Bacteria Posson Distribution Problem I'm having some difficulty two questions and was wondering if you could help me out. It goes something like this: a kind of bacteria is distributed in water according to a PPP (Poisson Point Process). It is known that the expected number of bacteria is 2 per liter. Samples of water are provided in bottles with volume 20 cc.
If two bottles have altogether 10 bacteria, what is the probability that the one of them contains less than 3 bacteria?
The second part of the question goes like this:
Find the probability that a bottle with more than 2 bacteria in it has exactly 5 bacteria.
Thanks in advance!
 A: For the first question, forget about Poisson processes. All you need to know about the Poisson process is that the positions of different bacteria are independent. It follows that the distribution conditioned on two bottles containing $10$ bacteria is the same as if you flipped $10$ coins independently to place the $10$ bacteria. Thus the probability that one of them contains less than $3$ bacteria is
$$
2\cdot2^{-10}\left(\binom{10}0+\binom{10}1+\binom{10}2\right)=\frac7{64}\;,
$$
where the factor $2$ accounts for the fact that either bottle can be the one with less than $3$ bacteria.
For the second question, the cases of $0$ and $1$ bacteria are excluded. The expected number of bacteria per bottle is $0.4$, so the number of bacteria per bottle is distributed according to a Poisson distribution with parameter $\lambda=0.4$. Then the probabilities for $0$ and $1$ bacteria are $0.4^0\mathrm e^{-0.4}/0!=\mathrm e^{-0.4}\approx0.67$ and $0.4^1\mathrm e^{-0.4}/1!=0.4\mathrm e^{-0.4}\approx0.27$, so the excluded event has a probability of $1.4\mathrm e^{-0.4}\approx0.94$. That raises the probability of there being $5$ bacteria to
$$
\frac{0.4^5\mathrm e^{-0.4}/5!}{1-1.4\mathrm e^{-0.4}}\approx\frac{0.0000572}{0.0616}\approx0.00093\;.
$$
