# 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.

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possible duplicate of Random Variables, Probability question –  Arkamis Nov 1 '12 at 21:52
Indeed, this duplicate is so uncanny that I am curious what textbook it came out of. –  Arkamis Nov 1 '12 at 21:54
@Ed: That's not a duplicate, let alone an uncanny one. It asks a different, easier question. It's obvious that it's not a duplicate simply from the fact that it asks only one question and not two. This reminds me of this meta thread. –  joriki Nov 1 '12 at 21:56
@joriki One of them asks the probability that a volume of water contains $> 3$ events, and this asks for $< 3$. They are essentially the same question, especially since the answer to the possible duplicate contains in its entirety the reasoning required to solve both parts of this problem. –  Arkamis Nov 1 '12 at 21:59
@Ed: Unless I'm missing something, that just emphasizes the point that questions should be read more carefully before being marked as duplicates. This one conditions on two bottles together having altogether $10$ bacteria, the other one doesn't. The conditioning makes this one a more interesting and slightly more difficult question. Also, even the first question were the same, it would be wrong to close this as an exact duplicate when it asks a second question; in that case, the answer should refer to the other thread for the first question and go on to answer the second one. –  joriki Nov 1 '12 at 22:19

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
$$\frac{0.4^5\mathrm e^{-0.4}/5!}{1-1.4\mathrm e^{-0.4}}\approx\frac{0.0000572}{0.0616}\approx0.00093\;.$$