# 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. $\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 8 \bold 4$ $\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 5$ $\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 0 \bold 8$ $\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 2 \bold 7$ $\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 0$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

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# 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? $\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 9}{\bold 2 \bold 5 \bold 5}$ $\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 2}{\bold 4 \bold 3}$ $\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 4}{\bold 8 \bold 5}$ $\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 5}{\bold 2 \bold 5 \bold 5}$ $\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1}{\bold 2}$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

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