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I am unsure how to calculate a poisson probablity.

There is a sample size of $n$. Test$_A$ detects $k$ samples that show a characteristic.

I want to create a Test$_B$ to detect the characteristic. I can only take a new set of samples $m$, the original $n$ samples are gone.

How big does sample size $m$ have to be to ensure at least one sample with the characteristic with $99%$ or $99.9%$ probability.

I can count if a characteristic is present or not so I assume poison distribution. I do not know poisson enough to solve this problem, can anyone help?

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There is nothing particularly Poisson about the problem. Since we have no other information, at first assume probability an object "shows a characteristic" is $k/n$. Then the probability none in a sample of $m$ shows the characteristic is $(1-k/n)^m$. – André Nicolas Aug 13 '13 at 16:02
@AndréNicolas Why would that not be poisson? I get 1000 consecutive units - three defective. What is the difference to poisson? – Johannes Aug 13 '13 at 16:15
It is binomial. Poisson is a good approximation. But we don't really need approximation, though with your numbers the approximation would be very good. – André Nicolas Aug 13 '13 at 16:31
up vote 1 down vote accepted

Let $p$ be the probabiity of a defective. Our estimate for $p$ is $\frac{k}{n}$. For clarity we use the numbers in your comment, where it seems that $k=3$ and $n=1000$. We want the sample size $m$ that will give us say a $99\%$ probability of at least one defective.

The binomial way: The probability of at least one defective is $1-(1-p)^m$. We want this to be say $\ge 0.99$. So we want $(1-p)^m \le 0.01$. This inequality can be solved by taking the logarithm of both sides.

So we want $m \ln(1-p) \le \ln(0.01)$, or equivalently, since $\ln(1-p)$ is negative, $m\ge \frac{\ln(0.01)}{\ln(1-p)}$. With $p=\frac{3}{1000}$, we get $m\approx1532.7$.

The Poisson way: We can also use the Poisson approximation to the binomial. Let $\lambda=pm=\frac{3}{1000}m$. The number of defectives in a sample of size $m$ has an approximately Poisson distribution with parameter $\lambda$. We want the probability of no defective to be $\le 0.01$.

The probability of no defective in a Poisson with parameter $\lambda$ is $e^{-\lambda}$. So we want to solve the inequality $$e^{-pm}\le 0.01.$$ Take natural logarithms. We get $-pm \le 0.01$, or equivalently $pm \ge \ln(100)$. Thus we want $m \ge \frac{\ln(100)}{p}$. With $p=\frac{3}{1000}$, that gives $m\approx 1535$.

Remark: The two estimates of the appropriate sample size $m$ are quite close to each other. Each is markedly unreliable. The problem is that the number of defectives in the initial sample of $1000$ is very small. In this kind of situation, the probability of say $2$ failures, or $4$, is quite close to the probability of $3$ failures. But the numbers $2$ out of $1000$, or $4$ out of $1000$, give estimates of $m$ that are quite far from our $1535$.

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