# Determining Type of Probability Distribution

Suppose that pollen spores are randomly scattered in a home, at a density of $8$ sports per cubic cm. What is the probability of finding at least two spores in a space of $0.2$ cubic cm?

Here is how I approached it:

Let's call the number of spores per cm$^3$ $n$. Then we are given $\mu=8$. Finding $2$ spores in an area of 0.2 cm$^3$ is equivalent to finding 10 spores in an area of $1$ $cm^3.$ So we want to find $\Pr(n\geq10),$ or $1 - \Pr(n \leq 9)$.

I'm confused about what sort of distribution this would be. Poisson?

And am I expected to then compute that distribution's values for values $n=0, 1, ..., 9?$

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"Finding 2 spores in an area of 0.2 cm^3 is equivalent to finding 10 spores in an area of 1 cm^3"... Nope. – Did Jul 18 '13 at 15:56
@Did why not? If there is a block of cheese with 10 things, and it's cut up into 5 equal pieces, each piece (should have??) 2 things, if the things inside are "randomly dispersed". Is that incorrect? – Moog Jul 18 '13 at 16:19
Yes it is. See my answer. – Did Jul 18 '13 at 16:47
@Moog: the expected number of spores is $1.6$ spores ($8$ spores/cc times $0.2$ cc). In a Poisson process, the probability of $k$ spores is $$\frac{1.6^k}{k!}e^{-1.6}$$ Note that if you sum the probabilities for $k\ge0$, you get $1$. What Did does in his answer (+1) is to subtract off the probabilities for $0$ and $1$ spores to get the probability of at least $2$ spores. – robjohn Jul 18 '13 at 21:40

The usual modelization of this kind of real-life (?) situation is that the spores are spread according to a Poisson process. The intensity $\varrho$ of the process corresponds to the mean number of spores per volume unit. Then the number of spores in any volume $v$ is Poisson with parameter $\lambda=v\varrho$ hence the probability to find at least $2$ spores in the volume $v$ is $p=P(X_\lambda\geqslant2)$ where $X_\lambda$ is Poisson of parameter $\lambda$, that is, $p=1-(1+\lambda)\mathrm e^{-\lambda}$.

Here, $\varrho=8\,\mathrm{cm}^{-3}$ and $v=0.2\,\mathrm{cm}^{3}$ hence $v\varrho=1.6$ and $p=1-2.6\,\mathrm e^{-1.6}=47.5\%$ (and the parameter $\lambda=v\varrho$ is unitless, as it should).

Note that asking to find at least $2$ spores in a volume $v$ or at least $20$ spores in a volume $10v$ are not equivalent. The second happens with probability $q=1-\mathrm e^{-16}\sum\limits_{n=0}^{19}(16)^n/n!$, hence $q\ne p$ (and additional considerations allow to show without computations that $q\lt p$).

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I think it would be poisson with $\mu = 8$.

$1 \ \text{cm}^{3}$ has a volume of $1$. 0.2 cm ^3 has volume 0.008. So $2/.008 = 250$.

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No, no, it's $0.2 cm^3$, not $(0.2 cm)^3$ – Moog Jul 18 '13 at 16:21