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A motor insurance company has sold $150$ insurance policies. Let $N_{i}$ represent the number of claims made on policy $i$. You may assume that $N_1, N_2,..., N_{150}$ is a sequence of independent Poisson random variables, each with mean $0.2$.

(i) Evaluate the probability that at least $2$ claims are made on a given policy.

I got - $0.0175$ (rounded)

(ii) A manager looks through the policies to find one on which two or more claims have been made. Calculate the probability that the manager has to look through at least $50$ policies before finding one.

Now here I put $0.0175$ as $\theta$ into this $\theta(1-\theta)^{49}$.

My answer - $0.0073677$ was not same as the answer booklet says - $0.4205$

What they have done is just consider $(1-\theta)^{49}$. Why have they not put the $\theta$ in front of the formula?

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What do geometric distributions have to do with geometric probability? Please change the tags to probability and possibly add probability-distributions. –  Dilip Sarwate Jan 16 '13 at 22:54

2 Answers 2

up vote 0 down vote accepted

$$P(N_i \ge 2) = \sum_{k=2}^{\infty} \frac{(0.2)^k}{k!} e^{-0.2} = \left ( e^{0.2} - 1 - \frac{0.2}{1!} \right )e^{-0.2} = 1 - (1.2) e^{-0.2} \approx 0.0175 $$

For the second part, the important word is "at least." The probability is

$$ \theta [(1-\theta)^{49} + (1-\theta)^{50} + (1-\theta)^{51} + \ldots] $$

$$ = \theta (1-\theta)^{49} [1 + (1-\theta) + (1-\theta)^2 + \ldots] $$

Using the fact that, for $|r| < 1$, $1 + r + r^2 + r^3 + \ldots = 1/(1-r)$, we get

$$\theta (1-\theta)^{49} [1 + (1-\theta) + (1-\theta)^2 + \ldots] = \theta (1-\theta)^{49} \frac{1}{1-(1-\theta)} = (1-\theta)^{49} $$

I hope this helps you see where the missing $\theta$ went.

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Thanks for the answer. How did you get $\frac{1}{1-(1-\theta)}$ ? –  neverloggedin Jan 16 '13 at 22:45
    
After you factor out the $(1-\theta)^{49}$, you get an infinite geometric series with ratio $1-\theta$, so the sum is $1/(1-(1-\theta))$ from which I was hoping you'd see is equal to $1/\theta$. I can fill in if you want. –  Ron Gordon Jan 16 '13 at 22:48
    
Thank you very much. That helped a lot! –  neverloggedin Jan 16 '13 at 22:55

Call finding a policy with $2$ or more claims a success. We have to go through $50$ or more policies before we get a success if and only if we get $49$ failures in a row.

The probability of failure ($0$ or $1$ claims) is $e^{-0.2}+0.2e^{-0.2}$. To find the probability of $49$ failures in a row, take the $49$-th power. I get about $0.420528$.

Remark: What you obtained was the probability that the first success was obtained on the $50$-th trial. So you found the probability that the manager had to look through exactly $50$ policies. The wording of the question ("at least") strongly indicates that this is not what is asked for.

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