If a condition is true for sample, is it possible to calculate the probably that it is true for the whole population I have spent some time Googling this, but I suspect I am using the wrong terminology, so please excuse me if this has been answered many times before.
Example: A computer system is designed to transform data records from system 1 to system 2 via a complex process that may have bugs. A human verifies the integrity of the records for a random sample of those that have been transferred.
If the human checks a random sample of records that have been transferred and finds that all these randomly chosen records are free of problems, what is the probably that this is true for all records?
 A: To handle this you may need some kind of Bayesian analysis, starting with a prior view of the distribution of the number of possible errors in the transfer of $N$ records, then combining this with the observation that $n$ selected at random without replacement do not have any errors, to get a posterior distribution for the number of remaining errors.
If $\mathbb P(X=x)$ is your prior probability of $x$ errors in total, and none are observed in the sample, then the posterior probability of the number of remaining errors also being $0$ would be $$\mathbb P(X=0 \mid \text{no errors in sample of }n)=\frac{\mathbb P(X=0) {N \choose n}}{\sum\limits_{x=0}^{N-n} \mathbb P(X=x) {N-y \choose n}}$$
Even then it might not be as helpful as you might hope, since the result is sensitive to your prior. Here are two examples of different prior distributions with the same prior mean:

*

*Suppose your prior probability that the total number of errors is $x$ is uniform, i.e. $\frac{1}{N+1}$ for $x \in \{0,1,2,3,\ldots,N\}$. Then having observed no errors in the sample of $n$, the posterior probability of no errors at all is $\frac{n+1}{N+1}$


*Suppose your prior probability that the total number of errors is $x$ is binomial with parameter $\frac12$, i.e. ${N \choose x}\frac{1}{2^{N}}$. Then having observed no errors in the sample of $n$, the posterior probability of no errors at all is $\frac{1}{2^{N-n}}$
These can be very different: suppose you transfer $10$ records and check $5$ of them as error-free. Then the first approach gives you a posterior probability of no errors at all of $\frac{6}{11} \approx 0.545$  while the second gives a posterior probability of no errors at all of $\frac{1}{2^{5}}=0.03125$
