Tag Archives: Geometric distribution

Exam P Practice Problem 98 – flipping coins

Problem 98-A

Coin 1 is an unbiased coin, i.e. when flipping the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when flipping the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained.

Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2.

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      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2856

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3060

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.3295

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.3564

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.3690

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Problem 98-B

Box 1 contains 3 red balls and 1 white ball while Box 2 contains 2 red balls and 2 white balls. The two boxes are identical in appearance. One of the boxes is chosen at random. A ball is sampled from the chosen box with replacement until a white ball is obtained.

Determine the probability that the chosen box is Box 1 if the first white ball is observed on the 6th draw.

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      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.7530

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.7632

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.7825

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.7863

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.7915

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probability exam P

actuarial exam

math

Daniel Ma

mathematics

geometric distribution

Bayes

Answers

\copyright 2017 – Dan Ma

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Exam P Practice Problem 71 – Estimating Claim Frequency

Problem 71-A

An auto insurer issued policies to a large group of drivers under the age of 40. These drivers are classified into five distinct groups by age. These groups are equal in size.

The annual claim count distribution for any driver being insured by this insurer is assumed to be a binomial distribution. The following table shows more information about these drivers.

      \displaystyle \begin{bmatrix} \text{Age}&\text{ }&\text{ }&\text{Mean} &\text{ }&\text{ }&\text{Variance} \\\text{Group}&\text{ }&\text{ }&\text{Of Claim Count} &\text{ }&\text{ }&\text{Of Claim Count} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{16-17}&\text{ }&\text{ }&\displaystyle \frac{5}{2}&\text{ }&\text{ }&\displaystyle \frac{5}{4} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{18-24}&\text{ }&\text{ }&\displaystyle 2&\text{ }&\text{ }&\displaystyle 1 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{25-29}&\text{ }&\text{ }&\displaystyle \frac{3}{2}&\text{ }&\text{ }&\displaystyle \frac{3}{4} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{30-34}&\text{ }&\text{ }&\displaystyle 1&\text{ }&\text{ }&\displaystyle \frac{1}{2} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{35-39}&\text{ }&\text{ }&\displaystyle \frac{1}{2} &\text{ }&\text{ }&\displaystyle \frac{1}{4}    \end{bmatrix}

An insured driver is randomly selected from this large pool of insured and is observed to have one claim in the last year.

What is the probability that the mean number of claims in a year for this insured driver is greater than 1.5?

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      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{67}

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{57}

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3}{5}

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{51}{67}

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{50}{64}

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Problem 71-B

An auto insurer issued policies to a large group of drivers under the age of 40. These drivers are classified into five distinct groups by age. These groups are equal in size.

The annual claim count distribution for any driver being insured by this insurer is assumed to be a geometric distribution. The following table shows more information about these drivers.

      \displaystyle \begin{bmatrix} \text{Age}&\text{ }&\text{ }&\text{Mean} &\text{ }&\text{ }&\text{Variance} \\\text{Group}&\text{ }&\text{ }&\text{Of Claim Count} &\text{ }&\text{ }&\text{Of Claim Count} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{35-39}&\text{ }&\text{ }&\displaystyle 1 &\text{ }&\text{ }&\displaystyle 2 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{30-34}&\text{ }&\text{ }&\displaystyle 2&\text{ }&\text{ }&\displaystyle 6 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{25-29}&\text{ }&\text{ }&\displaystyle 3&\text{ }&\text{ }&\displaystyle 12 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{18-24}&\text{ }&\text{ }&\displaystyle 4&\text{ }&\text{ }&\displaystyle 20 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{16-17}&\text{ }&\text{ }&\displaystyle 5&\text{ }&\text{ }&\displaystyle 30    \end{bmatrix}

An insured driver is randomly selected from this large pool of insured and is observed to have one claim in the last year.

What is the probability that the mean number of claims in a year for this insured driver is greater than 2.5?

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      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.49

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.51

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.55

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.57

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.60

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\copyright \ 2013 \ \ \text{Dan Ma}

Exam P Practice Problem 60 – Health Insurance Claim Frequency

Problem 60-A

An insurance company issued health insurance policies to individuals. The company determined that Y, the number of claims filed by an insured in a year, is a random variable with the following probability function.

      \displaystyle P(Y=y)=0.45 \ (0.55)^{\displaystyle y} \ \ \ \ \ \ y=0,1,2,3,\cdots

What is the probability that a random selected insured from this group of insured individuals will file more than 5 claims in a year?

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0226

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0277

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0357

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0503

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0749

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Problem 60-B

An insurance company issued health insurance policies to individuals. The company determined that Y, the number of claims filed by an insured in a year, is a random variable with the following probability function.

      \displaystyle P(Y=y)=0.45 \ (0.55)^{\displaystyle y} \ \ \ \ \ \ y=0,1,2,3,\cdots

The number of claims filed by one insured individual is independent of the number of claims filed by any other insured individual.

An actuary studied three randomly selected insured individuals from this group of individuals who purchased health policies from this company. What is the probability that these three insured individuals will file more than 6 claims in a year?

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0457

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0706

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1495

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2201

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2406

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\copyright \ 2013

Exam P Practice Problem 53 – Hospital Expense Plans

Problem 53-A

An insurer sells a hospital expense plan that pays a fixed sum per day of hospitalization. Suppose that the number of days of hospitalization in a year for someone insured under this plan has a Poisson distribution with mean 0.8.

In each calendar year, the plan pays 2,000 for each day of hospitalization subject to the condition that the first two days of hospitalization are the responsibilities of the insured.

What is the expected payment for hospitalization during a calendar year under this hospital expense plan?

      \displaystyle (A) \ \ \ \ \ \ \ 116.24

      \displaystyle (B) \ \ \ \ \ \ \ 244.75

      \displaystyle (C) \ \ \ \ \ \ \ 305.93

      \displaystyle (D) \ \ \ \ \ \ \ 785.26

      \displaystyle (E) \ \ \ \ \ \ \ 1600.00

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Problem 53-B

An insurer sells a hospital expense plan that pays a fixed sum per day of hospitalization. Suppose that the number of days of hospitalization in a year for someone insured under this plan has the following probability function.

      \displaystyle P(X=x)=\frac{3}{4^{x+1}} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots

In each calendar year, the plan pays 1,000 for each day of hospitalization subject to the condition that the first day of hospitalization is the responsibility of the insured.

What is the expected payment for hospitalization during a calendar year under this hospital expense plan?

      \displaystyle (A) \ \ \ \ \ \ \ 76.83

      \displaystyle (B) \ \ \ \ \ \ \ 83.33

      \displaystyle (C) \ \ \ \ \ \ \ 111.11

      \displaystyle (D) \ \ \ \ \ \ \ 145.83

      \displaystyle (E) \ \ \ \ \ \ \ 333.33

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\copyright \ 2013

Exam P Practice Problem 50 – Payments for Hospitalization

Problem 50-A

An insurer sells a hospital expense plan that pays a fixed sum per day of hospitalization. Suppose that the number of days of hospitalization in a year for someone insured under this plan has the following probability function.

      \displaystyle P(X=x)=\frac{3}{4^{x+1}} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots

The plan pays 1,000 for each day of hospitalization up to 4 days a year.

What is the expected payment for hospitalization under this hospital expense plan?

      \displaystyle (A) \ \ \ \ \ \ \ 250.00

      \displaystyle (B) \ \ \ \ \ \ \ 328.13

      \displaystyle (C) \ \ \ \ \ \ \ 332.03

      \displaystyle (D) \ \ \ \ \ \ \ 333.33

      \displaystyle (E) \ \ \ \ \ \ \ 444.44

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Problem 50-B

An insurer sells a hospital expense plan that pays cash for each day of hospitalization. Suppose that the number of days of hospitalization in a year for someone insured under this plan has the following probability function.

      \displaystyle P(X=x)=\frac{3}{4^{x+1}} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots

The plan pays for each day of hospitalization up to 4 days a year subject to the condition that the plan pays 500 for each of the first 3 days and 2500 for the fourth day.

What is the expected amount paid to the insured under this hospital expense plan?

      \displaystyle (A) \ \ \ \ \ \ \ 150.00

      \displaystyle (B) \ \ \ \ \ \ \ 166.67

      \displaystyle (C) \ \ \ \ \ \ \ 169.92

      \displaystyle (D) \ \ \ \ \ \ \ 173.83

      \displaystyle (E) \ \ \ \ \ \ \ 175.00

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Exam P Practice Problem 24 – Total Number of Claims

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Problem 24A
An insurance portfolio consists of four policyholders. The number of claims N for each policyholder in a calendar year follows a distribution with the following probability function:

    \displaystyle P(N=k)=\biggl( \frac{13}{25} \biggr)^k \ \frac{12}{25} \ \ \ \ \ \ \ \ k=0,1,2,3,\cdots

Assume that the number of claims for one policyholder is independent of the number of claims for any one of the other policyholders in the portfolio. What is the probability that the total number of claims in this portfolio in the upcoming calendar year is 4?

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Problem 24B
Use the same information as in Problem 24A. What is the probability that there will be at most 4 claims in the portfolio in the upcoming year?

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Exam P Practice Problem 5

Problem 5a
Suppose that X and Y are independent random variables with the following moment generating functions:

    \displaystyle M_X(t)=\frac{4}{9}+\frac{4}{9}e^t+\frac{1}{9}e^{2t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_Y(t)=e^{\displaystyle -3+3e^t}

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Find the probability \displaystyle P(\frac{Y}{2}=X).

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Problem 5b
Suppose that X and Y are independent random variables with the following moment generating functions:

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    \displaystyle M_X(t)=\frac{1}{3-2e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_Y(t)=\biggl(\frac{1+3e^t}{4}\biggr)^2

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Find the probability P(Y=X-2).

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Solution is found below.

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Solution to Problem 5a
The mgf M_X(t) is that of a binomial distribution with n=2 and p=\frac{1}{3}. The mgf M_Y(t) is that of a Poisson distribution with parameter \lambda=3.

    \displaystyle \begin{aligned}P(Y=2X)&=P(X=0,Y=0)+P(X=1,Y=2)+P(X=2,Y=4) \\&=\biggl(\frac{2}{3}\biggr)^2 \ e^{-3}+2 \biggl(\frac{1}{3}\biggr) \biggl(\frac{2}{3}\biggr) \ e^{-3} \ \frac{3^2}{2}+\biggl(\frac{1}{3}\biggr)^2 \ e^{-3} \ \frac{3^4}{4!} \\&=\frac{203}{72} \ e^{-3} \\&=0.1403718733  \end{aligned}

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Answer to Problem 5b

    \displaystyle \frac{1}{12}