Imagine that I have a region which is 1 meter in length. Let's say that I randomly pick 10 of the 1,000 millimeters (non-contiguous) within this region to count as "successes". If I drop 1 pebble (of negligible size) within the full region, there is a 1% chance that the pebble ends up within a "success" region. This means that I average 0.01 successes per pebble-drop. Now let's say that I drop 20 pebbles simultaneously. I'm just getting acquainted with basic concepts of probability, but I believe this means that I average 0.2 successes each time I drop 20 pebbles at once (a success is when a single pebble ends up within a "success" region).
Now I'm trying to use the poisson formula to calculate the probability that I will achieve 5 successes upon dropping 20 pebbles.
$\mu = 0.2$
$e = 2.718$
$x = 5$
$\frac{e^{-mu} \cdot \mu^x}{x!} = 2.18 \cdot 10^{-6}$
Have I thought about this correctly, or is there a flaw in my attempted application of the poisson probability formula in this situation?
Thanks very much.
The random number of successes is actually binomially distributed, namely $$X \sim \operatorname{Binomial}(n = 20, p = 0.01),$$ with PMF $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x} = \binom{20}{x} (0.01)^x (0.99)^{20-x}, \quad x \in \{0, 1, \ldots 20\}.$$ Thus $$\Pr[X = 5] = \binom{20}{5}(0.01)^5(0.99)^{15} \approx 1.3334344730358537290 \times 10^{-6}.$$ This allows for the possibility that more than one pebble could land in the same success region.
The reason why it is not exactly Poisson is because there are a fixed number of pebbles, thus your successes cannot exceed $20$. But since $p$ is small, the binomial probability can be approximated by a Poisson distribution with rate $\lambda = np = (20)(0.01) = 0.2$ and PMF $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}.$$ Thus $$\Pr[X = 5] = e^{-0.2} \frac{(0.2)^5}{5!} \approx 2.18328 \times 10^{-6},$$ as you computed. But while both results are very small, the Poisson approximation overestimates the true probability by a factor of about $1.6$.
In what kind of scenario would we use a Poisson distribution? One example is if instead of dropping a fixed number of pebbles, you drop some number of pebbles at some constant rate, and you want to compute the probability that there will be some number of successes in a given period of time. Specifically, we might say a machine drops pebbles randomly in such a way that on average, it releases $20$ pebbles a minute. Then the probability you will observe exactly $5$ successes in a one-minute period of time is $2.18328 \times 10^{-6}$, as we computed above. If you watch the pebbles drop over a $11$ minute period, the probability you observe exactly $5$ successes becomes $$\Pr[X = 5] = e^{-2.2} \frac{(2.2)^5}{5!} \approx 0.0475866.$$ This is because the rate parameter representing the mean number of successes in an $11$-minute period is $\lambda = 11(20)(0.01) = 2.2.$ $11$ minute observation window, average of $20$ pebbles dropped per minute, and $0.01$ probability of observing success for a given pebble. But note that in this situation, nowhere do we say that exactly $20$ pebbles are dropped in one minute, or that $220$ pebbles are dropped in $11$. These are average rates. The machine can drop fewer or more.