Poisson distribution and limit of detection (LoD)
Without any sampling error, a CFU of 1 per assay volume would be the limit of detection
Assay volume would be the total volume of the undiluted sample counted
For example, if you plate out 0.1 mL of a 100-fold diluted sample on three replicate plates, the total volume counted is 3 * 0.1/100 mL = 0.003 mL
Poisson distribution: P(k) = m^k/k! * exp(-m)
Where m is the mean CFU/assay volume and k is the number of actual events (counts)
Probability of not detecting CFUs:				P(k=0) = exp(-m)
Probability of detecting CFUs:				P(k>0) = 1 - P(k=0) = 1 - exp(-m)
m	P(k=0)	P(k>0)	cum.sum of P(k=0)	Weight = P(k=0)/cum.sum(0,10)	Contribution = Weight (probability true count is X if count was 0) * X	cum.sum of contribution (0,X)
0	1.00000	0.00000	1.00000	0.6321	0.00000	0.00000
1	0.36788	0.63212	1.36788	0.2325	0.23255	0.23255
2	0.13534	0.86466	1.50321	0.0855	0.17110	0.40365
3	0.04979	0.95021	1.55300	0.0315	0.09442	0.49806
4	0.01832	0.98168	1.57132	0.0116	0.04631	0.54437
5	0.00674	0.99326	1.57806	0.0043	0.02130	0.56567
6	0.00248	0.99752	1.58053	0.0016	0.00940	0.57507
7	0.00091	0.99909	1.58145	0.0006	0.00404	0.57911
8	0.00034	0.99966	1.58178	0.0002	0.00170	0.58080
9	0.00012	0.99988	1.58190	0.0001	0.00070	0.58151
10	0.00005	0.99995	1.58195	0.0000	0.00029	0.58179
If there are 3 CFUs in the assay volume, detection is highly likely (>0.95), so we can regard 3 CFU as limit of detection (LoD)
If nothing was detected, the true count could be anywhere between 0 when detection is impossible and up to 10 when detection is almost certain.
But the probabilities of non detection, given by P(k=0), are higher if the count is lower.
We could use these probabilities P(k=0), normalized by the cumulative sum of probabilities up to 10, to calculate "weights" giving the relative probability of the true count being X when the count was 0.
We then have to multiply these "weights" of X with X and integrate over X from 0 to 10 to get the weighted contribution of each value of X to the true count if the actual count was 0.
The result is about 0.6, which is just a bit higher than the midpoint between 0 and 1, which has been used to substitute counts of 0.
Conclusion: replace counts of 0 with 0.6 CFU/mL counted
Notes: cum.sum = cumulative sum

Poisson distribution and limit of quantification (LoQ)
The limit of quantification depends on how much of a Coefficient of Variation (CV) is regarded as acceptable
Due to property of the Poisson distribution that SD = sqrt(m) where m is the mean, the relative SD or CV = SD/m = 1/sqrt(m)
m	CV
0	0
1	1.000
2	0.707
3	0.577
4	0.5
5	0.447
10	0.316
20	0.224	Common rule is that the count should be at least 20
30	0.183	Also common rule is that the count should be at least 30, the difference in CV is not large
50	0.141
100	0.1
200	0.071
500	0.045
1000	0.032
