# Random Variables, Probability question

The question is: A type of algae is distributed in a liquid by the PPP (Poisson Point Process). We know that the number of algae is 2 per liter. Samples of these liquids are provided in containers with volume 20 cc.

I'm trying to answer: Find the probability that a container contains more than 3 bacteria

I have since 2 algae/liter hence 0.04/per 0.02 litres.. shouldn't the probability of a container contain more than 3 algae be 0, since we have a bottle of only 0.02 litres? .. I'm a little lost, could somewone clarify Thanks!

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how many liters would contain 3 algae? After that, you just need to enumerate the possibilities that one of that container will have all those algae. – jay-sun Oct 29 '12 at 21:26
Not $0$, but as you point out, for our small flasks we are dealing with a Poisson with very small mean. So the probability of more than $3$ bacteria (or is it algae?) is very close to $0$. You may not be quoting the problem quite correctly, because of the algae/bacteria shift. – André Nicolas Oct 29 '12 at 21:28

The Poisson distribution is a single-parameter distribution, the one parameter being the mean.

Basically, the general idea behind a Poisson distribution is that for any size area, you have the same process intensity. For example: let's say you're counting the number of people that come into a store. If the average number of people is 60 per hour, and the distribution is Poisson, this is exactly the same thing as saying that one person per minute enters the store, on average.

Therefore, you simply need to scale your average to match your units. The average number of algae is 2 per liter, which turns out to be 2 per 1000 cc, which means you have .004 algae per 20 cc, on average. So, you've done that part right.

Now, you can use your Poisson distribution formula using this process intensity parameter.

Another way of looking at it is this: if you have a 1 liter volume that has 2 algae in it, you can draw a neighborhood around one algae that has a volume of 20 cc; if you draw neighborhoods at random, what is the probability you enclose an algae?

So, to answer your question, you need to compute $P(X > 3) = 1- P(X \le 3) = 1- (P(X=0)+P(X=1)+P(X=2)+P(X=3))$.

Compute $P(X=k)$ using the PMF with $\lambda = 0.04$: $P(X = k) = \frac{\lambda^k}{k!}e^{-\lambda}$

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In other words, the Poisson distribution gives you a global average of events; since this is an average, any local neighborhood can contain more or fewer events. – Arkamis Oct 29 '12 at 21:31
Big X is also equal to 0.04 right? and isnt it e^(-lambda) ? – Thatdude1 Oct 29 '12 at 22:20
Yes to $-\lambda$, I will edit in the omission. No to big X. Big X is your random variable -- the number of algaes in a container. You are computing the probability that $X > 3$, which is equal to one minus the probability that $X \le 3$ (because a contain must contain either less than or equal to three algaes, or more than three algaes). – Arkamis Oct 29 '12 at 23:01
oh okay i see, another question... how would i approach the event where the probability that 3 selected containers have altogether more than 3 algae? ... Do i need to consider all the cases? – Thatdude1 Oct 30 '12 at 0:25
Start by examining all the cases, and see if you can simplify it. – Arkamis Oct 30 '12 at 3:02