# Exam P Practice Problem 105 – testing electronic devices

Problem 105-A

The length of operation (in years) for an electronic device follows an exponential distribution with mean 4. Ten such devices are being observed for one year for a quality control study.

The lengths of operation for these devices are independent.

Determine the probability that no more than three of the devices stop working before the end of the study.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 5 \bold 7$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 1 \bold 3 \bold 2$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 7 \bold 5 \bold 6 \bold 8$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 3 \bold 8 \bold 9$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 5 \bold 6 \bold 0$

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Problem 105-B

Twelve patients are randomly selected from a population of patients with history of heart disease to be tracked in a health study. The study begins with an initial assessment of health status. The participants are instructed to return for a follow up visit one year after the initial assessment.

For these patients, the time (in years) from the initial assessment to the next heart attack has an exponential distribution with mean 6.25 years. The times to the next heart attack for these patients are independent.

Determine the probability that ten or more patients experience no heart attack prior to the one-year follow up visit.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 9 \bold 1 \bold 3$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 1 \bold 9$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 5 \bold 4 \bold 8 \bold 1$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 4 \bold 5 \bold 5$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 7 \bold 4 \bold 3 \bold 2$

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probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

$\copyright$ 2019 – Dan Ma

# Exam P Practice Problem 100 – find the variance of loss in profit

Problem 100-A

The monthly amount of time $X$ (in hours) during which a manufacturing plant is inoperative due to equipment failures or power outage follows approximately a distribution with the following moment generating function.

$\displaystyle M(t)=\biggl( \frac{1}{1-7.5 \ t} \biggr)^2$

The amount of loss in profit due to the plant being inoperative is given by $Y=12 X + 1.25 X^2$.

Determine the variance of the loss in profit.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,927.20}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,608.20}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \text{475,693.76}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \text{583,358.20}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \text{601,769.56}$

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Problem 100-B

The weekly amount of time $X$ (in hours) that a manufacturing plant is down (due to maintenance or repairs) has an exponential distribution with mean 8.5 hours.

The cost of the downtime, due to maintenance and repair costs, is modeled by $Y=15+5 X+1.2 X^2$.

Determine the variance of the cost of the downtime.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \text{130,928.05}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \text{149,368.45}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \text{181,622.05}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \text{188,637.67}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \text{195,369.15}$

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probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

Daniel Ma actuarial

$\copyright$ 2017 – Dan Ma

Wording revised November 17, 2019

# Exam P Practice Problem 90 – Insurance Benefits

Problem 90-A

A random loss follows an exponential distribution with mean 20. An insurance reimburses this random loss up to a benefit limit of 30.

When a loss occurs, what is the expected value of the benefit not paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 4.5$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 5.1$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 6.3$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 8.5$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 11.2$

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Problem 90-B

A random loss follows an exponential distribution with mean 100. An insurance reimburses this random loss up to a benefit limit of 200.

When a loss occurs, what is the expected value of the benefit not paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 12.6$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 13.5$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 24.6$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 40.6$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 40.7$

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$\copyright \ 2014 \ \ \text{ Dan Ma}$

# Exam P Practice Problem 70 – Real Estate Sales Contest

Problem 70-A

A commercial real estate property company has three sales agents who are actively selling commercial real estate properties. The times (in days) to the next successful sale for these three agents are exponentially distributed with means 10 days, 15 days and 20 days.

These three agents work independently. So the time to the next successful sale for one agent is independent of the time to the next successful sale for any of the other agents.

To spur sales, the company has a contest among the three agents. Each agent produces a sale. The award will go to the first agent producing the first sale.

What is the probability that the winning sale will take place within one week?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.14$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.22$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.50$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.78$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.86$

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Problem 70-B

A commercial real estate property company has four sales agents who are actively selling commercial real estate properties. The times (in days) to the next successful sale for these four agents are exponentially distributed with means 10 days, 15 days and 20 days and 30 days.

These four agents work independently. So the time to the next successful sale for one agent is independent of the time to the next successful sale for any of the other agents.

To spur sales, the company has a contest among the four agents. Each agent produces a sale. The award will go to the first agent producing the first sale.

What is the expected waiting time (in days) from the beginning of the contest to the occurrence of the winning sale?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10$

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$\copyright \ 2013$

# Exam P Practice Problem 64 – Median Claim Size

Problem 64-A

The following is the density function of the claim size $X$ of an insurance policy from a block of property and casualty insurance policies:

$\displaystyle f(x)=\frac{1}{2} \biggl( 0.4 \ e^{-0.4 x} \biggr) + \frac{1}{2} \biggl( 0.2 \ e^{-0.2 x} \biggr) \ \ \ \ \ \ \ 0

Calculate the median claim size $X$.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.618$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.739$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.406$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.599$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.533$

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Problem 64-B

The following is the density function of the claim size $X$ of an insurance policy from a block of auto insurance policies:

$\displaystyle f(x)=\frac{3}{4} \biggl( \frac{1}{8} \ x \biggr) + \frac{1}{4} \biggl( \frac{1}{64} \ x^3 \biggr) \ \ \ \ \ \ \ 0

Calculate the median claim size $X$.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.2216$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.4972$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.8653$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.9622$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.9975$

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$\copyright \ 2013$

# Exam P Practice Problem 63 – Total Minutes of Telephone Calls

Problem 63-A

For a certain individual, the daily number of telephone calls (incoming or outgoing) has a Poisson distribution with mean 12. The length in time (in minutes) of each telephone call has an exponential distribution with mean 5 minutes.

The length of time of one telephone call is independent of the length of time of any other telephone call.

On a given day, this individual makes or receives 4 telephone calls. What is the probability that this person is on the telephone for more than half an hour?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1218$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1260$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1456$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1490$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1512$

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Problem 63-B

For a certain individual, the daily number of telephone calls (incoming or outgoing) has a Poisson distribution with mean 16. The length in time (in minutes) of each telephone call has an exponential distribution with mean 8 minutes.

The length of time of one telephone call is independent of the length of time of any other telephone call.

On a given day, this individual makes or receives 5 telephone calls. What is the probability that this person is on the telephone for more than 45 minutes?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2237$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2596$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.3384$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.3975$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.4085$

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$\copyright \ 2013$

# Exam P Practice Problem 62 – Waiting for Telephone Calls

Problem 62-A

An individual classifies the telephone calls he receives into the categories of Personal Calls (e.g. calls from friends and relatives) and Non-Personal Calls (all the other calls that are considered not Personal Calls).

Let $X$ be the time (in minutes) until the next Personal Call. Let $Y$ be the time (in minutes) until the next Non-Personal Call.

Suppose that $X$ and $Y$ are independent random variables and follow exponential distributions with means 8 minutes and 3 minutes, respectively.

What is the probability that the next incoming telephone call is a Personal Call?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2727$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.3735$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.5000$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.6265$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.7273$

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Problem 62-B

An individual classifies the telephone calls he receives into the categories of Personal Calls (e.g. calls from friends and relatives), Business Calls (calls related to his small business) and Other Calls (all the other calls not belonging to the Personal Call and Business Call categories).

Let $X$ be the time (in minutes) until the next Personal Call, let $Y$ be the time (in minutes) until the next Business Call and let $Z$ be the time (in minutes) until the next Other Call.

Suppose $X$, $Y$ and $Z$ are independent random variables and follow exponential distributions with means 12, 10 and 6 minutes, respectively.

What is the probability that the next telephone call this individual receives will be a Business Call?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3}{14}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{4}{14}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{3}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{14}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{6}{14}$

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$\copyright \ 2013$

# Exam P Practice Problem 61 – Claim Size of Auto Insurance Policies

Problem 61-A

An insurance company has a block of auto insurance policies. The claim size (in thousands) for a policy in this block of auto insurance policies is modeled by the random variable $Y=X^2$ where $X$ has an exponential distribution with mean 1.25.

What is the expected claim size for such an auto insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 1250$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 1563$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 2500$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 2755$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 3125$

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Problem 61-B

An insurance company has a block of auto insurance policies. The claim size (in thousands) for a policy in this block of auto insurance policies is modeled by the random variable $Y=X^2$ where $X$ has an exponential distribution with mean 1.6.

What is the standard deviation of the claim size for such an auto insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 1600$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 5120$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 9756.43$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 11448.67$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 12541.39$

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$\copyright \ 2013$

# Exam P Practice Problem 51 – Expected Claim Payment

Problem 51-A

The probability that a property will not be damaged in the upcoming year is 0.80. Assume that there is at most one incidence of damage in a year.

When there is a damage to the property, the amount of damage (in thousands) has an exponential distribution with mean 20.

The property owner insured the property against damage by purchasing an insurance policy with a deductible of 5.

What is the probability that the insurer’s payment to the owner will exceed 17.5?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ 0.0649$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ 0.0834$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ 0.3249$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ 0.4169$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ 0.8340$

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Problem 51-B

The probability that a property will not be damaged in the upcoming year is 0.80. Assume that there is at most one incidence of damage in a year.

When there is a damage to the property, the amount of damage (in thousands) has an exponential distribution with mean 20.

The property owner insured the property against damage by purchasing an insurance policy with a deductible of 5.

What is the expected payment made by the insurer to the owner of the property?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ 20.00$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ 4.00$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ 3.89$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ 3.50$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ 3.12$

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$\copyright \ 2013$

# Exam P Practice Problem 48 – Dental and Vision Expenses

Problem 48-A

An insurance company sells an ancillary health benefit plan that reimburses dental expenses and vision care expenses to the plan members. The plan provides no other benefits in addition to dental and vision care.

For the basic plan, the annual amount of dental reimbursement and the annual amount of vision care reimbursement are identically and exponentially distributed with mean 2 (in hundreds).

An actuary is designing a deluxe ancillary plan that provides similar but richer benefits. The annual amount of deluxe dental reimbursement is four times that of the basic plan. The annual amount of deluxe vision care reimbursement is two times that of the basic plan. Except for the richer benefit amounts, the actuary believes that the deluxe plan reimbursements have the same underlying probability distribution as the basic plan.

For both the basic plan and deluxe plan, the annual amount of dental reimbursement is independent of the annual amount of vision care reimbursement.

Which of the following is the probability density function of the total annual amount of expenses reimbursed by the deluxe plan?

$\displaystyle (A) \ \ \ \ \ \ f(x)=\frac{1}{64 \times 5!} \ x^5 \ e^{ -0.5 \ x}$

$\displaystyle (B) \ \ \ \ \ \ f(x)=4 \ e^{-0.125 \ x} - 4 \ e^{-0.25 \ x}$

$\displaystyle (C) \ \ \ \ \ \ f(x)=0.125 \ e^{-0.125 \ x} + 0.25 \ e^{-0.25 \ x}$

$\displaystyle (D) \ \ \ \ \ \ f(x)=0.25 \ e^{-0.125 \ x} - 0.25 \ e^{-0.25 \ x}$

$\displaystyle (E) \ \ \ \ \ \ f(x)=1.25 \ e^{-0.25 \ x} - 0.5 \ e^{-0.125 \ x}$

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Problem 48-B

When tornadoes occur, the total annual amount of property damages due to tornadoes (in millions) in area A has an exponential distribution with mean 20.

When tornadoes occur, the total annual amount of property damages due to tornadoes (in millions) in area B has an exponential distribution with mean 25.

Since area A and area B are sufficiently far apart, assume that the amount of tornado damages in one area is independent of the amount of damages in the other area.

What is the probability density function of the total annual amount of tornado damages for these two areas?

$\displaystyle (A) \ \ \ \ \ \ f(x)= e^{-0.04 \ x} - \ e^{-0.05 \ x}$

$\displaystyle (B) \ \ \ \ \ \ f(x)=0.2 \ e^{-0.04 \ x} - 0.2 \ e^{-0.05 \ x}$

$\displaystyle (C) \ \ \ \ \ \ f(x)=0.04 \ e^{-0.04 \ x} + 0.05 \ e^{-0.05 \ x}$

$\displaystyle (D) \ \ \ \ \ \ f(x)=\frac{1}{45} \ e^{- \frac{1}{45} \ x}$

$\displaystyle (E) \ \ \ \ \ \ f(x)=0.3 \ e^{-0.05 \ x} - 0.2 \ e^{-0.04 \ x}$

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