BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: A survey of a group's viewing habits over the last year revealed the following information:
(i) \(28 \%\) watched gymnastics
(ii) \(29 \%\) watched baseball
(iii) \(19 \%\) watched soccer
(iv) \(14 \%\) watched gymnastics and baseball
(v) \(12 \%\) watched baseball and soccer
(vi) \(10 \%\) watched gymnastics and soccer
(vii) \(8 \%\) watched all three sports.

Calculate the percentage of the group that watched none of the three sports during the last year.

Options:
A
\(\quad 24 \%\)
B
\(\quad 36 \%\)
C
\(41 \%\)
D
\(52 \%\)
E
\(60 \%\)

Correct answer: D

Solution: Let \(G=\) viewer watched gymnastics, \(B=\) viewer watched baseball, \(S=\) viewer watched soccer. Then we want to find
\[
\begin{aligned}
& \operatorname{Pr}\left[(G \cup B \cup S)^{c}\right]=1-\operatorname{Pr}(G \cup B \cup S) \\
& =1-[\operatorname{Pr}(G)+\operatorname{Pr}(B)+\operatorname{Pr}(S)-\operatorname{Pr}(G \cap B)-\operatorname{Pr}(G \cap S)-\operatorname{Pr}(B \cap S)+\operatorname{Pr}(G \cap B \cap S)] \\
& =1-(0.28+0.29+0.19-0.14-0.10-0.12+0.08)=1-0.48=0.52
\end{aligned}
\]
Question: You are given \(P[A \cup B]=0.7\) and \(P\left[A \cup B^{\prime}\right]=0.9\).

Calculate \(P[A]\).

Options:
A
0.2
B
0.3
C
0.4
D
0.6
E
0.8

Correct answer: D

Solution: First note
\[
\begin{aligned}
& P[A \cup B]=P[A]+P[B]-P[A \cap B] \\
& P\left[A \cup B^{c}\right]=P[A]+P\left[B^{c}\right]-P\left[A \cap B^{c}\right]
\end{aligned}
\]

Then add these two equations to get
\[
\begin{aligned}
& P[A \cup B]+P\left[A \cup B^{c}\right]=2 P[A]+\left(P[B]+P\left[B^{c}\right]\right)-\left(P[A \cap B]+P\left[A \cap B^{c}\right]\right) \\
& 0.7+0.9=2 P[A]+1-P\left[(A \cap B) \cup\left(A \cap B^{c}\right)\right] \\
& 1.6=2 P[A]+1-P[A] \\
& P[A]=0.6
\end{aligned}
\]
Question: An insurance company estimates that \(40 \%\) of policyholders who have only an auto policy will renew next year and \(60 \%\) of policyholders who have only a homeowners policy will renew next year. The company estimates that \(80 \%\) of policyholders who have both an auto policy and a homeowners policy will renew at least one of those policies next year.

Company records show that \(65 \%\) of policyholders have an auto policy, \(50 \%\) of policyholders have a homeowners policy, and 15\% of policyholders have both an auto policy and a homeowners policy.

Using the company's estimates, calculate the percentage of policyholders that will renew at least one policy next year.

Options:
A
\(20 \%\)
B
\(29 \%\)
C
\(\quad 41 \%\)
D
\(53 \%\)
E
\(70 \%\)

Correct answer: D

Solution: Let \(A=\) event that a policyholder has an auto policy and \(H=\) event that a policyholder has a homeowners policy. Then,
\[
\begin{aligned}
& \operatorname{Pr}(A \cap H)=0.15 \\
& \operatorname{Pr}\left(A \cap H^{c}\right)=\operatorname{Pr}(A)-\operatorname{Pr}(A \cap H)=0.65-0.15=0.50 \\
& \operatorname{Pr}\left(A^{c} \cap H\right)=\operatorname{Pr}(H)-\operatorname{Pr}(A \cap H)=0.50-0.15=0.35
\end{aligned}
\]
and the portion of policyholders that will renew at least one policy is given by
\[
\begin{aligned}
& 0.4 \operatorname{Pr}\left(A \cap H^{c}\right)+0.6 \operatorname{Pr}\left(A^{c} \cap H\right)+0.8 \operatorname{Pr}(A \cap H) \\
& =(0.4)(0.5)+(0.6)(0.35)+(0.8)(0.15)=0.53 \quad(=53 \%)
\end{aligned}
\]
Question: An actuary studying the insurance preferences of automobile owners makes the following conclusions:
(i) An automobile owner is twice as likely to purchase collision coverage as disability coverage.
(ii) The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage.
(iii) The probability that an automobile owner purchases both collision and disability coverages is 0.15 .

Calculate the probability that an automobile owner purchases neither collision nor disability coverage.

Options:
A
0.18
B
0.33
C
0.48
D
0.67
E
\(\quad 0.82\)

Correct answer: B

Solution: Let \(C=\) Event that a policyholder buys collision coverage and \(D=\) Event that a policyholder buys disability coverage. Then we are given that \(P[C]=2 P[D]\) and \(P[C \cap D]=0.15\).
By the independence of \(C\) and \(D\),
\(0.15=[C \cap D]=P[C] P[D]=2 P[D]^{2}\)
\(P[D]^{2}=0.15 / 2=0.075\)
\(P[D]=\sqrt{0.075}, P[C]=2 \sqrt{0.075}\).
Independence of \(C\) and \(D\) implies independence of \(C^{c}\) and \(D^{c}\). Then
\(P\left[C^{c} \cap D^{c}\right]=P\left[C^{c}\right] P\left[D^{c}\right]=(1-2 \sqrt{0.075})(1-\sqrt{0.075})=0.33\).
Question: A doctor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of her patients and notes their blood pressures (high, low, or normal) and their heartbeats (regular or irregular). She finds that:
(i) \(14 \%\) have high blood pressure.
(ii) \(22 \%\) have low blood pressure.
(iii) \(15 \%\) have an irregular heartbeat.
(iv) Of those with an irregular heartbeat, one-third have high blood pressure.
(v) Of those with normal blood pressure, one-eighth have an irregular heartbeat.

Calculate the portion of the patients selected who have a regular heartbeat and low blood pressure.

Options:
A
\(2 \%\)
B
\(\quad 5 \%\)
C
\(8 \%\)
D
\(\quad 9 \%\)
E
\(\quad 20 \%\)

Correct answer: E

Solution: "Boxed" numbers in the table below were computed.

\begin{tabular}{ll|l|l|l} 
& \multicolumn{1}{c}{ High BP } & \multicolumn{1}{c}{ Low BP } & \multicolumn{1}{c}{ Norm BP } & \multicolumn{1}{c}{ Total } \\
\cline { 2 - 5 } 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
\end{tabular}

From the table, \(20 \%\) of patients have a regular heartbeat and low blood pressure.
Question: An actuary is studying the prevalence of three health risk factors, denoted by \(A, B\), and \(C\), within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population has only this risk factor (and no others). For any two of the three factors, the probability is 0.12 that she has exactly these two risk factors (but not the other). The probability that a woman has all three risk factors, given that she has A and \(B\), is \(1 / 3\).

Calculate the probability that a woman has none of the three risk factors, given that she does not have risk factor A .

Options:
A
0.280
B
0.311
C
0.467
D
0.484
E
0.700

Correct answer: C

Solution: Let \(x\) be the probability of having all three risk factors.
\(\frac{1}{3}=P[A \cap B \cap C \mid A \cap B]=\frac{P[A \cap B \cap C]}{P[A \cap B]}=\frac{x}{x+0.12}\)
It follows that
\(x=\frac{1}{3}(x+0.12)=\frac{1}{3} x+0.04\)
\(\frac{2}{3} x=0.04\)
\(x=0.06\)
Now we want to find
\[
\begin{aligned}
P\left[(A \cup B \cup C)^{c} \mid A^{c}\right] & =\frac{P\left[(A \cup B \cup C)^{c}\right]}{P\left[A^{c}\right]} \\
& =\frac{1-P[A \cup B \cup C]}{1-P[A]} \\
& =\frac{1-3(0.10)-3(0.12)-0.06}{1-0.10-2(0.12)-0.06} \\
& =\frac{0.28}{0.60}=0.467
\end{aligned}
\]
Question: An insurance company pays hospital claims. The number of claims that include emergency room or operating room charges is \(85 \%\) of the total number of claims. The number of claims that do not include emergency room charges is \(25 \%\) of the total number of claims. The occurrence of emergency room charges is independent of the occurrence of operating room charges on hospital claims.

Calculate the probability that a claim submitted to the insurance company includes operating room charges.

Options:
A
0.10
B
0.20
C
0.25
D
0.40
E
0.80

Correct answer: D

Solution: Let \(O=\) event of operating room charges and \(E=\) event of emergency room charges. Then \(0.85=P(O \cup E)=P(O)+P(E)-P(O \cap E)\)
\(=P(O)+P(E)-P(O) \mathrm{P}(E)\) (Independence)
Because \(P\left(E^{c}\right)=0.25=1-P(E), P(E)=0.75\),
\(0.85=P(O)+0.75-P(O)(0.75)\)
\(P(O)(1-0.75)=0.85-0.75=0.10\)
\(P(O)=0.10 / 0.25=0.40\).
Question: The probability that a randomly chosen male has a blood circulation problem is 0.25 . Males who have a blood circulation problem are twice as likely to be smokers as those who do not have a blood circulation problem.

Calculate the probability that a male has a blood circulation problem, given that he is a smoker.

Options:
A
\(1 / 4\)
B
\(1 / 3\)
C
\(\quad 2 / 5\)
D
\(\quad 1 / 2\)
E
\(\quad 2 / 3\)

Correct answer: C

Solution: Let:
\(S=\) Event of a smoker
\(C=\) Event of a circulation problem
Then we are given that \(\mathrm{P}[C]=0.25\) and \(\mathrm{P}[S 1 / \mathscr{E}]=2 \mathrm{P}\left[S^{1} / \mathscr{X}^{c}\right]\)
Then,,
\[
\begin{aligned}
& P[C \mid S]=\frac{P[S \mid C] P[C]}{P[S \mid C] P[C]+P\left[S \mid C^{c}\right] P\left[C^{c}\right]} \\
& \quad=\frac{2 P\left[S \mid C^{c}\right] P[C]}{2 P\left[S \mid C^{c}\right] P[C]+P\left[S \mid C^{c}\right](1-P[C])}=\frac{2(0.25)}{2(0.25)+0.75}=\frac{2}{2+3}=\frac{2}{5}
\end{aligned}
\]
Question: A large pool of adults earning their first driver's license includes \(50 \%\) low-risk drivers, \(30 \%\) moderate-risk drivers, and \(20 \%\) high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool.

This month, the insurance company writes four new policies for adults earning their first driver's license.

Calculate the probability that these four will contain at least two more high-risk drivers than low-risk drivers.

Options:
A
0.006
B
0.012
C
0.018
D
0.049
E
0.073

Correct answer: D

Solution: 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
\[
\begin{aligned}
P[z \geq x+2] & =f(0,0,4)+f(1,0,3)+f(0,1,3)+f(0,2,2) \\
& =(0.20)^{4}+4(0.50)(0.20)^{3}+4(0.30)(0.20)^{3}+\frac{4!}{2!2!}(0.30)^{2}(0.20)^{2} \\
& =0.0488
\end{aligned}
\]
Question: The loss due to a fire in a commercial building is modeled by a random variable \(X\) with density function
\[
f(x)= \begin{cases}0.005(20-x), & 0<x<20 \\ 0, & \text { otherwise }\end{cases}
\]

Given that a fire loss exceeds 8, calculate the probability that it exceeds 16.

Options:
A
\(\quad 1 / 25\)
B
\(\quad 1 / 9\)
C
\(1 / 8\)
D
\(1 / 3\)
E
\(3 / 7\)

Correct answer: B

Solution: \[
\begin{aligned}
& P[X>x]=\int_{x}^{20} 0.005(20-t) d t=\left.0.005\left(20 t-\frac{1}{2} t^{2}\right)\right|_{x} ^{20} \\
& =0.005\left(400-200-20 x+\frac{1}{2} x^{2}\right)=0.005\left(200-20 x+\frac{1}{2} x^{2}\right)
\end{aligned}
\]
where \(0<x<20\). Therefore,
\[
P[X>16 \mid X>8]=\frac{P[X>16]}{P[X>8]}=\frac{200-20(16)+1 / 2(16)^{2}}{200-20(8)+1 / 2(8)^{2}}=\frac{8}{72}=\frac{1}{9} .
\]
Question: The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, a one-half refund if it fails during the second year, and no refund for failure after the second year.

Calculate the expected total amount of refunds from the sale of 100 printers.

Options:
A
\(\quad 6,321\)
B
7,358
C
\(\quad 7,869\)
D
10,256
E
12,642

Correct answer: D

Solution: Let \(T\) denote printer lifetime. The distribution function is \(F(t)=1-e^{-t / 2}\). The probability of failure in the first year is \(F(1)=0.3935\) and the probability of failure in the second year is \(F(2)- F(1)=0.6321-0.3935=0.2386\). Of 100 printers, the expected number of failures is 39.35 and 23.86 for the two periods. The total expected cost is \(200(39.35)+100(23.86)=10,256\).
Question: A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is 0.60 . The numbers of accidents that occur in different months are mutually independent.

Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.

Options:
A
0.01
B
0.12
C
0.23
D
0.29
E
\(\quad 0.41\)

Correct answer: D

Solution: One way to view this event is that in the first seven months there must be at least four with no accidents. These are binomial probabilities:
\(\binom{7}{4} 0.4^{4} 0.6^{3}+\binom{7}{5} 0.4^{5} 0.6^{2}+\binom{7}{6} 0.4^{6} 0.6+\binom{7}{7} 0.4^{7}\)
\(=0.1935+0.0774+0.0172+0.0016=0.2897\).
Alternatively, consider a negative binomial distribution where \(K\) is the number of failures before the fourth success (no accidents). Then
\(P[K<4]=0.4^{4}+\binom{4}{1} 0.4^{4} 0.6+\binom{5}{2} 0.4^{4} 0.6^{2}+\binom{6}{3} 0.4^{4} 0.6^{3}=0.2898\)
Question: A probability distribution of the claim sizes for an auto insurance policy is given in the table below:

\begin{tabular}{c|c}
\begin{tabular}{c} 
Claim \\
Size
\end{tabular} & Probability \\
\hline 20 & 0.15 \\
30 & 0.10 \\
40 & 0.05 \\
50 & 0.20 \\
60 & 0.10 \\
70 & 0.10 \\
80 & 0.30
\end{tabular}

Calculate the percentage of claims that are within one standard deviation of the mean claim size.

Options:
A
\(45 \%\)
B
\(\quad 55 \%\)
C
\(\quad 68 \%\)
D
\(\quad 85 \%\)
E
\(100 \%\)

Correct answer: A

Solution: The mean is \(20(0.15)+30(0.10)+40(0.05)+50(0.20)+60(0.10)+70(0.10)+80(0.30)=55\). The second moment is \(400(0.15)+900(0.10)+1600(0.05)+2500(0.20)+3600(0.10)+ 4900(0.10)+6400(0.30)=3500\). The variance is \(3500-55^{2}=475\). The standard deviation is 21.79. The range within one standard deviation of the mean is 33.21 to 76.79 , which includes the values \(40,50,60\), and 70 . The sum of the probabilities for those values is \(0.05+0.20+0.10+ 0.10=0.45\).
Question: An insurance company sells an auto insurance policy that covers losses incurred by a policyholder, subject to a deductible of 100 . Losses incurred follow an exponential distribution with mean 300.

Calculate the \(95{ }^{\text {th }}\) percentile of losses that exceed the deductible.

Options:
A
600
B
700
C
800
D
900
E
1000

Correct answer: E

Solution: This is a conditional probability. The solution is
\[
\begin{aligned}
0.95 & =P[X \leq p \mid X>100]=\frac{P[100 \leq X \leq p]}{P[X>100]}=\frac{F(p)-F(100)}{1-F(100)} \\
& =\frac{1-e^{-p / 300}-1+e^{-100 / 300}}{1-1+e^{-100 / 300}}=\frac{e^{-100 / 300}-e^{-p / 300}}{e^{-100 / 300}}=1-e^{-(p-100) / 300} \\
0.05 & =e^{-(p-100) / 300} \\
-2.9957 & =-(p-100) / 300 \\
p & =999
\end{aligned}
\]
Question: Claims filed under auto insurance policies follow a normal distribution with mean 19,400 and standard deviation 5,000.

Calculate the probability that the average of 25 randomly selected claims exceeds 20,000.

Options:
A
0.01
B
0.15
C
0.27
D
0.33
E
\(\quad 0.45\)

Correct answer: C

Solution: The average has the same mean as a single claim, 19,400 . The standard deviation is that for a single claim divided by the square root of the sample size, \(5,000 / 5=1,000\). The probability of exceeding 20,000 is the probability that a standard normal variable exceeds (20,000\(19,400) / 1,000=0.6\). From the tables, this is \(1-0.7257=0.2743\).
Question: A city has just added 100 new female recruits to its police force. The city will provide a pension to each new hire who remains with the force until retirement. In addition, if the new hire is married at the time of her retirement, a second pension will be provided for her husband. A consulting actuary makes the following assumptions:
(i) Each new recruit has a 0.4 probability of remaining with the police force until retirement.
(ii) Given that a new recruit reaches retirement with the police force, the probability that she is not married at the time of retirement is 0.25 .
(iii) The events of different new hires reaching retirement and the events of different new hires being married at retirement are all mutually independent events.

Calculate the probability that the city will provide at most 90 pensions to the 100 new hires and their husbands.

Options:
A
0.60
B
0.67
C
0.75
D
0.93
E
\(\quad 0.99\)

Correct answer: E

Solution: For a single recruit, the probability of 0 pensions is 0.6 , of 1 pension is \(0.4(0.25)=0.1\), and of 2 pensions is \(0.4(0.75)=0.3\). The expected number of pensions is \(0(0.6)+1(0.1)+2(0.3)=0.7\). The second moment is \(0(0.6)+1(0.1)+4(0.3)=1.3\). The variance is \(1.3-0.49=0.81\). For 100 recruits the mean is 70 and the variance is 81 . The probability of providing at most 90 pensions is (with a continuity correction) the probability of being below 90.5 . This is \((90.5-70) / 9=2.28\) standard deviations above the mean. From the tables, this probability is 0.9887 .
Question: A tour operator has a bus that can accommodate 20 tourists. The operator knows that tourists may not show up, so he sells 21 tickets. The probability that an individual tourist will not show up is 0.02 , independent of all other tourists.

Each ticket costs 50, and is non-refundable if a tourist fails to show up. If a tourist shows up and a seat is not available, the tour operator has to pay 100 (ticket cost +50 penalty) to the tourist.

Calculate the expected revenue of the tour operator.

Options:
A
955
B
962
C
967
D
976
E
985

Correct answer: E

Solution: The tour operator collects \(21 \times 50=1050\) for the 21 tickets sold. The probability that all 21 passengers will show up is \((1-0.02)^{21}=(0.98)^{21}=0.65\). Therefore, the tour operator's expected revenue is \(1050-100(065)=985\).
Question: A car dealership sells 0,1 , or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to buy an extended warranty for the car. Let \(X\) denote the number of luxury cars sold in a given day, and let \(Y\) denote the number of extended warranties sold.
\[
\begin{aligned}
& \mathrm{P}[X=0, Y=0]=1 / 6 \\
& \mathrm{P}[X=1, Y=0]=1 / 12 \\
& \mathrm{P}[X=1, Y=1]=1 / 6 \\
& \mathrm{P}[X=2, Y=0]=1 / 12 \\
& \mathrm{P}[X=2, Y=1]=1 / 3 \\
& \mathrm{P}[X=2, Y=2]=1 / 6
\end{aligned}
\]

Calculate the variance of \(X\).

Options:
A
0.47
B
0.58
C
0.83
D
1.42
E
2.58

Correct answer: B

Solution: \(P(X=0)=1 / 6\)
\(P(X=1)=1 / 12+1 / 6=3 / 12\)
\(P(X=2)=1 / 12+1 / 3+1 / 6=7 / 12\).
\(E[X]=(0)(1 / 6)+(1)(3 / 12)+(2)(7 / 12)=17 / 12\)
\(E\left[X^{2}\right]=(0)^{2}(1 / 6)+(1)^{2}(3 / 12)+(2)^{2}(7 / 12)=31 / 12\)
\(\operatorname{Var}[X]=31 / 12-(17 / 12)^{2}=0.58\).
Question: The profit for a new product is given by \(Z=3 X-Y-5 . X\) and \(Y\) are independent random variables with \(\operatorname{Var}(X)=1\) and \(\operatorname{Var}(Y)=2\).

Calculate \(\operatorname{Var}(Z)\).

Options:
A
1
B
5
C
7
D
11
E
16

Correct answer: D

Solution: Due to the independence of \(X\) and \(Y\)
\(\operatorname{Var}(Z)=\operatorname{Var}(3 X-Y-5)=3^{2} \operatorname{Var}(X)+(-1)^{2} \operatorname{Var}(Y)=9(1)+2=11\).
Question: In a small metropolitan area, annual losses due to storm, fire, and theft are assumed to be mutually independent, exponentially distributed random variables with respective means \(1.0,1.5\), and 2.4 .

Calculate the probability that the maximum of these losses exceeds 3 .

Options:
A
0.002
B
0.050
C
0.159
D
0.287
E
0.414

Correct answer: E

Solution: Let \(S, F\), and \(T\) be the losses due to storm, fire, and theft respectively. Let \(Y=\max (S, F, T)\). Then,
\[
\begin{aligned}
& P[Y>3]=1-P[Y \leq 3]=1-P[\max (S, F, T) \leq 3]=1-P[S \leq 3] P[F \leq 3] P[T \leq 3] \\
& \quad=1-\left(1-\mathrm{e}^{-3 / 1}\right)\left(1-e^{-3 / 1.5}\right)\left(1-e^{-3 / 2.4}\right)=0.414 .
\end{aligned}
\]
Question: Let \(X\) denote the size of a surgical claim and let \(Y\) denote the size of the associated hospital claim. An actuary is using a model in which
\(E(X)=5, E\left(X^{2}\right)=27.4, E(Y)=7, E\left(Y^{2}\right)=51.4, \operatorname{Var}(X+Y)=8\).

Let \(C_{1}=X+Y\) denote the size of the combined claims before the application of a \(20 \%\) surcharge on the hospital portion of the claim, and let \(C_{2}\) denote the size of the combined claims after the application of that surcharge.

Calculate \(\operatorname{Cov}\left(C_{1}, C_{2}\right)\).

Options:
A
8.80
B
9.60
C
\(\quad 9.76\)
D
11.52
E
12.32

Correct answer: A

Solution: First obtain \(\operatorname{Var}(X)=27.4-25=2.4, \operatorname{Var}(Y)=51.4-49=2.4, \operatorname{Cov}(X, Y)=(8-2.4-2.4) / 2=\) 1.6. Then,
\[
\begin{aligned}
\operatorname{Cov}\left(C_{1}, C_{2}\right) & =\operatorname{Cov}(X+Y, X+1.2 Y)=\operatorname{Cov}(X, X)+1.2 \operatorname{Cov}(X, Y)+\operatorname{Cov}(Y, X)+1.2 \operatorname{Cov}(Y, Y) \\
& =\operatorname{Var}(X)+1.2 \operatorname{Var}(Y)+2.2 \operatorname{Cov}(X, Y) \\
& =2.4+1.2(2.4)+2.2(1.6)=8.8
\end{aligned}
\]
Question: Two life insurance policies, each with a death benefit of 10,000 and a one-time premium of 500, are sold to a married couple, one for each person. The policies will expire at the end of the tenth year. The probability that only the wife will survive at least ten years is 0.025 , the probability that only the husband will survive at least ten years is 0.01 , and the probability that both of them will survive at least ten years is 0.96 .

Calculate the expected excess of premiums over claims, given that the husband survives at least ten years.

Options:
A
350
B
385
C
397
D
870
E
897

Correct answer: E

Solution: Because the husband has survived, the only possible claim payment is to the wife. So we need the probability that the wife dies within ten years given that the husband survives. The numerator of the conditional probability is the unique event that only the husband survives, with probability 0.01 . The denominator is the sum of two events, both survive ( 0.96 ) and only the husband survives ( 0.01 ). The conditional probability is \(0.01 /(0.96+0.01)=1 / 97\). The expected claim payment is \(10,000 / 97=103\) and the expected excess is \(1,000-103=897\).
Question: A diagnostic test for the presence of a disease has two possible outcomes: 1 for disease present and 0 for disease not present. Let \(X\) denote the disease state ( 0 or 1 ) of a patient, and let \(Y\) denote the outcome of the diagnostic test. The joint probability function of \(X\) and \(Y\) is given by:
\[
\begin{aligned}
& \mathrm{P}[X=0, Y=0]=0.800 \\
& \mathrm{P}[X=1, Y=0]=0.050 \\
& \mathrm{P}[X=0, Y=1]=0.025 \\
& \mathrm{P}[X=1, Y=1]=0.125
\end{aligned}
\]

Calculate \(\operatorname{Var}(Y \mid X=1)\).

Options:
A
0.13
B
0.15
C
0.20
D
0.51
E
\(\quad 0.71\)

Correct answer: C

Solution: \[
\begin{aligned}
& P[Y=0 \mid X=1]=\frac{P(X=1, Y=0)}{P(X=1)}=\frac{P(X=1, Y=0)}{P(X=1, Y=0)+P(X=1, Y=1)}=\frac{0.05}{0.05+0.125}=0.286 \\
& P[Y=1 \mid X=1]=1-0.286=0.714
\end{aligned}
\]

The conditional variable is Bernoulli with \(p=0.714\). The variance is \((0.714)(0.286)=0.204\).
Question: An actuary determines that the annual number of tornadoes in counties \(P\) and \(Q\) are jointly distributed as follows:

\begin{tabular}{cc|cccc} 
& & \multicolumn{4}{|c}{ Annual number of } \\
& & 0 & 1 & 2 & 3 \\
\hline Annual number & 0 & 0.12 & 0.06 & 0.05 & 0.02 \\
of tornadoes & 1 & 0.13 & 0.15 & 0.12 & 0.03 \\
in county P & 2 & 0.05 & 0.15 & 0.10 & 0.02
\end{tabular}

Calculate the conditional variance of the annual number of tornadoes in county Q , given that there are no tornadoes in county P .

Options:
A
0.51
B
0.84
C
0.88
D
0.99
E
\(\quad 1.76\)

Correct answer: D

Solution: With no tornadoes in County P the probabilities of \(0,1,2\), and 3 tornadoes in County Q are \(12 / 25,6 / 25,5 / 25\), and \(2 / 25\) respectively.
The mean is \((0+6+10+6) / 25=22 / 25\).
The second moment is \((0+6+20+18) / 25=44 / 25\).
The variance is \(44 / 25-(22 / 25)^{2}=0.9856\).