BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: Under an insurance policy, a maximum of five claims may be filed per year by a policyholder. Let \(p(n)\) be the probability that a policyholder files \(n\) claims during a given year, where \(n=0,1,2,3,4,5\). An actuary makes the following observations:
i) \(\quad p(n) \geq p(n+1)\) for \(n=0,1,2,3,4\).
ii) The difference between \(p(n)\) and \(p(n+1)\) is the same for \(n=0,1,2,3,4\).
iii) Exactly \(40 \%\) of policyholders file fewer than two claims during a given year.

Calculate the probability that a random policyholder will file more than three claims during a given year.

Options:
A
0.14
B
0.16
C
0.27
D
0.29
E
0.33

Correct answer: C

Solution: Due to the equal spacing of probabilities, \(p_{n}=p_{0}-n c\) for \(c=1,2,3,4,5\). Also, \(0.4=p_{0}+p_{1}=p_{0}+p_{0}-c=2 p_{0}-c\). Because the probabilities must sum to 1 , \(1=p_{0}+p_{0}-c+p_{0}-2 c+p_{0}-3 c+p_{0}-4 c+p_{0}-5 c=6 p_{0}-15 c\). This provides two equations in two unknowns. Multiplying the first equation by 15 gives \(6=30 p_{0}-15 c\). Subtracting the second equation gives \(5=24 p_{0} \Rightarrow p_{0}=5 / 24\). Inserting this in the first equation gives \(c=1 / 60\). The requested probability is \(p_{4}+p_{5}=5 / 24-4 / 60+5 / 24-5 / 60=32 / 120=0.267\).
Question: The amounts of automobile losses reported to an insurance company are mutually independent, and each loss is uniformly distributed between 0 and 20,000 . The company covers each such loss subject to a deductible of 5,000.

Calculate the probability that the total payout on 200 reported losses is between \(1,000,000\) and 1,200,000.

Options:
A
0.0803
B
0.1051
C
0.1799
D
0.8201
E
0.8575

Correct answer: D

Solution: Because the number of payouts (including payouts of zero when the loss is below the deductible) is large, apply the central limit theorem and assume the total payout \(S\) is normal. For one loss, the mean, second moment, and variance of the payout are
\[
\begin{aligned}
& \int_{0}^{5,000} 0 \frac{1}{20,000} d x+\int_{5,000}^{20,000}(x-5,000) \frac{1}{20,000}=0+\left.\frac{(x-5,000)^{2}}{40,000}\right|_{5,000} ^{20,000}=5,625 \\
& \int_{0}^{5,000} 0^{2} \frac{1}{20,000} d x+\int_{5,000}^{20,000}(x-5,000)^{2} \frac{1}{20,000}=0+\left.\frac{(x-5,000)^{3}}{60,000}\right|_{5,000} ^{20,000}=56,250,000 \\
& 56,250,000-5,625^{2}=24,609,375
\end{aligned}
\]

For 200 losses the mean, variance, and standard deviation are \(1,125,000,4,921,875,000\), and \(70,156.1,000,000\) is \(125,000 / 70,156=1.7817\) standard deviations below the mean and \(1,200,000\) is \(75,000 / 70,156=1.0690\) standard deviations above the mean. From the standard normal tables, the probability of being between these values is \(0.8575-(1-0.9626)=0.8201\).
Question: An insurance agent offers his clients auto insurance, homeowners insurance and renters insurance. The purchase of homeowners insurance and the purchase of renters insurance are mutually exclusive. The profile of the agent's clients is as follows:
i) \(17 \%\) of the clients have none of these three products.
ii) \(64 \%\) of the clients have auto insurance.
iii) Twice as many of the clients have homeowners insurance as have renters insurance.
iv) \(35 \%\) of the clients have two of these three products.
v) \(11 \%\) of the clients have homeowners insurance, but not auto insurance.

Calculate the percentage of the agent's clients that have both auto and renters insurance.

Options:
A
\(\quad 7 \%\)
B
\(\quad 10 \%\)
C
\(16 \%\)
D
\(\quad 25 \%\)
E
\(28 \%\)

Correct answer: B

Solution: Let H be the percentage of clients with homeowners insurance and R be the percentage of clients with renters insurance.

Because \(36 \%\) of clients do not have auto insurance and none have both homeowners and renters insurance, we calculate that \(8 \%(36 \%-17 \%-11 \%)\) must have renters insurance, but not auto insurance.
\((\mathrm{H}-11) \%\) have both homeowners and auto insurance, \((\mathrm{R}-8) \%\) have both renters and auto insurance, and none have both homeowners and renters insurance, so \((\mathrm{H}+\mathrm{R}-19) \%\) must equal \(35 \%\). Because \(\mathrm{H}=2 \mathrm{R}\), R must be \(18 \%\), which implies that \(10 \%\) have both renters and auto insurance.
Question: The cumulative distribution function for health care costs experienced by a policyholder is modeled by the function
\[
F(x)= \begin{cases}1-e^{-\frac{x}{100}}, & x>0 \\ 0, & \text { otherwise }\end{cases}
\]

The policy has a deductible of 20. An insurer reimburses the policyholder for \(100 \%\) of health care costs between 20 and 120. Health care costs above 120 are reimbursed at \(50 \%\).

Let \(G\) be the cumulative distribution function of reimbursements given that the reimbursement is positive.

Calculate \(G(115)\).

Options:
A
0.683
B
0.727
C
\(\quad 0.741\)
D
0.757
E
0.777

Correct answer: B

Solution: Let Y be the reimbursement. Then, \(G(115)=P[Y<115 \mid X>20]\). For \(Y\) to be 115 , the costs must be above 120 (up to 120 accounts for a reimbursement of 100 ). The extra 15 requires 30 in additional costs. Therefore, we need
\[
\begin{aligned}
P[X & \leq 150 \mid X>20]=\frac{P[X \leq 150]-P[X \leq 20]}{P[X>20]}=\frac{1-e^{-150 / 100}-1+e^{-20 / 100}}{1-1+e^{-20 / 100}} \\
& =\frac{-e^{-1.5}+{ }^{-0.2}}{e^{-0,2}}=1-e^{-1.3}=0.727
\end{aligned}
\]
Question: A man purchases a life insurance policy on his \(40^{\text {th }}\) birthday. The policy will pay 5000 if he dies before his \(50^{\text {th }}\) birthday and will pay 0 otherwise. The length of lifetime, in years from birth, of a male born the same year as the insured has the cumulative distribution function
\[
F(t)= \begin{cases}0, & t \leq 0 \\ 1-\exp \left(\frac{1-1.1^{t}}{1000}\right), & t>0\end{cases}
\]

Calculate the expected payment under this policy.

Options:
A
333
B
348
C
421
D
549
E
574

Correct answer: B

Solution: \(P(40\) year old man dies before age 50\()=\mathrm{P}(T<50 \mid T>40)\)
\(=\frac{\operatorname{Pr}(40<T<50)}{\operatorname{Pr}(T>40)}=\frac{F(50)-F(40)}{1-F(40)}\)
\(=\frac{1-\exp \left(\frac{1-1.1^{50}}{1000}\right)-1+\exp \left(\frac{1-1.1^{40}}{1000}\right)}{1-1+\exp \left(\frac{1-1.1^{40}}{1000}\right)}=\frac{\exp \left(\frac{1-1.1^{40}}{1000}\right)-\exp \left(\frac{1-1.1^{50}}{1000}\right)}{\exp \left(\frac{1-1.1^{40}}{1000}\right)}\)
\(=\frac{0.9567-0.8901}{0.9567}=0.0696\)
Expected Benefit \(=5000(0.0696)=348\).
Question: A fair die is rolled repeatedly. Let \(X\) be the number of rolls needed to obtain a 5 and \(Y\) the number of rolls needed to obtain a 6 .

Calculate \(\mathrm{E}(X \mid Y=2)\).

Options:
A
5.0
B
5.2
C
6.0
D
6.6
E
6.8

Correct answer: D

Solution: \(X\) follows a geometric distribution with \(p=1 / 6\) and \(Y=2\) implies the first roll is not a 6 and the second roll is a 6. This means a 5 is obtained for the first time on the first roll (probability \(=0.2\) ) or a 5 is obtained for the first time on the third or later roll (probability \(=0.8\) ).
\(E[X \mid X \geq 3]=\frac{1}{p}+2=6+2=8 .\), The expected value is \(0.2(1)+0.8(8)=6.6\).
Question: Each time a hurricane arrives, a new home has a 0.4 probability of experiencing damage. The occurrences of damage in different hurricanes are mutually independent.

Calculate the mode of the number of hurricanes it takes for the home to experience damage from two hurricanes.

Options:
A
2
B
3
C
4
D
5
E
6

Correct answer: B

Solution: Let \(X\) equal the number of hurricanes it takes for two losses to occur. Then \(X\) is negative binomial with "success" probability \(p=0.4\) and \(r=2\) "successes" needed.
\(P[X=n]=\binom{n-1}{r-1} p^{r}(1-p)^{n-r}=\binom{n-1}{2-1}(0.4)^{2}(1-0.4)^{n-2}=(n-1)(0.4)^{2}(0.6)^{n-2}\), for \(n \geq 2\).
We need to maximize \(P[X=n]\). Note that the ratio
\(\frac{P[X=n+1]}{P[X=n]}=\frac{n(0.4)^{2}(0.6)^{n-1}}{(n-1)(0.4)^{2}(0.6)^{n-2}}=\frac{n}{n-1}(0.6)\).
This ratio of "consecutive" probabilities is greater than 1 when \(n=2\) and less than 1 when \(n \geq 3\). Thus, \(P[X=n]\) is maximized at \(n=3\); the mode is 3 . Alternatively, the first few probabilities could be calculated.
Question: An auto insurance company is implementing a new bonus system. In each month, if a policyholder does not have an accident, he or she will receive a cash-back bonus of 5 from the insurer.

Among the 1,000 policyholders of the auto insurance company, 400 are classified as lowrisk drivers and 600 are classified as high-risk drivers.

In each month, the probability of zero accidents for high-risk drivers is 0.80 and the probability of zero accidents for low-risk drivers is 0.90 .

Calculate the expected bonus payment from the insurer to the 1000 policyholders in one year.

Options:
A
\(\quad 48,000\)
B
\(\quad 50,400\)
C
51,000
D
54,000
E
\(\quad 60,000\)

Correct answer: B

Solution: The expected bonus for a high-risk driver is \(0.8(12)(5)=48\).
The expected bonus for a low-risk driver is \(0.9(12)(5)=54\).
The expected bonus payment from the insurer is \(600(48)+400(54)=50,400\).
Question: Let \(X\) represent the number of customers arriving during the morning hours and let \(Y\) represent the number of customers arriving during the afternoon hours at a diner. You are given:
i) \(\quad X\) and \(Y\) are Poisson distributed.
ii) The first moment of \(X\) is less than the first moment of \(Y\) by 8 .
iii) The second moment of \(X\) is \(60 \%\) of the second moment of \(Y\).

Calculate the variance of \(Y\).

Options:
A
4
B
12
C
16
D
27
E
35

Correct answer: E

Solution: Let \(a\) be the mean and variance of \(X\) and \(b\) be the mean and variance of \(Y\). The two facts are \(a=b\) -8 and \(a+a^{2}=0.6\left(b+b^{2}\right)\). Substituting the first equation into the second gives
\(b-8+(b-8)^{2}=0.6 b+0.6 b^{2}\)
\(b-8+b^{2}-16 b+64=0.6 b+0.6 b^{2}\)
\(0.4 b^{2}-15.6 b+56=0\)
\(b=\frac{15.6 \pm \sqrt{15.6^{2}-4(0.4)(56)}}{2(0.4)}=\frac{15.6 \pm 12.4}{0.8}=4\) or 35 .
At \(b=4, a\) is negative, so the answer is 35 .
Question: The probability of \(x\) losses occurring in year 1 is \((0.5)^{x+1}\) for \(x=0,1,2, \ldots\).

The probability of \(y\) losses in year 2 given \(x\) losses in year 1 is given by the table:

\begin{tabular}{|c|c|c|c|c|c|}
\hline \multirow{3}{*}{\begin{tabular}{c} 
Number of \\
losses in \\
year 1 \((x)\)
\end{tabular}} & \multicolumn{5}{|c|}{\begin{tabular}{c} 
Number of losses in year 2 \((y)\) \\
given \(x\) losses in year 1
\end{tabular}} \\
\cline { 2 - 6 } & 0 & 1 & 2 & 3 & \(4+\) \\
\hline 0 & 0.60 & 0.25 & 0.05 & 0.05 & 0.05 \\
\hline 1 & 0.45 & 0.30 & 0.10 & 0.10 & 0.05 \\
\hline 2 & 0.25 & 0.30 & 0.20 & 0.20 & 0.05 \\
\hline 3 & 0.15 & 0.20 & 0.20 & 0.30 & 0.15 \\
\hline \(4+\) & 0.05 & 0.15 & 0.25 & 0.35 & 0.20 \\
\hline
\end{tabular}

Calculate the probability of exactly 2 losses in 2 years.

Options:
A
0.025
B
0.031
C
0.075
D
0.100
E
0.131

Correct answer: E

Solution: \(P(x=1, y=1)=P(y=1 \mid x=1) P(x=1)=0.3(0.5)^{2}=0.075\)
\(P(x=2, y=0)=P(y=0 \mid x=2) P(x=2)=0.25(0.5)^{3}=0.03125\)
\(P(x=0, y=2)=P(y=2 \mid x=0) P(x=0)=0.05(0.5)^{1}=0.025\)
The total is 0.13125 .
Question: An airport purchases an insurance policy to offset costs associated with excessive amounts of snowfall. For every full ten inches of snow in excess of 40 inches during the winter season, the insurer pays the airport 300 up to a policy maximum of 700.

The following table shows the probability function for the random variable \(X\) of annual (winter season) snowfall, in inches, at the airport.

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline Inches & {\([0,20)\)} & {\([20,30)\)} & {\([30,40)\)} & {\([40,50)\)} & {\([50,60)\)} & {\([60,70)\)} & {\([70,80)\)} & {\([80,90)\)} & {\([90\), inf \()\)} \\
\hline Probability & 0.06 & 0.18 & 0.26 & 0.22 & 0.14 & 0.06 & 0.04 & 0.04 & 0.00 \\
\hline
\end{tabular}

Calculate the standard deviation of the amount paid under the policy.

Options:
A
134
B
235
C
271
D
313
E
352

Correct answer: B

Solution: The payments are 0 with probability 0.72 (snowfall up to 50 inches), 300 with probability 0.14 , 600 with probability 0.06 , and 700 with probability 0.08 . The mean is \(0.72(0)+0.14(300)+ 0.06(600)+0.08(700)=134\) and the second moment is \(0.72\left(0^{\wedge} 2\right)+0.14\left(300^{\wedge} 2\right)+0.06\left(600^{\wedge} 2\right)+ 0.08\left(700^{\wedge} 2\right)=73,400\). The variance is \(73,400-134^{\wedge} 2=55,444\). The standard deviation is the square root, 235.
Question: Damages to a car in a crash are modeled by a random variable with density function
\[
f(x)= \begin{cases}c\left(x^{2}-60 x+800\right), & 0<x<20 \\ 0, & \text { otherwise }\end{cases}
\]
where \(c\) is a constant.
A particular car is insured with a deductible of 2 . This car was involved in a crash with resulting damages in excess of the deductible.

Calculate the probability that the damages exceeded 10.

Options:
A
0.12
B
0.16
C
0.20
D
0.26
E
0.78

Correct answer: D

Solution: \(1=\int_{0}^{20} c\left(x^{2}-60 x+800\right) d x=\left.c\left(x^{3} / 3-30 x^{2}+800 x\right)\right|_{0} ^{20}=c 20,000 / 3 \Rightarrow c=3 / 20,000\)
\(P(X>d)=\int_{d}^{20} c\left(x^{2}-60 x+800\right) d x=\left.c\left(x^{3} / 3-30 x^{2}+800 x\right)\right|_{d} ^{20}=1-\frac{3}{20,000}\left(d^{3} / 3-30 d^{2}+800 d\right)\)
\(P(X>10 \mid X>2)=\frac{P(X>10)}{P(X>2)}=\frac{0.2}{0.7776}=0.2572\)
Question: In a group of 25 factory workers, 20 are low-risk and five are high-risk.

Two of the 25 factory workers are randomly selected without replacement.
Calculate the probability that exactly one of the two selected factory workers is low-risk.

Options:
A
0.160
B
0.167
C
0.320
D
0.333
E
0.633

Correct answer: D

Solution: This is a hypergeometric probability, \(\frac{\binom{20}{1}\binom{5}{1}}{\binom{25}{2}}=\frac{20(5)}{25(24) / 2}=\frac{100}{300}=0.333\),
Alternatively, the probability of the first worker being high risk and the second low risk is \((5 / 25)(20 / 24)=100 / 600\) and of the first being low risk and the second high risk is \((20 / 25)(5 / 24) =100 / 600\) for a total probability of \(200 / 600=0.333\).
Question: The proportion \(X\) of yearly dental claims that exceed 200 is a random variable with probability density function
\(f(x)= \begin{cases}60 x^{3}(1-x)^{2}, & 0<x<1 \\ 0, & \text { otherwise } .\end{cases}\)
Calculate \(\operatorname{Var}[X /(1-X)]\)

Options:
A
\(149 / 900\)
B
\(10 / 7\)
C
6
D
8
E
10

Correct answer: C

Solution: \[
\begin{aligned}
& E\left(\frac{X}{1-X}\right)=60 \int_{0}^{1} \frac{x}{1-x} x^{3}(1-x)^{2} d x=60 \int_{0}^{1} x^{4}(1-x) d x=\left.60\left(x^{5} / 5-x^{6} / 6\right)\right|_{0} ^{1}=60(1 / 5-1 / 6)=2 \\
& E\left[\left(\frac{X}{1-X}\right)^{2}\right]=60 \int_{0}^{1} \frac{x^{2}}{(1-x)^{2}} x^{3}(1-x)^{2} d x=60 \int_{0}^{1} x^{5} d x=\left.60\left(x^{6} / 6\right)\right|_{0} ^{1}=60(1 / 6)=10 \\
& \operatorname{Var}\left(\frac{X}{1-X}\right)=10-2^{2}=6
\end{aligned}
\]
Question: A company has five employees on its health insurance plan. Each year, each employee independently has an \(80 \%\) probability of no hospital admissions. If an employee requires one or more hospital admissions, the number of admissions is modeled by a geometric distribution with a mean of 1.50 . The numbers of hospital admissions of different employees are mutually independent.

Each hospital admission costs 20,000.
Calculate the probability that the company's total hospital costs in a year are less than 50,000.

Options:
A
0.41
B
0.46
C
0.58
D
0.69
E
0.78

Correct answer: E

Solution: A geometric probability distribution with mean 1.5 will have \(p=2 / 3\). So \(\operatorname{Pr}(1\) visit \()=2 / 3, \mathrm{P}(\) two visits) \(=2 / 9\), etc. There are four disjoint scenarios in which total admissions will be two or less.

Scenario 1: No employees have hospital admissions. Probability \(=0.8^{5}=0.32768\).
Scenario 2: One employee has one admission and the other employees have none. Probability = \(\binom{5}{1}(0.2)(0.8)^{4}(2 / 3)=0.27307\).

Scenario 3: One employee has two admissions and the other employees have none. Probability = \(\binom{5}{1}(0.2)(0.8)^{4}(2 / 9)=0.09102\).

Scenario 4: Two employees each have one admission and the other three employees have none.
Probability \(=\binom{5}{2}(0.2)^{2}(0.8)^{3}(2 / 3)(2 / 3)=0.09102\).
The total probability is 0.78279 .
Question: In a certain group of cancer patients, each patient's cancer is classified in exactly one of the following five stages: stage 0 , stage 1 , stage 2 , stage 3 , or stage 4 .
i) \(75 \%\) of the patients in the group have stage 2 or lower.
ii) \(80 \%\) of the patients in the group have stage 1 or higher.
iii) \(80 \%\) of the patients in the group have stage \(0,1,3\), or 4 .

One patient from the group is randomly selected.
Calculate the probability that the selected patient's cancer is stage 1 .

Options:
A
0.20
B
0.25
C
0.35
D
0.48
E
0.65

Correct answer: C

Solution: Let \(p_{i}\) represent the probability that the patient's cancer is in stage \(i\), for \(i=0,1,2,3\), or 4 . The probabilities must sum to 1 . That fact and the three facts given the question produce the following equations.
\(p_{0}+p_{1}+p_{2}+p_{3}+p_{4}=1\)
\(p_{0}+p_{1}+p_{2}=0.75\)
\(p_{1}+p_{2}+p_{3}+p_{4}=0.8\)
\(p_{0}+p_{1}+p_{3}+p_{4}=0.8\)
Therefore, we have
\(p_{0}=\left(p_{0}+p_{1}+p_{2}+p_{3}+p_{4}\right)-\left(p_{1}+p_{2}+p_{3}+p_{4}\right)=1-0.8=0.2\)
\(p_{2}=\left(p_{0}+p_{1}+p_{2}+p_{3}+p_{4}\right)-\left(p_{0}+p_{1}+p_{3}+p_{4}\right)=1-0.8=0.2\).
\(p_{1}=\left(p_{0}+p_{1}+p_{2}\right)-p_{0}-p_{2}=0.75-0.2-0.2=0.35\).
Question: A theme park conducts a study of families that visit the park during a year. The fraction of such families of size \(m\) is \(\frac{8-m}{28}, m=1,2,3,4,5,6\), and 7 .

For a family of size \(m\) that visits the park, the number of members of the family that ride the roller coaster follows a discrete uniform distribution on the set \(\{1, \ldots, m\}\).

Calculate the probability that a family visiting the park has exactly six members, given that exactly five members of the family ride the roller coaster.

Options:
A
0.17
B
0.21
C
0.24
D
0.28
E
0.31

Correct answer: E

Solution: Let \(M\) be the size of a family that visits the park and let \(N\) be the number of members of that family that ride the roller coaster. We want \(P(M=6 \mid N=5)\). By Bayes theorem
\(P(M=6 \mid N=5)\)
\(=\frac{P(N=5 \mid M=6) P(M=6)}{\sum_{m=1}^{7} P(N=5 \mid M=m) P(M=m)}\)
\(=\frac{\frac{1}{6} \frac{2}{28}}{0+0+0+0+\frac{1}{5} \frac{3}{28}+\frac{1}{6} \frac{2}{28}+\frac{1}{7} \frac{1}{28}}=\frac{\frac{1}{3}}{\frac{3}{5}+\frac{1}{3}+\frac{1}{7}}=\frac{35}{63+35+15}=\frac{35}{113} \approx 0.3097\).
Question: An insurance company issues policies covering damage to automobiles. The amount of damage is modeled by a uniform distribution on \([0, b]\).

The policy payout is subject to a deductible of \(b / 10\).
A policyholder experiences automobile damage.
Calculate the ratio of the standard deviation of the policy payout to the standard deviation of the amount of the damage.

Options:
A
0.8100
B
0.9000
C
0.9477
D
0.9487
E
\(\quad 0.9735\)

Correct answer: E

Solution: Without the deductible, the standard deviation is, from the uniform distribution, \(b / \sqrt{12}=0.28868 b\). Let \(Y\) be the random variable representing the payout with the deductible.
\(E(Y)=\int_{0.1 b}^{b}(y-0.1 b) \frac{1}{b} d y=\frac{y^{2}}{2 b}-\left.0.1 y\right|_{0.1 b} ^{b}=0.5 b-0.1 b-0.005 b+0.01 b=0.405 b\)
\(E\left(Y^{2}\right)=\int_{0.1 b}^{b}(y-0.1 b)^{2} \frac{1}{b} d y=\frac{y^{3}}{3 b}-0.1 y^{2}+\left.0.01 b y\right|_{0.1 b} ^{b}\)
\(=b^{2} / 3-0.1 b^{2}+0.01 b^{2}-0.001 b^{2} / 3+0.001 b^{2}-0.001 b^{2}=0.243 b^{2}\)
\(\operatorname{Var}(Y)=0.243 b^{2}-(0.405 b)^{2}=0.078975 b^{2}\)
\(S D(Y)=0.28102 b\).
The ratio is \(0.28102 / 0.28868=0.97347\).
Question: On a block of ten houses, \(k\) are not insured. A tornado randomly damages three houses on the block.

The probability that none of the damaged houses are insured is 1/120.
Calculate the probability that at most one of the damaged houses is insured.

Options:
A
\(1 / 5\)
B
\(\quad 7 / 40\)
C
\(\quad 11 / 60\)
D
\(\quad 49 / 60\)
E
\(119 / 120\)

Correct answer: C

Solution: The probability that none of the damaged houses are insured is
\(\frac{1}{120}=\frac{\binom{10-k}{0}\binom{k}{3}}{\binom{10}{3}}=\frac{k(k-1)(k-2)}{720}\).
\(k(k-1)(k-2)=6\)
This cubic equation could be solved by expanding, subtracting 6, and refactoring. However, because \(k\) must be an integer, the three factors must be integers and thus must be \(3(2)(1)\) for \(k=\) 3.

The probability that at most one of the damaged houses is insured equals
\(\frac{1}{120}+\frac{\binom{10-3}{1}\binom{3}{2}}{\binom{10}{3}}=\frac{1}{120}+\frac{7(3)}{120}=\frac{22}{120}=\frac{11}{60}\).
Question: The distribution of the size of claims paid under an insurance policy has probability density function
\[
f(x)= \begin{cases}c x^{a}, & 0<x<5 \\ 0, & \text { otherwise }\end{cases}
\]

Where \(a>0\) and \(c>0\).
For a randomly selected claim, the probability that the size of the claim is less than 3.75 is 0.4871 .

Calculate the probability that the size of a randomly selected claim is greater than 4 .

Options:
A
0.404
B
0.428
C
0.500
D
0.572
E
0.596

Correct answer: B

Solution: Because the density function must integrate to \(1,1=\int_{0}^{5} c x^{a} d x=c \frac{5^{a+1}}{a+1} \Rightarrow c=\frac{a+1}{5^{a+1}}\).
From the given probability,
\(0.4871=\int_{0}^{3.75} c x^{a} d x=c \frac{3.75^{a+1}}{a+1}=\frac{a+1}{5^{a+1}} \frac{3.75^{a+1}}{a+1}=\left(\frac{3.75}{5}\right)^{a+1}\)
\(\ln (0.4871)=-0.71929=(a+1) \ln (3.75 / 5)=-0.28768(a+1)\)
\(a=(-0.71929) /(-0.28768)-1=1.5\).
The probability of a claim exceeding 4 is,
\(\int_{4}^{5} c x^{a} d x=c \frac{5^{a+1}-4^{a+1}}{a+1}=\frac{a+1}{5^{a+1}} \frac{5^{a+1}-4^{a+1}}{a+1}=1-\left(\frac{4}{5}\right)^{1.5+1}=0.42757\).
Question: The working lifetime, in years, of a particular model of bread maker is normally distributed with mean 10 and variance 4 .

Calculate the 12th percentile of the working lifetime, in years.

Options:
A
5.30
B
7.65
C
8.41
D
12.35
E
14.70

Correct answer: B

Solution: Let \(X\) be normal with mean 10 and variance 4 . Let \(Z\) have the standard normal distribution. Let \(p=12\) th percentile. Then
\(0.12=P(X \leq p)=P\left(\frac{X-10}{2} \leq \frac{p-10}{2}\right)=P\left(Z \leq \frac{p-10}{2}\right)\).
From the tables, \(P(Z \leq-1.175)=0.12\). Therefore,
\(\frac{p-10}{2}=-1.175 ; p-10=-2.35 ; p=7.65\).
Question: An insurance company will cover losses incurred from tornadoes in a single calendar year. However, the insurer will only cover losses for a maximum of three separate tornadoes during this timeframe. Let \(X\) be the number of tornadoes that result in at least 50 million in losses, and let \(Y\) be the total number of tornadoes. The joint probability function for \(X\) and \(Y\) is
\[
p(x, y)= \begin{cases}c(x+2 y), & \text { for } x=0,1,2,3, y=0,1,2,3, x \leq y \\ 0, & \text { otherwise }\end{cases}
\]
where \(c\) is a constant.
Calculate the expected number of tornadoes that result in fewer than 50 million in losses.

Options:
A
0.19
B
0.28
C
0.76
D
1.00
E
\(\quad 1.10\)

Correct answer: E

Solution: The possible events are \((0,0),(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,2),(2,3)\), and \((3,3)\). The probabilities (without \(c\) ) sum to \(0+2+4+6+3+5+7+6+8+9=50\). Therefor \(c=1 / 50\). The number of tornadoes with fewer than 50 million in losses is \(Y-X\). The expected value is \((1 / 50)[0(0)+1(2)+2(4)+3(6)+0(3)+1(5)+2(7)+0(6)+1(8)+0(9)]=55 / 50=1.1\).
Question: The random variables \(X\) and \(Y\) have joint probability function \(p(x, y)\) for \(x=0,1\) and \(y=0,1,2\).

Suppose \(3 p(1,1)=p(1,2)\), and \(p(1,1)\) maximizes the variance of \(X Y\).
Calculate the probability that \(X\) or \(Y\) is 0 .

Options:
A
\(11 / 25\)
B
\(\quad 23 / 50\)
C
\(\quad 23 / 49\)
D
\(\quad 26 / 49\)
E
\(\quad 14 / 25\)

Correct answer: C

Solution: Let \(Z=X Y\). Let \(a, b\), and \(c\) be the probabilities that \(Z\) takes on the values 0,1 , and 2 , respectively. We have \(b=p(1,1)\) and \(c=p(1,2)\) and thus \(3 b=c\). And because the probabilities sum to 1 , \(a=1-b-c=1-4 b\). Then, \(E(Z)=b+2 c=7 b, E(Z * Z)=b+4 c=13 b\). Then, \(\operatorname{Var}(Z)=13 b-49 b^{2}\)
\((d / d b) \operatorname{Var}(Z)=13-98 b=0 \Rightarrow b=13 / 98\).
The probability that either \(X\) or \(Y\) is zero is the same as the probability that \(Z\) is 0 which is \(a=1-4 b=46 / 98=23 / 49\).