# 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 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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$\copyright \ 2013$

# “General Binomial” distribution

Consider the following problems.

Problem A
Four biased coins are tossed where the probabilities of head are 0.53, 0.55, 0.57, 0.60. What is the probability of obtaining exactly $j$ heads where $j=1,2,3,4$?

Problem B
Three independent insurance policies are randomly selected from a portfolio of insurance policies. Each policyholder has at most 1 claim in a policy period. The probabilities of incurring a claim in a policy period are 0.10, 0.15 and 0.20 for these three policyholders. What are the mean number of claims and the standard deviation of the number of claims among these three policyholders in the next policy period?

The problem at hand is an independent sum of Bernoulli distributions where the probability of success is not constant across the trials. Hence we cannot use the usual binomial distribution. For the lack of a better term, we call this “general binomial” distribution. There does not appear to be any quick route to finding the probability of exactly $j$ successes, except to sum all possible combinations. This is a nice exercise for thinking about independent trials.

Problem A
Let $X$ be the number of heads in the 4 trials. The following shows the calculation. The numbers in bold face are the probabilities of success (head).

\displaystyle \begin{aligned}P(X=0)&=0.47 \times 0.45 \times 0.43 \times 0.40 \\&=0.036378 \end{aligned}

\displaystyle \begin{aligned}P(X=1)&=\bold{0.53} \times 0.45 \times 0.43 \times 0.40 \\&\ \ \ +0.47 \times \bold{0.55} \times 0.43 \times 0.40 \\&\ \ \ + 0.47 \times 0.45 \times \bold{0.57} \times 0.40 \\&\ \ \ + 0.47 \times 0.45 \times 0.43 \times \bold{0.60} \\&=0.188273 \end{aligned}

\displaystyle \begin{aligned}P(X=2)&=\bold{0.53} \times \bold{0.55} \times 0.43 \times 0.40 \\&\ \ \ +\bold{0.53} \times 0.45 \times \bold{0.57} \times 0.40 \\&\ \ \ + \bold{0.53} \times 0.45 \times 0.43 \times \bold{0.60} \\&\ \ \ + 0.47 \times \bold{0.55} \times \bold{0.57} \times 0.40 \\&\ \ \ + 0.47 \times \bold{0.55} \times 0.43 \times \bold{0.60} \\&\ \ \ + 0.47 \times 0.45 \times \bold{0.57} \times \bold{0.60}\\&=0.364013 \end{aligned}

\displaystyle \begin{aligned}P(X=3)&=0.47 \times \bold{0.55} \times \bold{0.57} \times \bold{0.60} \\&\ \ \ +\bold{0.53} \times 0.45 \times \bold{0.57} \times \bold{0.60} \\&\ \ \ + \bold{0.53} \times \bold{0.55} \times 0.43 \times \bold{0.60} \\&\ \ \ + \bold{0.53} \times \bold{0.55} \times \bold{0.57} \times 0.40 \\&=0.3111643 \end{aligned}

\displaystyle \begin{aligned}P(X=4)&=\bold{0.53} \times \bold{0.55} \times \bold{0.57} \times \bold{0.60} \\&=0.099693 \end{aligned}

\displaystyle \begin{aligned}E(X)&=0 \times P(X=0) + 1 \times P(X=1)+ 2 \times P(X=2) + 3 \times P(X=3) \\&=0 \times 0.612 + 1 \times 0.329+ 2 \times 0.056 + 3 \times 0.003 \\&=0.45 \end{aligned}
\displaystyle \begin{aligned}E(X^2)&=0^2 \times P(X=0) + 1^2 \times P(X=1)+ 2^2 \times P(X=2) + 3^2 \times P(X=3) \\&=0 \times 0.612 + 1 \times 0.329+ 4 \times 0.056 + 9 \times 0.003 \\&=0.58 \end{aligned}
\displaystyle \begin{aligned}Var(X)&=E(X^2)-E(X)^2 \\&=0.58-0.45^2 \\&=0.3775 \end{aligned}
$\displaystyle \sqrt{Var(X)}=\sqrt{0.3775}=0.61441$