### Practice Problem Set 6 – Negative Binomial Distribution

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This post has exercises on negative binomial distributions, reinforcing concepts discussed in
this previous post. There are several versions of the negative binomial distribution. The exercises are to reinforce the thought process on how to use the versions of negative binomial distribution as well as other distributional quantities.

 Practice Problem 6A The annual claim frequency for an insured from a large population of insured individuals is modeled by the following probability function. $\displaystyle P(X=x)=\binom{1.8+x}{x} \ \biggl(\frac{5}{6}\biggr)^{2.8} \ \biggl(\frac{1}{6}\biggr)^x \ \ \ \ \ \ x=0,1,2,3,\cdots$ Determine the following: The percentage of the population of insureds that are expected to have exactly 2 claims during a year. The mean annual claim frequency of a randomly selected insured. The variance of the number of claims in a year for a randomly selected insured.
 Practice Problem 6B The number of claims in a year for an insured from a large group of insureds is modeled by the following model. $\displaystyle P(X=x \lvert \lambda)=\frac{e^{-\lambda} \lambda^x}{x!} \ \ \ \ \ x=0,1,2,3,\cdots$ The parameter $\lambda$ varies from insured to insured. However, it is known that $\lambda$ is modeled by the following density function. $\displaystyle g(\lambda)=62.5 \ \lambda^2 \ e^{-5 \lambda} \ \ \ \ \ \ \lambda>0$ Given that a randomly selected insured has at least one claim, determine the probability that the insured has more than one claim.
 Practice Problem 6C Suppose that the number of accidents per year per driver in a large group of insured drivers follows a Poisson distribution with mean $\lambda$. The parameter $\lambda$ follows a gamma distribution with mean 0.6 and variance 0.24. Determine the probability that a randomly selected driver from this group will have no more than 2 accidents next year.
 Practice Problem 6D Suppose that the random variable $X$ follows a negative binomial distribution such that $P(X=0)=0.2397410$ $P(X=1)=0.1038878$ $P(X=2)=0.0398236$ Determine the mean and variance of $X$.
 Practice Problem 6E Suppose that the random variable $X$ follows a negative binomial distribution with mean 0.36 and variance 1.44. Determine $P(X=3)$.
 Practice Problem 6F A large group of insured drivers is divided into two classes – “good” drivers and “bad”drivers. Seventy five percent of the drivers are considered “good” drivers and the remaining 25% are considered “bad”drivers. The number of claims in a year for a “good” driver is modeled by a negative binomial distribution with mean 0.5 and variance 0.625. On the other hand, the number of claims in a year for a “bad” driver is modeled by a negative binomial distribution with mean 2 and variance 4. For a randomly selected driver from this large group, determine the probability that the driver will have 3 claims in the next year.
 Practice Problem 6G The number of losses in a year for one insurance policy is the random variable $X$ where $X=0,1,2,\cdots$. The random variable $X$ is modeled by a geometric distribution with mean 0.4 and variance 0.56. What is the probability that the total number of losses in a year for three randomly selected insurance policies is 2 or 3?
 Practice Problem 6H The random variable $X$ follows a negative binomial distribution. The following gives further information. $E(X)=3$ $\displaystyle P(X=0)=\frac{4}{25}$ $\displaystyle P(X=1)=\frac{24}{125}$ Determine $P(X=2)$ and $P(X=3)$.
 Practice Problem 6I Coin 1 is an unbiased coin, i.e. when tossing the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when tossing 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.
 Practice Problem 6J In a production process, the probability of manufacturing a defective rear view mirror for a car is 0.075. Assume that the quality status of any rear view mirror produced in this process is independent of the status of any other rear view mirror. A quality control inspector is to examine rear view mirrors one at a time to obtain three defective mirrors. Determine the probability that the third defective mirror is the 10th mirror examined.

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6A
• 8.8696%
• 0.56
• 0.672
6B $\displaystyle \frac{171}{546}$
6C 0.9548
6D mean = 0.65, variance = 0.975
6E 0.016963696
6F 0.04661
6G $\displaystyle \frac{31000}{117649}$
6H
• $\displaystyle P(X=2)=\frac{108}{625}$
• $\displaystyle P(X=3)=\frac{432}{3125}$
6I 0.329543
6J 0.008799914

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