Exam P Practice Problem 109 – counting insurance payments

Problem 109-A

Amounts of damages due to auto collision accidents follow a probability distribution whose density function is given by the following.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{8000} \ (400-40x+x^2) &\ \ \ \ \ \ 0 < x < 20 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

When occurred, the collision damages are reimbursed by an insurance coverage subject to a deductible of 4.

Fifteen unrelated auto collision accidents have been reported. Determine the probability that exactly nine or ten of the accidents will be reimbursed by the insurance coverage.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 4 \bold 2$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 6 \bold 3$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 2$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 6 \bold 6$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 8 \bold 9$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

Problem 109-B

Amounts of damages due to auto collision accidents follow a probability distribution whose density function is given by the following.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{4000} \ (400-80x+4 x^2) &\ \ \ \ \ \ 0 < x < 10 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

When occurred, the damages are reimbursed by an insurance coverage subject to a deductible of 2.

Twelve unrelated auto collision accidents have been reported. Determine the probability that exactly six or seven of the accidents will not be reimbursed by the insurance coverage.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 8 \bold 4$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 5$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 0 \bold 8$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 2 \bold 7$

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

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

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

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?

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

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

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

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

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

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

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