I have this problem which I'm struggling with:
We've got 50.000 people insured, each one has to pay \$40 at the start of the year, and if they die the family gets \$5.000. The odds for a death is 0,006. Now I gotta find the odds of the company losing money.
My attempt: model the sum $$ \sum^ {50000}_k {{5000} \choose k} 0,006^k0,994^{50000-k}$$ as a Poisson distribution so I get $e^{-300}\frac{300^x}{x!}$ with $x$ being the bankruptcy number (it's 400), so the odds of the company going broke should be $$1-\sum^{399}_k e^{-300}\frac{300^k}{k!}$$ if I haven't done anything wrong so far. This is where I'm stuck, because can't compute this sum. Any ideas?
First of all, you use the word "odds" but your computation suggests you actually mean "probability." Odds and probability have different meanings and you must not confuse the two. For instance, rolling a fair six-sided die has a $1/6$ probability of showing the outcome $3$, because there is one desired outcome out of six equally likely outcomes in total; but the odds of this outcome is $1/5$, because there is one desired outcome out of five undesired outcomes. If $p$ is the probability of an event, then $p/(1-p)$ is the odds of the same event.
Having said this, I will assume you mean "probability" wherever you used the word "odds" in your question.
Let $n = 50000$ be the size of the insured cohort. The random number of deaths is modeled as a binomial random variable $$X \sim \operatorname{Binomial}(n, p = 0.006).$$ The total premiums paid is $40n$. The total loss random variable is $L = 5000X$. Then the event that the insurer loses money is $5000X > 40n$, or equivalently, $X > 400$; that is to say, if more than $400$ policyholders die, then the total death benefit paid will exceed the total premiums. The probability of such an outcome is
$$\Pr[X > 400] = 1 - \Pr[X \le 400] = 1 - \sum_{x=0}^{400} \binom{n}{x} p^x (1-p)^{n-x}. \tag{1}$$
However, as you observed, this sum is computationally intensive. We therefore employ a normal approximation to the binomial, with continuity correction. First we check that the variance is sufficiently small: $$\operatorname{Var}[X] = np(1-p) = 298.2, \\ \sigma = \sqrt{\operatorname{Var}[X]} \approx 17.2685, \\ 3\sigma \approx 51.8054.$$ Since the binomial mean is $\mu = \operatorname{E}[X] = np = 300$, the normal approximation will be a good fit, since $3\sigma < \mu$.
Thus let $$Y \sim \operatorname{Normal}(\mu = 300, \sigma = 17.2685), \tag{2}$$ and the desired probability (with continuity correction) is approximately $$\Pr[X > 400] \approx \Pr[Y > 400.5] = \Pr\left[\frac{Y - \mu}{\sigma} > \frac{400.5 - 300}{17.2685}\right] = \Pr[Z > 5.81986], \tag{3}$$ where $Z \sim \operatorname{Normal}(0,1)$ is standard normal. This is a very small probability: $$\Pr[Z > 5.81986] = 1 - \Phi(5.81986) \approx 2.94492 \times 10^{-9}, \tag{4}$$ which is about $1$ in $339$ million. This makes intuitive sense, because on average, $300$ policyholders die, and the chance that more than $400$ die means that the insurer would need to see an excess of more than $100$ deaths when the standard deviation of the number of deaths is only about $17$.
Using a computer, the exact probability based on Equation $(1)$ is $1.4768 \times 10^{-8}$.