How to calculate the compound distribution for this actuarial science question? Suppose you work for an insurance company. Your company offers a certain type of insurance policy, which pays the insured party $10,000 in case of a house fire. Suppose that 1,000,000 people have bought this insurance. Further suppose that each of these people, independent of the other policy holders, has a 0.0002 probability of having a house fire in the next year.
This year, your company has \$3,000,000 to cover these policies. If its total payments exceed this amount, they must take out a loan to cover the difference. Let random variable $Y$ be the size of the loan your company takes out at the end of the year.
(a) We can write $Y$ as a function of another random variable $X$, such that $X$ has a distribution. Give the distribution of $X$ (including all relevant parameters), and write $Y$ as a function of $X$.
(b) What is the probability that your company must take out a loan of at least $1,000,000?
 A: Let $W$ be the number of fires. Then the technical distribution of $W$ is Binomial, with parameters $n=10^6$ and $p=2\times 10^{-4}$. The total payments $X$ are $10^4W$. The random variable $X$ is well-approximated by the random variable $X'$ which is normally distributed, mean $10^4np$, that is, $2\times 10^6$, and variance $(10^4)^2np(1-p)$, which for all practical purposes is $2\times 10^{10}$. 
The random variable $Y$ is $0$ if $X\le 3\times 10^6$, and $X-3\times 10^{6}$ if $X\gt 3\times 10^6$.
As to the probability that $Y\ge 10^6$, we want the probability that $X\ge 4\times 10^6$. The mean of $X$ is $2\times 10^6$, and the standard deviation is about $1.4\times 10^5$. The number $4\times 10^6$ is many many standard deviation units away from the mean, so the probability that $X\ge 4\times 10^6$ is virtually $0$. There is no practical reason to give an estimate, since uncertainties in the model would make such an estimate meaningless. 
Remark: I do not work for an insurance company, and find it hard to suppose that I do. 
A: Remark: I used to work for a reinsurance company, does that count?
We can view this as an example of an aggregate loss distribution with variable frequency (number of events) and fixed severity per event (\$10,000). 
The frequency distribution, as André said, is binomial. However, as we are dealing with large numbers of people, we can use the property that the binomial with fixed $np$ converges to the Poisson with $\lambda = np$, so, for frequency, we have a Poisson with $\lambda = 200$. Therefore the expected value of of the insurer's payments is $200\cdot \$10000 = \$2,000,000$. As the severity is fixed, the probability of the insurance company requiring additional capital depends solely on the frequency, and would require more than 300 events. Now, to answer the questions.
(a)
Let $X$ represent the random variable representing the number of events as Poisson(200). So we can write:
$$
P(X = x) = \frac{e^{-200}200^x}{x!}
$$
and we can define $Y$ in terms of $X$ as:
$$
Y = 
\begin{cases}
0 &X \leq 300\\
10000\cdot(X - 300) &X > 300
\end{cases}
$$ 
(b)
The probability that we need to take out a loan of at least \$1,000,000 is the probability that there are more than 300 claims. This would be 1 - P(300 or fewer claims). Technically, this should be calculated as:
$$
1 - \sum_{k=0}^{300}\frac{e^{-200}200^k}{k!}
$$
which is completely unreasonable to calculate manually, so we can use the normal approximation. The mean and variance of the frequency is distribution is 200, as the Poisson's mean and variance are equal. So, we have $\hat{\mu} = 200, \hat{\sigma} = \sqrt{200} \approx 14.142$. Using the continuity correction, we want:
$$
1 - P(X \leq 300.5)\\
1 - P\left(\frac{X-\hat{\mu}}{\hat{\sigma}}\right) \leq \frac{300.5 - 200}{14.142}\\
1 - \Phi(7.11)
$$
Which is 0 for all intents and purposes.
