# 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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# Exam P Practice Problem 41 – Conditional Expected Number of Balls

Problem 41-A

An insurer has a block of business where the number of claims in a year for a policyholder in the block has the following probability distribution.

$\displaystyle \begin{bmatrix} \text{Number of Claims}&\text{ }&\text{Probability} \\\text{ }&\text{ }&\text{ } \\ 0&\text{ }&\displaystyle 0.08 \\\text{ }&\text{ }&\text{ } \\ 1&\text{ }&\displaystyle 0.35 \\\text{ }&\text{ }&\text{ } \\ 2&\text{ }&\displaystyle 0.40 \\\text{ }&\text{ }&\text{ } \\ 3&\text{ }&\displaystyle 0.12 \\\text{ }&\text{ }&\text{ } \\ 4&\text{ }&\displaystyle 0.0375 \\\text{ }&\text{ }&\text{ } \\ 5&\text{ }&\displaystyle 0.0125 \end{bmatrix}$

Two policyholders from this block are randomly selected and observed for a year. It is found that there are exactly four claims for these two policyholders in the past year. What is the probability that one of the policyholders has exactly three claims?

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Problem 41-B

There are two bowls containing red balls and white balls. Bowl 1 contains 5 red balls and 5 white balls. Five balls are randomly selected from Bowl 1 without replacement. Bowl 2 also contains 5 red balls and 5 white balls. Five balls are randomly selected from Bowl 2 with replacement.

If the total number of red balls drawn from the two bowls is eight, what is the expected number of red balls from Bowl 2?

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# Exam P Practice Problem 34 – Number of Claims

Problem 34A

An actuary is asked to review the claim history in a two-year period for one policyholder. The number of claims in any year for this policyholder has a Poisson distribution with mean 0.5 and is independent of the number of claims in any previous year.

The actuary finds that the policyholder has made 5 claims in the two-year period.

What is the probability that the policyholder has made more than three claims in the first year of the review period?

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Problem 34B

An actuary is asked to review the claim history in a two-year period for one policyholder. In the first year of the review period, the number of claims for this policyholder has a Poisson distribution with mean 0.3. In the second year, the number of claims for this policyholder has a Poisson distribution with mean 0.6.

The number of claims in one year is independent of the number of claims in any previous year.

The actuary finds that the policyholder has made 5 claims in the review period. What is the probability that the policyholder has made at least three claims in the first year of the review period?

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# Exam P Practice Problem 16 – Conditional Probability and Exponential Distribution

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Practice Problem 16A
Losses $X$ and $Y$ due to two different perils are independently and identically distributed according to the density function $f(x)=0.25 \ e^{-0.25 \ x}$ where $x>0$. If the total loss for the two perils is greater than $2$, what is the probability that $X$ is greater than $1$?

Practice Problem 16B
Losses $X$ and $Y$ due to two different perils are independently and identically distributed according to the density function $f(x)=0.125 \ e^{-0.125 \ x}$ where $x>0$. If the total loss for the two perils is greater than $6$, what is the probability that $X$ is less than $4$?

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# Continuous Convolution – A Guided Example

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Thia post is a guided example of a practice problem (Problem 15A found in Exam P Practice Problem 15 – Still More Convolution Practice). The exposition is to make clear the thought process on how to set up and evaluate the integral to find the pdf of an independent sum. The following is the statement of the problem:

Problem 15A
Find the pdf of $Z=X+Y$ where $X$ and $Y$ are independent random variables such that $X$ is uniformly distributed on the interval $0 and $Y$ is exponentially distributed with mean 10.

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Setting up the Scene

The pdfs of $X$ and $Y$ are:

$\displaystyle f(x)=\frac{1}{10} \ \ \ \ 0

$\displaystyle g(y)=0.1 \ e^{-0.1 \ y} \ \ \ \ y>0$

Since $X$ and $Y$ are independent, the joint pdf of $X$ and $Y$ is $f_{X,Y}(x,y)=f(x) \ g(y)$. The following figure shows the support of this joint distribution.

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Figure 1 – Support of the Joint Distribution

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Let $h(z)$ denote the pdf of the independent sum $Z=X+Y$. To calculate $h(z)$, we need to sum the joint density $f_{X,Y}(x,y)$ over the entire green region in Figure 1. Each value of $h(z)$ is the sum of the joint density $f_{X,Y}(x,y)$ on a particular line $x+y=z$ (see Figure 2 below).

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Figure 2

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For each point of the line $x+y=z$, the joint density $f_{X,Y}(x,y)$ at that point is $f(x) \ g(z-x)$ or $f(z-y) \ g(y)$. Thus we have:

$\displaystyle h(z)=\int_{-\infty}^{+\infty} f(x) \ g(z-x) \ dx \ \ \ \ \ \ \ \ (1)$

$\displaystyle h(z)=\int_{-\infty}^{+\infty} f(z-y) \ g(y) \ dy \ \ \ \ \ \ \ \ \ (2)$

Let’s focus on integral $(1)$. Since the support of the $X$ variable is a bounded interval, there are two cases to consider: $0 and $10. So we calculate $h(z)$ separately in each case. Let's look at the following figures.

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Looking at Two Cases

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Figure 3 – Case 1

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Figure 4 – Case 2

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Case 1 covers all the lines $x+y=z$ from the red line to the blue line in Figure 3. Case 2 covers all lines $x+y=z$ above the blue line in Figure 4. These considerations will dictate the limits for the integral in $(1)$.

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Setting Up Integrals

Here’s the integrals for the two cases.
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Case 1

$0
$\displaystyle h(z)=\int_{0}^{z} f(x) \ g(z-x) \ dx \ \ \ \ \ \ \ \ \ \ \ (3)$

Case 2

$10
$\displaystyle h(z)=\int_{0}^{10} f(x) \ g(z-x) \ dx \ \ \ \ \ \ \ \ \ \ (4)$

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The following two figures show how integrals $(3)$ and $(4)$ are set up. For a typical line $x+y=z$ in Case 1 (Figure 5), the range for $x$ is $0. For a typical line $x+y=z$ in Case 2 (Figure 6), the range for $x$ is $0.

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Figure 5 – Typical Line in Case 1

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Figure 6 – Typical Line in Case 2

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Calculation

Case 1

\displaystyle \begin{aligned}h(z)&=\int_{0}^{z} f(x) \ g(z-x) \ dx \\&=\int_0^z \frac{1}{10} \ 0.1 \ e^{-0.1 \ (z-x)} \ dx \\&=\int_0^z \frac{1}{10} \ e^{-0.1 \ z} \ 0.1 \ e^{0.1 \ x} \ dx \\&=\frac{1}{10} \ e^{-0.1 \ z} \ (e^{0.1 \ z}-1) \\&= \frac{1}{10}\ (1-e^{-0.1 \ z})\\&=0.1-0.1e^{-0.1 \ z} \end{aligned}

Case 2

\displaystyle \begin{aligned}h(z)&=\int_{0}^{10} f(x) \ g(z-x) \ dx \\&=\int_0^{10} \frac{1}{10} \ 0.1 \ e^{-0.1 \ (z-x)} \ dx \\&=\int_0^{10} \frac{1}{10} \ e^{-0.1 \ z} \ 0.1 \ e^{0.1 \ x} \ dx \\&=\frac{1}{10} \ e^{-0.1 \ z} \ (e-1) \\&=\frac{e-1}{10} \ e^{-0.1 \ z} \end{aligned}

Thus we have the following pdf:

$\displaystyle h(z)=\left\{\begin{matrix}0.1-0.1 \ e^{-0.1 \ z}& \ \ \ \ \ \ 0

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Additional practice problems can be found in:

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# Exam P Practice Problem 15 – Still More Convolution Practice

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Problem 15A
Find the pdf of $W=X+Y$ where $X$ and $Y$ are independent random variables such that $X$ is uniformly distributed on the interval $0 and $Y$ is exponentially distributed with mean 10.

Problem 15B
Find the pdf of $W=X+Y$ where $X$ and $Y$ are independent random variables such that the pdf of $X$ is $f(x)=2x$ on the interval $0 and $Y$ is exponentially distributed with mean 5.

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Problem 15A is explained in Continuous Convolution – A Guided Example

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# Exam P Practice Problem 14 – More Convolution Practice

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Problem 14A
Suppose $X$ and $Y$ are independent random variables such that $X$ is exponentially distributed with mean 10 and $Y$ has the following density function:

$\displaystyle g(y)=e^{-0.2 \ y}-e^{-0.25 \ y}$ where $y>0$

Find the pdf of $W=X+Y$.

Problem 14B
Suppose $X$ and $Y$ are independent random variables such that $X$ is exponentially distributed with mean 20 and $Y$ has the following density function:

$\displaystyle g(y)=\frac{1}{3} \ e^{-0.1 \ y}-e^{-0.2 \ y}+\frac{2}{3} \ e^{-0.25 \ y}$ where $y>0$

Find the pdf of $W=X+Y$.

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The two problems here provide more practice for using convolution method to find the pdf of an independent sum. An actual Exam P question may not be this straightforward. But using convolution to find the pdf of a sum is a good skill to have. The problems here can help solidify the thought process behind the method of convolution.

For a discussion of the convolution technique, see the blog post Examples of convolution (continuous case).

For additional practice, see Exam P Practice Problem 13 – Convolution Technique.

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# Exam P Practice Problem 13 – Convolution Technique

Problem 13A
Suppose that $X$ and $Y$ are independent random variables where their probability density functions are $f(x)=0.1 \ e^{-0.1 \ x}$ and $g(y)=0.05 \ e^{-0.05 \ y}$, respectively. Find the pdf of $W$ where $W=X+Y$.

Problem 13B
Suppose that $X$ and $Y$ are independent random variables where their probability density functions are $f(x)=0.2 \ e^{-0.2 \ x}$ and $g(y)=0.25 \ e^{-0.25 \ y}$, respectively. Find the pdf of $W$ where $W=X+Y$.

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These are two practice problems on finding the density function of an independent sum using the convolution approach. For a discussion of the convolution technique, see the blog post Examples of convolution (continuous case).

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# Exam P Practice Problem 11 – Binomial Sums

Problem 11A
Suppose $X$ and $Y$ are independent binomial variables with the following probability functions:

$\displaystyle P(X=k)=\binom{5}{k} \ \biggl[\frac{6}{10}\biggr]^k \ \biggl[\frac{4}{10}\biggr]^{10-k} \text{ where } k=0.1,2,3,4,5$

$\displaystyle P(Y=k)=\binom{5}{k} \ \biggl[\frac{6}{10}\biggr]^k \ \biggl[\frac{4}{10}\biggr]^{10-k} \text{ where } k=0.1,2,3,4,5$

Calculate the following sum:

$\displaystyle P(X=0) \ P(Y=4)+P(X=1) \ P(Y=3)+P(X=2) \ P(Y=2)$

$\displaystyle +P(X=3) \ P(Y=1) + P(X=4) \ P(Y=0)$

Problem 11B
Suppose $X$ and $Y$ are independent binomial variables with the following probability functions:

$\displaystyle P(X=k)=\binom{6}{k} \ \biggl[\frac{1}{2}\biggr]^6 \text{ where } k=0.1,2,3,4,5,6$

$\displaystyle P(Y=k)=\binom{4}{k} \ \biggl[\frac{1}{2}\biggr]^4 \text{ where } k=0.1,2,3,4$

Calculate the following sum:

$\displaystyle P(X=1) \ P(Y=4)+P(X=2) \ P(Y=3)+P(X=3) \ P(Y=2)$

$\displaystyle +P(X=4) \ P(Y=1) + P(X=5) \ P(Y=0)$

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Hint

Performing the calculation as the problem is stated is a trap. Falling for this trap could be costly. Note that the sum of the two random variables is also a binomial distribution and that the required calculation is merely a probability calculation of the sum. To see this, write out the probability function of the independent sum using the convolution method.

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# Examples of convolution (discrete case)

Consider the following problems.

Problem 1
Roll a fair die two times. What is the probability that the sum of the two rolls is 5?

Problem 2
There are two independent multiple choice quizzes where each quiz has 5 questions. Each question on the first quiz has 4 choices and each question on the second quiz has 5 choices. Suppose a student answers the questions in the quizzes by pure guessing. What is the probability that the student obtains 5 correct answers in these two quizzes?

Problem 3
For the two independent quizzes in Problem 2, find the probability function for the total number of correct answers. What is the probability that the student get 6 or more correct answers in the two quizzes?

Problem 4
There are two independent multiple choice quizzes where each quiz has 5 questions. There are 5 choices for all questions in these two quizzes. Suppose the student answers the questions in the quizzes by pure guessing. Find the probability function for the total number of correct answers.

All of the above problems are about the independent sum of discrete random variables. We demonstrate the convolution technique using Problem 2.

The Convolution Formula (Discrete Case)
Let $X$ and $Y$ be independent discrete random variables with probability functions $p_X(x)=P(X=x)$ and $p_Y(y)=P(Y=y)$, respectively. Then the following is the probability function of $Z=X+Y$.

\displaystyle \begin{aligned}p_Z(z)&=\sum \limits_{x+y=z} p_X(x) \ p_Y(y) \\&\text{ } \\&=\sum \limits_{x} p_X(x) \ p_Y(z-x) \\&\text{ } \\&=\sum \limits_{y} p_X(z-y) \ p_Y(y) \end{aligned}

Note that the joint probability function of $X$ and $Y$ is $p_{X,Y}(x,y)=p_X(x) \ p_Y(y)$. The convolution formula says that the probability function of the independent sum $Z=X+Y$ is obtained by summing the joint probabiity over the line $x+y=z$.

Problem 2
Let $X$ be the number of correct guesses in quiz 1 and let $Y$ be the number of correct guesses in quiz 2. The variable $X$ has a binomial distribution with parameters $n=5$ and $p=\frac{1}{4}$. The variable $Y$ has a binomial distribution with parameters $n=5$ and $p=\frac{1}{5}$. Let $Z=X+Y$.

\displaystyle \begin{aligned}p_Z(5)&=P(X=0)P(Y=5)+P(X=1)P(Y=4)+P(X=2)P(Y=3) \\&\text{ } \\&\ \ \ +P(X=3)P(Y=2)+P(X=4)P(Y=1)+P(X=5)P(Y=0) \\&\text{ } \\&=\binom{5}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^5 \ \ \binom{5}{5} \biggl(\frac{1}{5}\biggr)^5 \biggl(\frac{4}{5}\biggr)^0 \\&\ \ \ +\binom{5}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^4 \ \ \binom{5}{4} \biggl(\frac{1}{5}\biggr)^4 \biggl(\frac{4}{5}\biggr)^1 \\&\ \ \ +\binom{5}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^3 \ \ \binom{5}{3} \biggl(\frac{1}{5}\biggr)^3 \biggl(\frac{4}{5}\biggr)^2 \\&\ \ \ +\binom{5}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^2 \ \ \binom{5}{2} \biggl(\frac{1}{5}\biggr)^2 \biggl(\frac{4}{5}\biggr)^3 \\&\ \ \ +\binom{5}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^1 \ \ \binom{5}{1} \biggl(\frac{1}{5}\biggr)^1 \biggl(\frac{4}{5}\biggr)^4 \\&\ \ \ +\binom{5}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^0 \ \ \binom{5}{0} \biggl(\frac{1}{5}\biggr)^0 \biggl(\frac{4}{5}\biggr)^5 \\&\text{ } \\&=\frac{129367}{3200000}=0.0404271875 \end{aligned}

Problem 3
Let $Z=X+Y$ where $X$ and $Y$ are as in Problem 2. The following probabilities are obtained by applying the convolution formula.

$\displaystyle P(Z=0)=\frac{248832}{3200000}=0.07776$

$\displaystyle P(Z=1)=\frac{725760}{3200000}=0.2268$

$\displaystyle P(Z=2)=\frac{950400}{3200000}=0.297$

$\displaystyle P(Z=3)=\frac{735840}{3200000}=0.22995$

$\displaystyle P(Z=4)=\frac{373020}{3200000}=0.11656875$

$\displaystyle P(Z=5)=\frac{129367}{3200000}=0.040427$

$\displaystyle P(Z=6)=\frac{31085}{3200000}=0.009714$

$\displaystyle P(Z=7)=\frac{5110}{3200000}=0.001596875$

$\displaystyle P(Z=8)=\frac{550}{3200000}=0.000171875$

$\displaystyle P(Z=9)=\frac{35}{3200000}=0.0000109375$

$\displaystyle P(Z=10)=\frac{1}{3200000}$

$\displaystyle P(Z \ge 6)=\frac{36781}{3200000}=0.011494$

Problem 4
Let $Z=X+Y$ where both $X$ and $Y$ are binomial with parameter $n=5$ and $p=\frac{1}{5}$. The independent sum $Z$ is binomial with $n=10$ and $p=\frac{1}{5}$. The following probabilities are also exercises for using the convolution formula.

$\displaystyle P(Z=0)=\frac{1048576}{9765625}=0.10737$

$\displaystyle P(Z=1)=\frac{2621440}{9765625}=0.26844$

$\displaystyle P(Z=2)=\frac{2949120}{9765625}=0.30199$

$\displaystyle P(Z=3)=\frac{1966080}{9765625}=0.20133$

$\displaystyle P(Z=4)=\frac{860160}{9765625}=0.08808$

$\displaystyle P(Z=5)=\frac{258048}{9765625}=0.02642$

$\displaystyle P(Z=6)=\frac{53760}{9765625}=0.00551$

$\displaystyle P(Z=7)=\frac{7680}{9765625}=0.000786$

$\displaystyle P(Z=8)=\frac{720}{9765625}=0.0000737$

$\displaystyle P(Z=9)=\frac{40}{9765625}$

$\displaystyle P(Z=10)=\frac{1}{9765625}$