Tag Archives: Bayes’ Theorem

Exam P Practice Problem 110 – likelihood of auto accidents

Problem 110-A

An actuary studied the likelihood of accidents in a one-year period among a large group of insured drivers. The following table gives the results.

Age Group Percent of Drivers Probability of 0 Accidents Probability of 1 Accident
16-20 15% 0.20 0.25
21-30 25% 0.35 0.40
31-50 35% 0.60 0.30
51-70 20% 0.67 0.23
71+ 5% 0.50 0.35

Suppose that a randomly selected insured driver in the studied group had at least 2 accidents in the past year. Calculate the probability that the insured driver is in the age group 21-30.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \  \bold 0 \bold . \bold 1 \bold 7

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 4

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 3 \bold 0

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 0

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5

\text{ }

\text{ }

\text{ }

\text{ }

Problem 110-B

An auto insurance company performed a study on the frequency of accidents of its insured drivers in a one-year period. The following table gives the results of the study.

Age Group Percent of Drivers Probability of At Least 1 Accident
16-20 10% 0.30
21-40 20% 0.20
41-65 35% 0.10
66+ 35% 0.12

A randomly selected insured driver from the study was found to have no accidents in the one-year period.

Calculate the probability that the insured driver is from the age group 16-20.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \  \bold 0 \bold . \bold 0 \bold 8

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 2

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 5

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 9

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 0

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers

exam P practice problem

probability exam P

actuarial exam

actuarial practice problem

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

\copyright 2019 – Dan Ma

Advertisements

Exam P Practice Problem 108 – random selection of balls

Both 108-A and 108-B use the following information.

Bowl One contains 1 blue ball and 4 orange balls. Bowl Two contains 3 blue balls and 2 orange balls. A bowl is chosen at random. Balls are randomly chosen one at a time from the chosen bowl, with each chosen ball returning to the bowl.

.

Problem 108-A

What is the probability that four of the first six selections are blue ball?

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \  \frac{\bold 4 \bold 8 \bold 6}{\bold 3 \bold 1 \bold 2 \bold 5}

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 0 \bold2}{\bold 6 \bold 2 \bold 5}

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 3 \bold 4}{\bold 2 \bold 0 \bold 5}

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 2 \bold 1}{\bold 1 \bold 2 \bold 5}

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 0 \bold 2 \bold 0}{\bold 3 \bold 1 \bold 2 \bold 5}

\text{ }

\text{ }

\text{ }

\text{ }

Problem 108-B

If four of the first six selections are blue balls, what is the probability that the balls are selected from Bowl One?

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \  \frac{\bold 9}{\bold 2 \bold 5 \bold 5}

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 2}{\bold 4 \bold 3}

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 4}{\bold 8 \bold 5}

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 5}{\bold 2 \bold 5 \bold 5}

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1}{\bold 2}

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

dan ma statistical actuarial

daniel ma statistical actuarial

\copyright 2019 – Dan Ma

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.

\text{ }

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

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

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

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

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

\text{ }

\text{ }

\text{ }

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.

\text{ }

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

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

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

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

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

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

probability exam P

actuarial exam

math

Daniel Ma

mathematics

geometric distribution

Bayes

Answers

\copyright 2017 – Dan Ma

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?

\text{ }

      \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}

\text{ }

\text{ }

\text{ }

\text{ }

______________________________________________________________________

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?

\text{ }

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

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

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

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

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

______________________________________________________________________

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

______________________________________________________________________

______________________________________________________________________

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

Exam P Practice Problem 67 – Statistical Studies of Insured Drivers

Problem 67-A

An auto insurance company performed a statistical study on its insured drivers. The following table shows the results.

            \displaystyle \begin{bmatrix} \text{Age Group}&\text{ }&\text{Percentage}&\text{ }&\text{Annual Probability of} \\ \text{ }&\text{ }&\text{of its Drivers}&\text{ }&\text{At Least One Claim}  \\\text{ }&\text{ }&\text{ } \\ \text{16-20}&\text{ }&15 \% &\text{ }&0.18  \\\text{ }&\text{ }&\text{ } \\ \text{21-30}&\text{ }&20 \% &\text{ }&0.12 \\\text{ }&\text{ }&\text{ } \\ \text{31-50}&\text{ } &30 \% &\text{ }&0.08 \\\text{ }&\text{ }&\text{ } \\ \text{51-70}&\text{ }&25 \% &\text{ }&0.09 \\\text{ }&\text{ }&\text{ } \\ \text{71 and up}&\text{ }&10 \% &\text{ }&0.11  \end{bmatrix}

The authors of the statistical study also found that for any insured driver in the study, the annual number of claims follows a Poisson distribution.

Suppose that an insured driver in the study had exactly 2 claims in the past year. What is the probability that the insured driver is from the age group 16-20?

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.150

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.223

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.249

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.376

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.415

\text{ }

\text{ }

\text{ }

\text{ }

Problem 67-B

An auto insurance company performed a statistical study on its younger insured drivers (under 35 years of age). The following table shows the results.

            \displaystyle \begin{bmatrix} \text{Age Group}&\text{ }&\text{Percentage}&\text{ }&\text{Annual Probability of} \\ \text{ }&\text{ }&\text{of its Drivers}&\text{ }&\text{At Least One Claim}  \\\text{ }&\text{ }&\text{ } \\ \text{16-17}&\text{ }&12 \% &\text{ }&0.18  \\\text{ }&\text{ }&\text{ } \\ \text{18-24}&\text{ }&38 \% &\text{ }&0.10 \\\text{ }&\text{ }&\text{ } \\ \text{25-34}&\text{ } &50 \% &\text{ }&0.06  \end{bmatrix}

The authors of the statistical study also found that for any insured driver in the study, the annual number of claims follows a Poisson distribution. Furthermore, for any insured driver in the study, the number of claims in one year is independent of the number of claims in any other year.

Suppose that in a 2-year period, an insured driver in the study had exactly 1 claim in year 1 and exactly 2 claims in year 2. What is the probability that the insured driver is from the age group 16-17?

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.120

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.229

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.241

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.329

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.576

_____________________________________________________________________

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

_____________________________________________________________________

Answers

_____________________________________________________________________

\copyright \ 2013

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?

\text{ }

Soultion is found below.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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}

\text{ }

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

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