BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: An insurance company has an equal number of claims in each of three territories. In each territory, only three claim amounts are possible: 100, 500, and 1000. Based on the company's data, the probabilities of each claim amount are:

\begin{tabular}{|l|c|c|c|}
\cline { 2 - 4 } \multicolumn{1}{c|}{} & \multicolumn{3}{c|}{ Claim Amount } \\
\cline { 2 - 4 } \multicolumn{1}{c|}{} & 100 & 500 & 1000 \\
\hline Territory 1 & 0.90 & 0.08 & 0.02 \\
\hline Territory 2 & 0.80 & 0.11 & 0.09 \\
\hline Territory 3 & 0.70 & 0.20 & 0.10 \\
\hline
\end{tabular}

Calculate the standard deviation of a randomly selected claim amount.

Options:
A
254
B
291
C
332
D
368
E
396

Correct answer: A

Solution: Because the territories are evenly distributed, the probabilities can be averaged. Thus the probability of a 100 claim is 0.80 , of a 500 claim is 0.13 , and of a 1000 claim is 0.07 . The mean is \(0.80(100)+0.13(500)+0.07(1000)=215\). The second moment is \(0.80(10,000)+ 0.13(250,000)+0.07(1,000,000)=110,500\). The variance is \(110,500-(215)(215)=64,275\). The standard deviation is 253.53 .
Question: An insurance policy covers losses incurred by a policyholder, subject to a deductible of 10,000. Incurred losses follow a normal distribution with mean 12,000 and standard deviation \(c\). The probability that a loss is less than \(k\) is 0.9582 , where \(k\) is a constant. Given that the loss exceeds the deductible, there is a probability of 0.9500 that it is less than \(k\).

Calculate \(c\).

Options:
A
2045
B
2267
C
2393
D
2505
E
2840

Correct answer: A

Solution: We have
\(0.95=P(X<k \mid X>10,000)=\frac{P(X<k)-P(X \leq 10,000)}{1-P(\mathrm{X} \leq 10,000)}\)
\(0.95[1-P(X \leq 10,000)]=0.9582-P(X \leq 10,000)\)
\(P(X \leq 10,000)=\frac{0.9582-0.95}{1-0.95}=0.164\)
\(0.164=\Phi\left(\frac{10,000-12,000}{c}\right)\).
The \(z\)-value that corresponds to 0.164 is between -0.98 and -0.97 . Interpolating leads to \(z=-0.978\). Then,
\(0.164=\Phi\left(\frac{10,000-12,000}{c}\right) \Rightarrow-0.978=\frac{-2,000}{c} \Rightarrow c=2045\).
Question: An insurance agent's files reveal the following facts about his policyholders:
i) 243 own auto insurance.
ii) 207 own homeowner insurance.
iii) 55 own life insurance and homeowner insurance.
iv) 96 own auto insurance and homeowner insurance.
v) 32 own life insurance, auto insurance and homeowner insurance.
vi) 76 more clients own only auto insurance than only life insurance.
vii) 270 own only one of these three insurance products.

Calculate the total number of the agent's policyholders who own at least one of these three insurance products.

Options:
A
389
B
407
C
423
D
448
E
483

Correct answer: B

Solution: 32 own L/A/H
55 own L/H so \(55-32=23\) own L/H/notA
96 own A/H so \(96-32=64\) own A/H/notL
207 own H so \(207-32-23-64=88\) own H only
L only \(=X\), A only \(=X+76\)
\(88+X+(X+76)=270\) so \(X=53\) so L only \(=53\), A only \(=129\)
\(129+64+32+\mathrm{L} / \mathrm{A} / \operatorname{notH}=243\) so L/A/notH = 18
Total clients \(=53+129+88+18+64+23+32=407\)
Alternatively, a Venn diagram could be used to guide the calculations.
Question: A health insurance policy covers visits to a doctor's office. Each visit costs 100 . The annual deductible on the policy is 350 . For a policy, the number of visits per year has the following probability distribution:

\begin{tabular}{|l|c|c|c|c|c|c|c|}
\hline Number of Visits & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline Probability & 0.60 & 0.15 & 0.10 & 0.08 & 0.04 & 0.02 & 0.01 \\
\hline
\end{tabular}

A policy is selected at random from those where costs exceed the deductible.
Calculate the probability that this policyholder had exactly five office visits.

Options:
A
0.050
B
0.133
C
0.286
D
0.333
E
0.429

Correct answer: C

Solution: The deductible is exceeded for 4,5 or 6 office visits. Therefore, the requested probability is \(0.02 /(0.04+0.02+0.01)=0.286\).
Question: A machine has two parts labelled A and B. The probability that part A works for one year is 0.8 and the probability that part B works for one year is 0.6 . The probability that at least one part works for one year is 0.9 .

Calculate the probability that part B works for one year, given that part A works for one year.

Options:
A
\(\quad 1 / 2\)
B
\(\quad 3 / 5\)
C
\(\quad 5 / 8\)
D
\(\quad 3 / 4\)
E
\(\quad 5 / 6\)

Correct answer: C

Solution: Let \(A\) be the event that part A is working after one year and \(B\) be the event that part B is working after one year. Then,
\(P(B \mid A)=\frac{P(A \text { and } B)}{P(A)}=\frac{P(A)+P(B)-P(A \text { or } B)}{P(A)}=\frac{0.8+0.6-0.9}{0.8}=5 / 8\).
Question: An insurance company studies back injury claims from a manufacturing company. The insurance company finds that \(40 \%\) of workers do no lifting on the job, \(50 \%\) do moderate lifting and 10\% do heavy lifting.

During a given year, the probability of filing a claim is 0.05 for a worker who does no lifting, 0.08 for a worker who does moderate lifting and 0.20 for a worker who does heavy lifting.

A worker is chosen randomly from among those who have filed a back injury claim.
Calculate the probability that the worker's job involves moderate or heavy lifting.

Options:
A
0.75
B
0.81
C
0.85
D
0.86
E
0.89

Correct answer: A

Solution: Define the events as follows:
\(C=\) files a claim
\(N=\) no lifting
\(M=\) moderate lifting
\(H=\) heavy lifting
Then, using Bayes' Theorem,
\[
\begin{aligned}
& P(M \text { or } H \mid C)=1-P(N \mid C)=1-\frac{P(C \mid N) P(N)}{P(C \mid N) P(N)+P(C \mid \mathrm{M}) P(\mathrm{M})+P(C \mid \mathrm{H}) P(\mathrm{H})} \\
& =1-\frac{0.05(0.4)}{0.05(0.4)+0.08(0.5)+0.2(0.1)}=1-0.25=0.75
\end{aligned}
\]
Question: The number of traffic accidents occurring on any given day in Coralville is Poisson distributed with mean 5 . The probability that any such accident involves an uninsured driver is 0.25 , independent of all other such accidents.

Calculate the probability that on a given day in Coralville there are no traffic accidents that involve an uninsured driver.

Options:
A
0.007
B
0.010
C
0.124
D
0.237
E
0.287

Correct answer: E

Solution: From the Law of Total Probability, the required probability is \(\sum_{k=0}^{\infty} P(0\) accidents with an uninsured driver \(\mid k\) accidents \() P(k\) accidents \()\)
\(=\sum_{k=0}^{\infty}(0.75)^{k} \frac{e^{-5} 5^{k}}{k!}=\frac{e^{-5}}{e^{-3.75}} \sum_{k=0}^{\infty} \frac{e^{-3.75}(3.75)^{k}}{k!}=e^{-1.25}=0.287\).
Question: Losses due to burglary are exponentially distributed with mean 100.

The probability that a loss is between 40 and 50 equals the probability that a loss is between 60 and \(r\), with \(r>60\).

Calculate \(r\).

Options:
A
68.26
B
70.00
C
70.51
D
72.36
E
75.00

Correct answer: D

Solution: The cumulative distribution function for the exponential distribution is \(F(x)=1-e^{-\lambda x}=1-e^{-x / \mu}=1-e^{-x / 100}, \mathrm{x}>0\).

From the given probability data,
\[
\begin{aligned}
& F(50)-F(40)=F(r)-F(60) \\
& 1-e^{-50 / 100}-\left(1-e^{-40 / 100}\right)=1-e^{-r / 100}-\left(1-e^{-60 / 100}\right) \\
& e^{-40 / 100}-e^{-50 / 100}=e^{-60 / 100}-e^{-r / 100} \\
& e^{-r / 100}=e^{-60 / 100}-e^{-40 / 100}+e^{-50 / 100}=0.4850 \\
& -r / 100=\ln (0.4850)=-0.7236 \\
& r=72.36
\end{aligned}
\]
Question: The lifetime of a machine part is exponentially distributed with a mean of five years. Calculate the mean lifetime of the part, given that it survives less than ten years.

Options:
A
0.865
B
1.157
C
2.568
D
2.970
E
3.435

Correct answer: E

Solution: The given mean of 5 years corresponds to the pdf \(f(t)=0.2 e^{-0.2 t}\) and the cumulative distribution function \(F(t)=1-e^{-0.2 t}\). The conditional pdf is
\[
g(t)=\frac{f(t)}{F(10)}=\frac{0.2 e^{-0.2 t}}{1-e^{-2}}, 0<t<10
\]

The conditional mean is (using integration by parts)
\[
\begin{aligned}
& E(T \mid T<10)=\int_{0}^{10} \operatorname{tg}(t) d t=\int_{0}^{10} t \frac{0.2 e^{-0.2 t}}{1-e^{-2}} d t=0.2313 \int_{0}^{10} t e^{-0.2 t} d t \\
& \left.=0.2313\left[\left.t\left(-5 e^{-0.2 t}\right)\right|_{0} ^{10}-\int_{0}^{10}-5 e^{-0.2 t}\right) d t\right]=0.2323\left[-6.7668+0-\left.25 e^{-0.2 t}\right|_{0} ^{10}\right] \\
& =0.2313[-6.7668-3.3834+25]=3.435
\end{aligned}
\]
Question: A factory tests 100 light bulbs for defects. The probability that a bulb is defective is 0.02 . The occurrences of defects among the light bulbs are mutually independent events.

Calculate the probability that exactly two are defective given that the number of defective bulbs is two or fewer.

Options:
A
0.133
B
0.271
C
0.273
D
0.404
E
0.677

Correct answer: D

Solution: The number of defects has a binomial distribution with \(n=100\) and \(p=0.02\).
\[
\begin{aligned}
& P[X=2 \mid X \leq 2]=\frac{P[X=2]}{P[X \leq 2]}=\frac{\binom{100}{2}(0.02)^{2}(0.98)^{98}}{\binom{100}{0}(0.02)^{0}(0.98)^{100}+\binom{100}{1}(0.02)^{1}(0.98)^{99}+\binom{100}{2}(0.02)^{2}(0.98)^{98}} \\
& =\frac{0.27341}{0.13262+0.27065+0.27341}=0.404 .
\end{aligned}
\]
Question: A company has purchased a policy that will compensate for the loss of revenue due to severe weather events. The policy pays 1000 for each severe weather event in a year after the first two such events in that year. The number of severe weather events per year has a Poisson distribution with mean 1.

Calculate the expected amount paid to this company in one year.

Options:
A
80
B
104
C
368
D
512
E
632

Correct answer: B

Solution: The payment random variable is \(1000(X-2)\) if positive, where \(X\) has a Poisson distribution with mean 1 . The expected value is
\[
\begin{aligned}
& 1000 \sum_{x=3}^{\infty}(x-2) \frac{e^{-1}}{x!}=1000\left(\sum_{x=0}^{\infty}(x-2) \frac{e^{-1}}{x!}-\sum_{x=0}^{2}(x-2) \frac{e^{-1}}{x!}\right) \\
& =1000\left(1-2-\left[-2 e^{-1}-e^{-1}\right]\right)=1000\left(-1+3 e^{-1}\right)=104
\end{aligned}
\]

Note the first sum splits into the expected value of \(X\), which is 1 , and 2 times the sum of the probabilities (also 1).
Question: A company provides each of its employees with a death benefit of 100 . The company purchases insurance that pays the cost of total death benefits in excess of 400 per year. The number of employees who will die during the year is a Poisson random variable with mean 2.

Calculate the expected annual cost to the company of providing the death benefits, excluding the cost of the insurance.

Options:
A
171
B
189
C
192
D
200
E
208

Correct answer: C

Solution: Let \(X\) be the number of employees who die. The expected cost to the company is
\[
\begin{aligned}
& 100 P[Y=1]+200 P[Y=2]+300 P[Y=3]+400 P[Y>3] \\
& =100(2) e^{-2}+200(2) e^{-2}+300(4 / 3) e^{-2}+400\left[1-(1+2+2+4 / 3) e^{-2}\right] \\
& =400-1533.33 e^{-2}=192
\end{aligned}
\]
Question: A motorist just had an accident. The accident is minor with probability 0.75 and is otherwise major.

Let \(b\) be a positive constant. If the accident is minor, then the loss amount follows a uniform distribution on the interval \([0, b]\). If the accident is major, then the loss amount follows a uniform distribution on the interval \([b, 3 b]\).

The median loss amount due to this accident is 672 .
Calculate the mean loss amount due to this accident.

Options:
A
392
B
512
C
672
D
882
E
1008

Correct answer: D

Solution: Let \(X\) represent the loss due to the accident.
From the given information, the probability that \(X\) is \(\mathrm{in}[0, b]\) is 0.75 , which is larger than 0.5 . So the median, 672 , must lie in the interval \([0, b]\).

Note that in a uniform distribution over an interval \(I\), the probability of landing in an interval \(J\) is the length of the intersection of \(J\) and \(I\), divided by the length of \(I\).
Therefore, we have \(0.5=\mathrm{P}[X \leq 672]=0.75\left(\frac{672-0}{b-0}\right)\) which gives \(b=672\left(\frac{0.75}{0.5}\right)=1008\).
From the law of total probability applied to means, the mean loss due to the accident is
\[
\begin{aligned}
\mathrm{E}(X) & =\mathrm{P}[\text { minor acc. }] \mathrm{E}(X \mid \text { minor acc. })+\mathrm{P}[\text { major acc. }] \mathrm{E}(X \mid \text { major acc. }) \\
& =0.75 \mathrm{E}(X \mid X \text { is uniform on }[0, b])+0.25 \mathrm{E}(X \mid X \text { is uniform on }[b, 3 b]) . \\
& =0.75\left(\frac{0+1008}{2}\right)+0.25\left(\frac{1008+3(1008)}{2}\right)=882 .
\end{aligned}
\]
Question: The loss \(L\) due to a boat accident is exponentially distributed.

Boat insurance policy A covers up to 1 unit for each loss. Boat insurance policy B covers up to 2 units for each loss.

The probability that a loss is fully covered under policy B is 1.9 times the probability that it is fully covered under policy A.

Calculate the variance of \(L\).

Options:
A
0.1
B
0.4
C
2.4
D
9.5
E
\(\quad 90.1\)

Correct answer: E

Solution: The distribution function of \(L\) is \(F(x)=1-e^{-\lambda x}\) and its variance is \(1 / \lambda^{2}\). We are given \(F(2)=1.9 F(1)\) and therefore,
\(1-e^{-2 \lambda}=1.9\left(1-e^{-\lambda}\right)\)
\(\left(1-e^{-\lambda}\right)\left(1+e^{-\lambda}\right)=1.9\left(1-e^{-\lambda}\right)\)
\(e^{-\lambda}=0.9\)
\(\lambda=-\ln (0.9)=0.10536\)
\(\operatorname{Var}(L)=1 / 0.10536^{2}=90.1\).
Question: The number of claims \(X\) on a health insurance policy is a random variable with \(\mathrm{E}\left[X^{2}\right]=61\) and \(E\left[(X-1)^{2}\right]=47\).

Calculate the standard deviation of the number of claims.

Options:
A
2.18
B
2.40
C
7.31
D
7.50
E
7.81

Correct answer: A

Solution: \(E\left[(X-1)^{2}\right]=E\left[X^{2}\right]-2 E[X]+1=47\) so \(E[X]=(61+1-47) / 2=7.5\). The standard deviation is \(\sqrt{E\left[X^{2}\right]-E[X]^{2}}=\sqrt{61-7.5^{2}}=2.18\).
Question: Ten cards from a deck of playing cards are in a box: two diamonds, three spades, and five hearts. Two cards are randomly selected without replacement.

Calculate the variance of the number of diamonds selected, given that no spade is selected.

Options:
A
0.24
B
0.28
C
0.32
D
0.34
E
\(\quad 0.41\)

Correct answer: D

Solution: Let \(D\) be the number of diamonds selected and \(S\) be the number of spades. First obtain the hypergeometric probability \(S=0\) :
\(P(S=0)=\frac{\binom{3}{0}\binom{7}{2}}{\binom{10}{2}}=\frac{1(21)}{45}=\frac{7}{15}\).
The required probability distribution is:
\(P(D=0 \mid S=0)=\frac{P(D=0, S=0)}{P(S=0)}=\frac{1}{7 / 15} \frac{\binom{2}{0}\binom{3}{0}\binom{5}{2}}{\binom{10}{2}}=\frac{15}{7} \frac{1(1)(10)}{45}=\frac{10}{21}\)
\(P(D=1 \mid S=0)=\frac{P(D=1, S=0)}{P(S=0)}=\frac{1}{7 / 15} \frac{\binom{2}{1}\binom{3}{0}\binom{5}{1}}{\binom{10}{2}}=\frac{15}{7} \frac{2(1)(5)}{45}=\frac{10}{21}\)
\(P(D=2 \mid S=0)=\frac{P(D=2, S=0)}{P(S=0)}=\frac{1}{7 / 15} \frac{\binom{2}{2}\binom{3}{0}\binom{5}{0}}{\binom{10}{2}}=\frac{15}{7} \frac{1(1)(1)}{45}=\frac{1}{21}\).
Then,
\(E(D \mid S=0)=0(10 / 21)+1(10 / 21)+2(1 / 21)=12 / 21=4 / 7\)
\(E\left(D^{2} \mid S=0\right)=0^{2}(10 / 21)+1^{2}(10 / 21)+2^{2}(1 / 21)=14 / 21=2 / 3\)
\(\operatorname{Var}(D \mid S=0)=2 / 3-(4 / 7)^{2}=50 / 147=0.34\).
Question: A car and a bus arrive at a railroad crossing at times independently and uniformly distributed between 7:15 and 7:30. A train arrives at the crossing at 7:20 and halts traffic at the crossing for five minutes.

Calculate the probability that the waiting time of the car or the bus at the crossing exceeds three minutes.

Options:
A
0.25
B
0.27
C
0.36
D
0.40
E
0.56

Correct answer: A

Solution: To be delayed over three minutes, either the car or the bus must arrive between 7:20 and 7:22. The probability for each is \(2 / 15\). The probability they both arrive in that interval is \((2 / 15)(2 / 15)\). Thus, the probability of at least one being delayed is \(2 / 15+2 / 15-(2 / 15)(2 / 15)=56 / 225=0.25\).
Question: The number of minor surgeries, \(X\), and the number of major surgeries, \(Y\), for a policyholder, this decade, has joint cumulative distribution function
\(F(x, y)=\left[1-(0.5)^{x+1}\right]\left[1-(0.2)^{y+1}\right]\),
for nonnegative integers \(x\) and \(y\).
Calculate the probability that the policyholder experiences exactly three minor surgeries and exactly three major surgeries this decade.

Options:
A
0.00004
B
0.00040
C
0.03244
D
0.06800
E
0.12440

Correct answer: B

Solution: \(P(X=3, Y=3))=F(3,3)-F(2,3)-F(3,2)+F(2,2)=0.9360-0.8736-0.9300+0.8680=0.0004\)
Question: On Main Street, a driver's speed just before an accident is uniformly distributed on [5,20]. Given the speed, the resulting loss from the accident is exponentially distributed with mean equal to three times the speed.

Calculate the variance of a loss due to an accident on Main Street.

Options:
A
525
B
1463
C
1575
D
1632
E
1744

Correct answer: E

Solution: Let \(S\) be the speed and \(X\) be the loss. Given \(S, X\) has an exponential distribution with mean \(3 X\). Then, noting that the variance of an exponential random variable is the square of the mean, the variance of a uniform random variable is the square of the range divided by 12 , and for any random variable the second moment is the variance plus the square of the mean:
\[
\begin{aligned}
\operatorname{Var}(X) & =\operatorname{Var}[E(X \mid S)]+E[\operatorname{Var}(X \mid S) \\
& =\operatorname{Var}[3 S]+E\left(9 S^{2}\right) \\
& =9(20-5)^{2} / 12+9\left[(20-5)^{2} / 12+12.5^{2}\right] \\
& =1743.75
\end{aligned}
\]
Question: A loss under a liability policy is modeled by an exponential distribution. The insurance company will cover the amount of that loss in excess of a deductible of 2000 . The probability that the reimbursement is less than 6000, given that the loss exceeds the deductible, is 0.50 .

Calculate the probability that the reimbursement is greater than 3000 but less than 9000, given that the loss exceeds the deductible.

Options:
A
0.28
B
0.35
C
0.50
D
0.65
E
\(\quad 0.72\)

Correct answer: B

Solution: Due to the memoryless property of the exponential distribution, the distribution of the reimbursement given that there is a payment is exponential with the same parameter. Thus \(0.5=F(6000)=1-e^{-6000 / \lambda}\) which implies that \(\lambda=8656.17\). The solution is \(F(9000)-F(3000)=\left(1-e^{-9000 / 8656.17}\right)-\left(1-e^{-3000 / 8656.17}\right)=0.35\).
Question: A manufacturer produces computers and releases them in shipments of 100 . From a shipment of 100, the probability that exactly three computers are defective is twice the probability that exactly two computers are defective. The events that different computers are defective are mutually independent.

Calculate the probability that a randomly selected computer is defective.

Options:
A
0.040
B
0.042
C
0.058
D
0.060
E
0.072

Correct answer: C

Solution: The number of defective computers has a binomial distribution with \(n=100\) and \(p\) unknown. We have
\(P(X=3)=2 P(X=2)\)
\(\binom{100}{3} p^{3}(1-p)^{97}=2\binom{100}{2} p^{2}(1-p)^{98}\)
\(161,700 p=2(4,950)(1-p)\)
\(171,600 p=9,900\)
\(p=9,900 / 171,600=0.058\)
Question: In any 12 -month period, the probability that a home is damaged by fire is \(20 \%\) and the probability of a theft loss at a home is \(30 \%\). The occurrences of fire damage and theft loss are independent events.

Calculate the probability that a randomly selected home will either be damaged by fire or will have a theft loss, but not both, during the next year.

Options:
A
0.30
B
0.38
C
0.44
D
0.50
E
0.56

Correct answer: B

Solution: \(\mathrm{P}(\) fire damage and no theft \()=0.2(1-0.3)=0.14\)
\(\mathrm{P}(\) no fire damage and theft \()=(1-0.2)(0.3)=0.24\)
\(\mathrm{P}(\) exactly one \()=0.14+0.24=0.38\)
Question: The lifetime of a certain electronic device has an exponential distribution with mean 0.50 . Calculate the probability that the lifetime of the device is greater than 0.70 , given that it is greater than 0.40 .

Options:
A
0.203
B
0.247
C
0.449
D
0.549
E
0.861

Correct answer: D

Solution: The distribution function of an exponential distribution with mean 0.5 is \(F(x)=1-e^{-2 x}\). \(\operatorname{Pr}(X>0.7 \mid X>0.4)=\frac{\operatorname{Pr}(X>0.7)}{\operatorname{Pr}(X>0.4)}=\frac{e^{-1.4}}{e^{-0.8}}=e^{-0.6}=0.549\). This can be more efficiently solved using the memoryless property: \(\operatorname{Pr}(X>0.7 \mid X>0.4)=\operatorname{Pr}(X>0.3)=e^{-0.6}=0.549\).
Question: For a pregnant woman, a certain test will give the outcome "not pregnant" with probability 0.10 . For a non-pregnant woman, the test will give the outcome "pregnant" with probability 0.20 . Of women who take the test, \(20 \%\) are pregnant.

Calculate the probability that a woman is pregnant, given her test outcome is "pregnant."

Options:
A
0.10
B
0.20
C
0.50
D
0.53
E
\(\quad 0.90\)

Correct answer: D

Solution: Let WP denote the event "Woman is Pregnant", WNP the event "Woman is Not Pregnant", and TP the event "Test Shows Pregnancy". Using Bayes Theorem:
\[
\begin{aligned}
& \mathrm{P}(W P \mid T P)=\frac{\mathrm{P}(T P \mid W P) * \mathrm{P}(W P)}{\mathrm{P}(T P \mid W P) * \mathrm{P}(W P)+\mathrm{P}(T P \mid W N P) * \mathrm{P}(W N P)} \\
& =\frac{(1-0.1)(0.2)}{(1-0.1)(0.2)+(0.2)(1-0.2)}=\frac{0.18}{0.34}=0.5294
\end{aligned}
\]
Question: Let \(X\) and \(Y\) be discrete random variables with joint probability function
\(p(x, y)= \begin{cases}0.250, & \text { for } x=0, y=0 \\ 0.250, & \text { for } x=1, y=0 \\ 0.125, & \text { for } x=0, y=1 \\ 0.375, & \text { for } x=1, y=1 .\end{cases}\)
Calculate Corr ( \(X, Y\) ), the correlation coefficient of \(X\) and \(Y\).

Options:
A
0.06
B
0.23
C
0.26
D
0.38
E
0.63

Correct answer: C

Solution: \(\mathrm{E}(X)=0.25 * 1+0.375 * 1=0.625\)
\(\mathrm{E}(Y)=0.125 * 1+0.375=0.500\)
\(\mathrm{E}(X Y)=\mathrm{P}[X=1, Y=1]=0.375\)
\(\operatorname{Var}(X)=0.625-0.625^{2}=0.234\)
\(\operatorname{Var}(Y)=0.5-0.5^{2}=0.25\)
\(\operatorname{Cov}(X, Y)=\mathrm{E}(X Y)-\mathrm{E}(X) \mathrm{E}(Y)=0.375-0.625^{*} 0.500=0.0625\)
\(\operatorname{Corr}(X, Y)=0.0625 /(0.234 * 0.25)^{1 / 2}=0.258\)
Question: An insurance company sells a one-year insurance policy that covers fire and theft losses. The variance of the number of fire losses is 5 . The variance of the number of theft losses is 8 . The covariance between the number of fire and theft losses is 3 .

Calculate the variance of the total number of fire and theft losses covered by this policy.

Options:
A
7
B
10
C
13
D
16
E
19

Correct answer: E

Solution: \(\operatorname{Var}(\) Total \()=\operatorname{Var}(\) Fire \()+\operatorname{Var}(\) Theft \()+2 \operatorname{Cov}(\) Fire, Theft \()=5+8+2(3)=19\).
Question: Companies \(\mathrm{P}, \mathrm{Q}\), and R use routes that take their trucks through a common inspection checkpoint each day. The number of trucks for each company that pass the checkpoint each day is as follows:

\begin{tabular}{|c|c|}
\hline Company & \begin{tabular}{c} 
Number \\
of Trucks
\end{tabular} \\
\hline P & 4 \\
Q & 3 \\
R & 2 \\
\hline Total & 9 \\
\hline
\end{tabular}

Calculate the probability that at least one of two randomly chosen trucks is from Company P.

Options:
A
0.28
B
0.31
C
0.56
D
0.69
E
\(\quad 0.72\)

Correct answer: E

Solution: \(\mathrm{P}[\) At least one Company P truck]:
\(\mathrm{P}[\mathrm{P}\) and \((\mathrm{Q}\) or R\()]+\mathrm{P}[(\mathrm{R}\) or Q\()\) and P\(]+\mathrm{P}[\mathrm{P}\) and P\(]\)
\(=2\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)=0.72\).
Alternatively, take the complement of having no Company P trucks:
\(1-(5 / 9)(4 / 8)=0.72\).
Question: The probability that a homeowners policyholder reports a property claim in a year increases by \(25 \%\) per year. Conversely, the probability that a homeowners policyholder reports a liability claim in a year decreases by \(25 \%\) per year.

The probability that a homeowners policyholder reports both a property claim and a liability claim in Year 1 is 0.01 . The event that a homeowners policyholder reports a property claim is independent of the event that the policyholder reports a liability claim.

Calculate the probability that a homeowners policyholder reports both a property claim and a liability claim in Year 9.

Options:
A
0.005
B
0.006
C
0.010
D
0.014
E
0.015

Correct answer: B

Solution: \(\mathrm{P}\left[\right.\) Reports Property Claim in Year 9] \(=1.25^{8} * \mathrm{P}[\) Reports Property Claim in Year 1] \(\mathrm{P}\left[\right.\) Reports Liability Claim in Year 9] \(=0.75^{8} * \mathrm{P}[\) Reports Liability Claim in Year 1]
\(0.01 *(1.25)^{8} \times(0.75)^{8}=0.005967\)