Tag Archives: The law of total probability

Exam P Practice Problem 80 – Total Insurance Payment

Problem 80-A

An individual purchases an insurance policy to cover a random loss. If a random loss occurs during the year, the amount of loss is at least 1. Once a random loss occurs, the insurance payment to the insured is modeled by the random variable X with the following density function

    \displaystyle f(x)=\frac{1}{x^2} \ \ \ \ \ 1<x<\infty

If there is a loss, there is only one loss in each year. In each year, the probability of a loss is 0.25. What is the probability that the annual amount paid to the policyholder under this policy is less than 2?

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.250

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.500

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.750

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.875

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.925

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Problem 80-B

An individual purchases an insurance policy to cover a random loss. If a random loss occurs during the year, the loss amount is at least 1. Once a loss occurs, the insurance payment to the insured is modeled by the random variable X with the following density function

    \displaystyle f(x)=\frac{1}{30} \ x(1+3x) \ \ \ \ \ 1<x<3

If there is a loss, there is only one loss in each year. In each year, the probability of a loss is 0.15. What is the probability that the annual amount paid to the policyholder under this policy is less than 2?

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.1500

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.2833

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.8500

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.8735

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.8925

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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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Exam P Practice Problem 40 – Total Claim Amount

Problem 40-A

The number of claims in a calendar year for an insured has a probability function indicated below.

            \displaystyle \begin{bmatrix} \text{Number of Claims}&\text{ }&\text{Probability}  \\\text{ }&\text{ }&\text{ } \\ 0&\text{ }&\displaystyle \frac{27}{64} \\\text{ }&\text{ }&\text{ }  \\ 1&\text{ }&\displaystyle \frac{27}{64} \\\text{ }&\text{ }&\text{ }  \\ 2&\text{ }&\displaystyle \frac{9}{64} \\\text{ }&\text{ }&\text{ }  \\ 3&\text{ }&\displaystyle \frac{1}{64}     \end{bmatrix}

When a claim occurs, the claim amount X, regardless of how many claims the insured will have in the calendar year, has probabilities P(X=1)=0.8 and P(X=2)=0.2. The claim amounts in a calendar year for this insured are independent.

Let T be the total claim amount for this insured in a calendar year. Calculate P(3 \le T \le 4).

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Problem 40-B

A bowl has 3 red balls and 6 white balls. Select two balls at random from this bowl with replacement. Let N be the number of red balls found in the two selected balls. When N=n where n>0, roll a fair die n times.

Let W be the sum of the rolls of the die. Calculate P(4 \le W \le 5).

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Exam P Practice Problem 10

Problem 10a
An individual is facing an outcome of an annual financial loss X (in tens of thousands of dollars) whose probability density function is given by

\displaystyle f(x)=0.003 x^2, \ \ \ \ 0<x<10

The probability of a loss in the next year is 0.08. If there is a loss, there is only one loss in any given year. An insurance policy is available to protect against the financial loss by paying in full when a loss occurs.

  1. What is the probability that the insurer’s payment to the insured will exceed $50,000?
  2. What is the mean payment made by the insurer to the insured?
  3. What is the variance of the amount of payment made by the insurer?

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Problem 10b
Suppose that instead of buying a policy that pays the loss in full, the individual buys a policy that has a 80/20 coinsurance provision, i.e., the insurance company pays 80% of the loss and the insured retains the remaining 20% of a loss. Answer the same three questions.

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Solution is found below.

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Solution to Problem 10a
Answers
\displaystyle 10a.1 \ \ \ \ \ 0.07

\displaystyle 10a.2 \ \ \ \ \ 0.6

\displaystyle 10a.3 \ \ \ \ \ 4.44

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Let X be the loss variable as described in the problem. Then the following is the probability P(X>t).

\displaystyle \begin{aligned}P(X>t)&=\int_t^{10} 0.003 x^2 \ dx \\&=1-0.001 t^3   \end{aligned}

One important thing to keep in mind is that the occurrence of a financial loss is not certain. So the answer to question #1 is not P(X>5). Let Y be the insurance payment to the insured. Note that Y is conditional on the occurrence of a loss. If the loss does not happen, Y=0. If the loss does happen, Y=X. Likewise, P(Y>t)=0 in case of no loss and P(Y>t)=P(X>t) in case of a loss. So we can use the law of total probability to obtain P(Y>5).

\displaystyle \begin{aligned}P(Y>5)&=0 \times 0.92+P(X>5) \times 0.08  \\&=(1-0.001 5^3) \times 0.08 \\&=0.07   \end{aligned}

The answers to the other two questions can also be obtained by using the law of total probability.

\displaystyle \begin{aligned}E(Y)&=0 \times 0.92+\int_0^{10} x 0.003 x^2 \ dx \times 0.08  \\&=\int_0^{10} 0.003 x^3 \ dx \times 0.08 \\&=7.5 \times 0.08 \\&=0.6 \\&=\$6000   \end{aligned}

\displaystyle \begin{aligned}E(Y^2)&=0 \times 0.92+\int_0^{10} x^2 0.003 x^2 \ dx \times 0.08  \\&=\int_0^{10} 0.003 x^4 \ dx \times 0.08 \\&=60 \times 0.08 \\&=4.8   \end{aligned}

\displaystyle Var(Y)=4.8-0.6^2=4.44

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Answers to Problem 10b
\displaystyle 10b.1 \ \ \ \ \ 0.06046875

\displaystyle 10b.2 \ \ \ \ \ 0.48

\displaystyle 10b.3 \ \ \ \ \ 2.8416

Exam P Practice Problem 7

Problem 7a
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by

      \displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0

The property owner purchased an insurance policy that pays the amount of the damage in full during the next year.

  1. What is the probability that the insurer’s payment to the owner will exceed $17,500?
  2. What is the mean payment made by the insurer to the owner of the property?
  3. What is the variance of the amount of payment made by the insurer?

Problem 7b
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by

      \displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0

The property owner purchased an insurance policy with a coinsurance provision that pays 80% of the amount of the damage during the next year. The remaining 20% of the amount of the damage is retained by the property owner.

  1. What is the probability that the insurer’s payment to the owner will exceed $17,500?
  2. What is the mean payment made by the insurer to the owner of the property?
  3. What is the variance of the amount of payment made by the insurer?

Solution is found below.

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Solution to Problem 7a
Let X be the loss amount (i.e. the pdf is the one given in the problem). One important thing to keep in mind is that a loss to the property is not certain. So the answer for #1 is not P(X>17.5). Let Y be the payment made by the insurer to the property owner. The answer for #1 is P(Y>17.5). One way to look at the problem is through the law of total probability.

\displaystyle \begin{aligned}P(Y>17.5)&=P(Y>17.5 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +P(Y>17.5 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+e^{-0.05 (17.5)} \times 0.2 \\&=0.2 \times e^{-0.875} \\&=0.08337  \end{aligned}

The following provides the answers to the rest of the problem:

\displaystyle \begin{aligned}E(Y)&=E(Y \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{1}{0.05} \times 0.2 \\&=4 \\&=\$4000  \end{aligned}

\displaystyle \begin{aligned}E(Y^2)&=E(Y^2 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y^2 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{2}{0.05^2} \times 0.2 \\&=160  \end{aligned}

\displaystyle \begin{aligned}Var(Y)&=E(Y^2)-E(Y)^2 \\&=160-4^2 \\&=144  \end{aligned}

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Answer to Problem 7b
\displaystyle 7b.1 \ \ \ \ \ \ 0.2 e^{-1.09375}=0.0669916086

\displaystyle 7b.2 \ \ \ \ \ \ \$3200

\displaystyle 7b.3 \ \ \ \ \ \ 92.16

Exam P Practice Problem 6

Problem 6a
An auto insurer offers collison coverage to two large groups of policyholders, Group 1 and Group 2. On the basis of historical data, the insurer has determined that the loss due to collision for a policyholder in Group 1 has an exponential distribution with mean 5. On the other hand, the loss due to collision for a policyholder in Group 2 has an exponential distribution with mean 10.

Considering the two groups as one block, about 75% of the losses are from Group 1.

  1. Given a randomly selected loss in this block, what is the probability that the loss is greater than 15?
  2. If a randomly selected loss is greater than 15, what is the probability that it is a from a policyholder in Group 1?

Problem 6b
An auto insurer has two groups of policyholders – those considered good risks and those considered bad risks. On the basis of historical data, the insurer has determined that the number of car accidents during a policy year for a policyholder classified as good risk follows a binomial distribution with n=2 and p=\frac{1}{10}. The number of car accidents for a policyholder classified as bad risk follows a binomial distribution with n=2 and p=\frac{3}{10}. In this block of policies, 75% are classified as good risks and 25% are classified as bad risks. A new customer, whose risk class is not yet known with certainty, has just purchased a new policy.

  1. What is the probability that this new policyholder is not accident-free in the upcoming policy year?
  2. By the end of the policy year, it is found that this policyholder is not accident-free, what is the probability that the policyholder is a “good risk” policyholder?

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Solution to Problem 6a
Let X be the loss amount of a randomly selected policyholder. The conditional probabilities of a loss greater than 7.5 are:

\displaystyle P(X>15 \lvert \text{ Group 1 Policyholder})=e^{-\frac{15}{5}}
\displaystyle P(X>15 \lvert \text{ Group 2 Policyholder})=e^{-\frac{15}{10}}

By the law of total probability, the unconditional probability is:

\displaystyle \begin{aligned}P(X>15)&=P(X>15 \lvert \text{ Group 1 Policyholder}) \times P(\text{ Group 1 Policyholder}) \\&\ \ \ +P(X>15 \lvert \text{ Group 2 Policyholder}) \times P(\text{ Group 2 Policyholder}) \\&=\frac{3}{4} \times e^{-\frac{15}{5}}+\frac{1}{4} \times e^{-\frac{15}{10}} \\&=\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5} \\&=0.0931228413  \end{aligned}

The above calculation indicates that the unconditional probability is the weighted average of the conditional probabilities. The answer to the second question is obtained by applying the Bayes’ theorem:

\displaystyle \begin{aligned}P(\text{Group 1 Policyholder } \lvert X>15)&=\frac{P[(\text{Group 1 Policyholder}) \cap (X>15)]}{P(X>15)} \\&=\frac{\frac{3}{4} \times e^{-3}}{\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5}}  \\&=0.400978973 \end{aligned}

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Answer to Problem 6b
\displaystyle 6b.1 \ \ \ \ \frac{108}{400}=0.27

\displaystyle 6b.2 \ \ \ \ \frac{57}{108}=0.5278