Probability of 4 or fewer errors in 100,000 messages The probability of an error occurring in a message is 10^-5. The probability is independent for different messages. There are 100,000 messages sent. What is the probability that 4 or fewer errors occur?
 A: In principle, the number $X$ of errors in $100000$ messages has binomial distribution.  But in this kind of situation (probability $p$ of an "error" small, number $n$ of trials large, $np$ of moderate size) it is standard to approximate the distribution of $X$ by using the Poisson distribution with parameter $\lambda=np$.
In our case we have $\lambda=np=(10^{-5})(100000)=1$. The probability of $4$ or fewer errors is approximately 
$$\sum_{k=0}^4 e^{-1} \frac{1^k}{k!}.$$ 
A: It’s nothing tricky, just a slightly messy calculation. Calculate separately the probability that $0,1,2,3$, and $4$ of the messages contain errors. For instance, the probability that exactly $2$ messages contain errors is computed as follows. Let $p=10^{-5}$.


*

*The probability that a particular $2$ messages, like the third and fiftieth, contain errors and the other $99,998$ don’t is $p^2(1-p)^{99,998}$: you just multiply the probabilities of the desired outcomes of each of the $100,000$ messages, which are $p$ for $2$ of them and $1-p$ for the rest.

*There are $\binom{100,000}2$ pairs of messages.
The total probability of getting exactly $2$ messages with errors is therefore
$$\binom{100,000}2p^2(1-p)^{99,998}\;.$$
Make a similar calculation for each of the other three cases, and add the results.
You can get a rough check on your work by noticing that when $k$ is small,
$$\binom{100,000}kp^k=\frac{10^5(10^5-1)(10^5-2)\ldots(10^5-k+1)}{4!10^{5k}}\approx\frac1{k!}\;.$$
