------------------------------------------------------------------------- 


The diagram below illustrates the domain of the joint density 

We are told that the marginal density function of X is 

while 

yx 


Therefore, 

( ) = 1, < < + 1 

f yx x y x 

x 

f ( xy , ) of Xand Y. 

( ) = 1,0< < 1 

f x x 

⎧1 

if 0< x < 1, x< y< x+ 

1 

f ( x, y) = fx( x) fyx( y x) 

=⎨ 

⎩0 

otherwise 

[ ] [ ] 

1 1 

2 2 

∫ ∫ 

Pr Y > 0.5 = 1−Pr Y ≤ 0.5 = 1− 

dydx 

0 

1 1 1 

2 2 

1 

2 ⎛1 ⎞ ⎡1 1 2 ⎤ 1 1 7 

2 

= 1− ∫ y 1 1 

0 

x dx = −∫ x dx x x 

0 ⎜ − ⎟ = − − 0 = 1− 

+ = 

⎝2 ⎠ ⎣ 

⎢2 2 ⎦ 

⎥ 4 8 8 

[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.] 

Page 52 of 55 

x


We are given that X denotes loss. In addition, denote the time required to process a claim 

by T. 

⎧3 

2 1 3 

⎪ x ⋅ = x, x< t < 2 x,0≤ x≤ 

2 

Then the joint pdf of X and T is f( x, t) = ⎨8 x 8 

⎪⎩ 0, otherwise. 

Now we can find P[T ≥ 3] = 

4 2 4 2 

3 ⎡ 3 2⎤ ∫∫ xdxdt = 

8 ∫⎢ x 

16 3 t /2 3⎣ ⎥ 

⎦t/2 = 11/64 = 0.17 . 

4 

4 

⎛12 3 2⎞ ⎡1213⎤ dt = ∫⎜ 

− t ⎟dt= − t 

16 64 16 64 3⎝ 

⎠ ⎣ 

⎢ 

⎦ 

⎥ 

3 

12 ⎛36 

27 ⎞ 

= −1−⎜ − ⎟ 

4 ⎝16 

64 ⎠ 

t t=2x 

4 

3 

2 

1 

1 2 

t=x 

x 

-------------------------------------------------------------------------------------------------------- 


The marginal density of X is given by 

1 

3 

1 1 1 

1 

2 ⎛ xy ⎞ ⎛ x⎞ 

fx( x) = ∫ ( 10 − xy ) dy = ⎜10y− ⎟ = 10 

0 

⎜ − ⎟ 

64 64 ⎝ 3 ⎠ 64 ⎝ 3 ⎠ 

Then 

2 

10 10 1 ⎛ x ⎞ 

∫2 x ∫ ⎜ ⎟ x 

2 

E( X) = xf ( x) dx= 10x− 

d 

64 ⎝ 3 ⎠ 

= 1 1000 8 

⎡⎛ ⎞ ⎛ ⎞⎤ 

500 20 

64 

⎢⎜ − ⎟−⎜ − ⎟ 

9 9 

⎥ = 5.778 

⎣⎝ ⎠ ⎝ ⎠ ⎦ 

0 

Page 53 of 55 

10 

3 

⎛ 2 x ⎞ 

1 

= ⎜5x−⎟ 64 ⎝ 9 ⎠ 

2


x 

6 

e − 

The marginal distribution of Y is given by f2(y) 

= e e dx = 6 e dx 

–2y –y –2y –2y –3y 

= −6 e e + 6e = 6 e – 6 e , 0 < y < ∞ 

∞ 

−2y −3y 

Therefore, E(Y) = ∫ y f2(y) 

dy = ∫ (6ye − 6 ye ) dy = 6 

∞ 

6 –2y 6 –3y 

2 

2 ∫ ye dy − 3 

3 ∫ y e dy 

0 

0 

∞ 

0 

∞ 

0 

y 

∫ 

0 

–x –2y –2y 

∞ 

∫ 

0 

ye 

−2 

y 

y 

∫ 

0 

∞ 

∫ 

–3y 

dy – 6 y e dy = 

∞ 

∞ 

But 

–2y 

2 y e dy and 

–3y 

3y 

e dy are equivalent to the means of exponential random 

∫ 

0 

∫ 

0 

∞ 

–2y 

variables with parameters 1/2 and 1/3, respectively. In other words, 2 y e dy = 1/2 

∞ 

–3y 

and 3y 

e 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 . 

-------------------------------------------------------------------------------------------------------- 


Observe 

Pr 4< S < 8 = Pr⎡ ⎣4< S < 8 N = 1⎤ ⎦Pr N = 1 + Pr⎡ ⎣4< S < 8 N > 1⎤ ⎦Pr 

1 −4 −8 

1 −1 

5 5 2 −1 

= 

3( e − e ) + 

6( 

e −e 

) * 

= 0.122 

N > 1 

*Uses that if X has an exponential distribution with mean μ 

a b 

1 t 1 t 

Pr ( a X b) Pr ( X a) Pr ( X b) e dt e dt e e μ 

∞ ∞ 

− − 

− μ − μ μ 

≤ ≤ = ≥ − ≥ = ∫ − = − 

μ ∫ μ 

[ ] [ ] [ ] 

Page 54 of 55 

a b 

∫ 

0 

0


− x −2 

y 

Because f(x,y) can be written as f ( x) 

f ( y) 

= e 2e 

and the support of f(x,y) is a cross 

product, X and Y are independent. Thus, the condition on X can be ignored and it suffices 

−2 

y 

to just consider f ( y) 

= 2e 

. 

Because of the memoryless property of the exponential distribution, the conditional 

density of Y is the same as the unconditional density of Y+3. 

Because a location shift does not affect the variance, the conditional variance of Y is 

equal to the unconditional variance of Y. 

Because the mean of Y is 0.5 and the variance of an exponential distribution is always 

equal to the square of its mean, the requested variance is 0.25. 

Page 55 of 55

EXAM P SAMPLE SOLUTIONS

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