Sampling from a liquid containing suspended particles

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If the sample is small (say less than 10% of the volume of the total liquid), and the particles are randomly distributed in the liquid, we can use the Poisson process to model the particles in our sample. If the sample is large, we will need to use the binomial process. The two approaches are discussed below:

Poisson modelling

If l is the concentration of particles in the liquid, and t is the volume of liquid in the sample, then Poisson mathematics gives us the following results:

Probability no particles in the sample: = EXP(-l*t), or =POISSON(0,l*t,0)
Probability at least one particle in the sample: = 1-EXP(-l*t), or =1-POISSON(0,l*t,0)
Simulation of number of particles in the sample: =VosePoisson(l*t)

So, for example:

100 bacteria are randomly distributed in a vat of 1000 liters of wine. If a sample of two liters of wine is taken from the vat, what is the probability that there will be at least one bacterium? What is the distribution of the number of bacteria in that sample?

Answer:

l = 100/1000 = 0.1 bacteria per litre

t = 2 liters

Probability at least one bacterium in sample = 1-EXP(-0.1*2) = 18.1269...%

Number of bacteria in sample = VosePoisson(0.1*2)

The problem with this approach is that the Poisson process potentially allows an infinite number of particles to exist. Once our sample is large compared to the volume of liquid, we could start generating numbers of bacteria greater than are actually in the liquid. For example, if the sample was 800 liters, the above approach would model the number of bacteria as: =VosePoisson(80). A plot of this graph below shows that the distribution exceeds the total number of bacteria (=100).

It might look like the problem will only be important when we get close to the total volume, but a comparison of the binomial and Poisson methods below show that there are significant difference at much smaller samples.

Binomial modelling

We can think of each bacterium as a trial, and that being in the liquid sample is a success. If the bacteria are randomly distributed in the liquid body, then each of the n trials has a probability v/V of being in the sample, where v = the sample volume and V is the volume of the whole liquid body. We now see that this is a binomial process:

n trials = 100 bacteria

Probability of success p = v/V = 2/1000 = 0.2%

Then:

Probability no particles in the sample: = (1-p)n, or =BINOMDIST(0,n,p,0)
Probability at least one particle in the sample: = 1-(1-p)n, or =1-BINOMDIST(0,n,p,0)
Simulation of number of particles in the sample: =VoseBinomial(n,p)

and the answers to the questions above are:

Probability at least one bacterium in sample = 1-(1-0.2%)100 = 18.1433...%

Number of bacteria in sample = VoseBinomial(100,0.2%)

A comparison of a Poisson(0.2) and a Binomial(100,0.2%) shows that the Poisson (skinny light red columns) is a very good approximation because the sample is so small.

But if the sample had been just, say, 30% of the volume the two modelling approaches (Binomial(100,30%) and Poisson(30)) would have already started to give different answers;