# Combining two success runs in parallel

The goal is to calculate the reliability of a process. Here reliability is defined as follows:

Definitions and tests I used

Let $$X$$ be a random variable that is equal to $$1$$ when no defect is present on a fabricated piece and is equal to $$0$$ when a defect is present. Then, the reliability $$R$$ of the process is the probability that $$X = 1$$: $$R = \mathbb{P}(X = 1)$$.

To compute $$R$$, I use the Bayesian success run theorem, it says that the $$(1-\alpha)$$-lower bound $$r$$ for $$R$$ after having test $$N$$ pieces in a row without having detected any defaults is given by:

$$r = \alpha^\frac{1}{n+1}.$$

The problem

Two tests in parallel are carried out for the same process on different samples: test A and test B to obtain respectively the reliability $$R_A$$ and $$R_B$$. Test A respectively test B consists in observing $$N_A$$ respectively $$N_B$$ pieces in a row and then applying the success run theorem:

$$R_A = \alpha^\frac{1}{N_A+1},$$

and

$$R_B = \alpha^\frac{1}{N_B+1}.$$

The question Since these two tests are runned on different samples coming from the same process, can I combined the sample size to obtain the following results for the overall reliability $$R$$ of the process:

$$R = \alpha^\frac{1}{N_A + N_B+1}.$$ Does it make sense to do it?

The Bayesian success run theorem is based on the assumption that your observations are independent Bernoulli with parameter $$p$$, where $$p$$ is the probability that a piece is defect-free. Our prior distribution for $$p$$ is a $$\mathrm{Unif}(0,1)$$, which is meant to convey that we have no prior information about the value of $$p$$.
In this model, examining a sample of $$N_A$$ pieces then another sample of $$N_B$$ pieces and finding them all defect-free is the same as examining a sample of $$N_A+N_B$$ pieces. The lower bound for the reliability is, thus, exactly as you stated it.
The key is that (conditional on $$p$$) these are independent Bernoulli trials.