# Health Insurance question

An insurance company issued health insurance policies to individuals. The company determined that Y, the number of claims filed by an insured in a year, is a random variable with the following probability function.
$$P(Y=y)=0.45(0.55)^{y},\;\;\; y=0,1,2,3,\ldots$$ The number of claims filed by one insured individual is independent of the number of claims filed by any other insured individual.

An actuary studied three randomly selected insured individuals from this group of individuals who purchased health policies from this company. What is the probability that these three insured individuals will file more than 6 claims in a year?

My thinking was $$1-P(Y=0)-P(Y=1)-P(Y=2)-\dotsb-P(Y=6)$$ and than raising to the third power but this doesn't give me the answer. What am I missing?

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It sounds to me as though the question is asking for the probability that the three people file a total of six claims in a year. Have you already tried calculating that (and seeing if that gives the answer)? –  dmk Feb 23 '14 at 1:02
Also, I just want to make sure: Is that p.d.f. correct? According to that definition, $P(Y = 0) = 0$; usually they set aside the values whose probability is $0$. (And anyway, if you take out that $y$, it looks sorta like a geometric distribution...) –  dmk Feb 23 '14 at 1:16
@dmk it is a geometric distribution with support 0,1,2,3... There should be a space between the $^y$ and the $y$. I'll edit... –  TooTone Feb 23 '14 at 1:18
sorry the $y$ is to indicate what the values of $y$ can be and is not part of the p.d.f –  adam Feb 23 '14 at 1:18
@dmk are you going to answer the question? (I started typing s.thing out but am happy to wait) –  TooTone Feb 23 '14 at 1:19

## 3 Answers

Here's another way; it's the opposite of elegant, but if brute force'll do it, why not try?

If the question is, indeed, asking for the probability that the three customers make a total of more than six claims, you can find

$$1 - P(\text{They file at most six claims)} = 1 - P(Y_1 = a)P(Y_2 = b)P(Y_3 = c)$$

where $Y_i$ has p.d.f. $f(y) = 0.45(0.55)^y$ and $0 \leq a + b + c \leq 6$.

At first glance, this would seem enormously tedious. There are a lot of possible strings for $a, b, c$, and order matters. On further inspection, though, it turns out to be only sorta kinda tedious. Consider two possibilities: Let $A$ be the event in which $Y_1 = 4, Y_2 = 1, Y_3 = 1$, and $B$ the event in which $Y_1 = 0, Y_2 = 6, Y_3 = 0$. Then

\begin{aligned} \ P(A) &= P(Y_1 = 4)P(Y_2 = 1)P(Y_3 = 1) \\ \ &= (0.45)(0.55)^4(0.45)(0.55)(0.45)(0.55) \\ \ &= (0.45)^3(0.55)^6 \\ \end{aligned}

and

\begin{aligned} \ P(B) &= P(Y_1 = 0)P(Y_2 = 6)P(Y_3 = 0) \\ \ &= (0.45)(0.55)^0(0.45)(0.55)^6(0.45)(0.55)^0 \\ \ &= (0.45)^3(0.55)^6 \\ \end{aligned}

It's not too much of a stretch to see that any string $a, b, c$ such that $a + b + c = 6$ has the same probability. How many such strings are there? For $a + b + c = 6$, I get $28$. The number of strings summing to $5$ (or any number), of course, will be different, but the observation that each event corresponding to a given sum has the same probability will still hold. Therefore, you should be able to get something of the form

$$1 - (0.45)^3\left[\phi\cdot(0.55)^6 + \epsilon\cdot(0.55)^5 + \delta\cdot(0.55)^4 + \gamma\cdot(0.55)^3 + \beta\cdot(0.55)^2 + \alpha\cdot(0.55) + 1 \right]$$

where $\phi$ is the number of strings adding to $6$, $\epsilon$ the number of strings adding to $5$, etc. (Of course, there's only one way for no one to make a claim.) Using this method I get an answer of $0.1495$.

This is not the best way to do it — it's been a while since I've seen a negative binomial distribution — but it should show you that you can do stuff with very limited information. (And that you can probably do that stuff in something like ten minutes.)

If someone could comment on the combinatorial aspect of this, I'd be grateful. That is, where does $28$ come from? I have a suspicion it's $\binom{6 + 3 - 1}{3 - 1}$ — and if we simplify, the triangular numbers come in throughout my solution — but regardless of how much informal study I've done (probably not that much), I've never gotten comfortable with the twelvefold way.

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My thinking was $1-P(Y=0)-P(Y=1)-P(Y=2)...P(Y=6)$ and than raising to the third power but this doesn't give me the answer. What am I missing?

What you calculated is the probability that $P(Y_1 > 6 \cap Y_2 > 6 \cap Y_3 > 6)$, i.e. the probability that all of them filed more than 6 claims.

What the question probably wants is the probability that between them they filed more than 6 claims. Note that you have a geometric distribution with pmf $(1-p)^yp$ and cdf $1 - (1-p)^y$ with $p=0.45$, i.e. $Y_i\sim G_0(p)$ (this would have saved you looking at individual probabilities in your question because you could have used the known cdf.)

For a random variable $Y=Y_1 + Y_2 + Y_3$ that is the sum of independent geometric random variables you can use the negative binomial distribution. I.e., $Y\sim \mathrm{NB}(3,p)$.

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Answer:

Edit: If indeed the solution looks for total claims made by a randomly selected three people with n=3, then the following procedure will get you to a closer answer. Sorry for the initial misunderstanding.

The probability of a claim from an insured looks like geometric distribution with p =0.45 and q = 0.55

$$E(Y) = \frac{(1-p)}{p} = \frac{(1-0.45)}{.45} = 1.222$$

Standard Deviation = $$\sqrt{\frac{q}{p^{2}}} = \sqrt{\frac{.55}{.45^{2}}} = 1.648$$

Using normal approximation with continuity correction, $Mean = 3*1.222 =$3.667

Standard Deviation = $1.648*sqrt(3) = 2.8544$

$$P(Y>6) = 1-P(Y<=6) = 1-P(z<\frac{(6.5-3.667)}{2.8544})=1- 0.83952316 = 0.16047684$$

Satish

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First, I've already butchered this one once tonight, so maybe I should just quit now. But second, I'm not sure you should use the central limit theorem when $n=3$. Third, if you are going to use it, then you want to subtract $n\mu = 20/3$ in the numerator, not $\mu = 20/9$, and divide by $\sqrt{n}\sigma$. Finally, to approximate a discrete variable (again, usually with much larger $n$), you take $P(X \leq 6) = P(X \leq 6.5)$ and then standardize. Going through these steps (and again, I don't think any of them apply in this case), I get $0.5233$ for the probability. –  dmk Feb 23 '14 at 3:11
Normal Approximation to the Negative Binomial Distribution Suppose now that Yk has the negative binomial distribution with trial parameter k and success parameter p. Then $$Z_k=(pY_{k}−k)/\sqrt{(k(1−p)}$$ In the negative binomial experiment, vary k and p and note the shape of the probability density function. With k=5 and p=0.4, run the experiment 1000 times and note the agreement between the empirical density function and the true probability density function. –  satish ramanathan Feb 23 '14 at 3:47
I agree with most of your points and have corrected my solution to take care of it and has posted some information that I researched which is not new to you I suppose. Here in this example k = 5 and in our case, it is k = 3. You will see the closeness of our solution. Sorry for relying on the mind than to refer the notes. –  satish ramanathan Feb 23 '14 at 3:50
math.uah.edu/stat/sample/CLT.html. This is what I as referring to. Let me know if I could be of help. –  satish ramanathan Feb 23 '14 at 3:53