Imagine two experiments:
1) A group of $n=100$ individuals are under condition $A$, and they perform some test. Among these individuals, $y=5$ individuals succeed in the test.
2) A group of $m=1000$ individuals are under condition $\Delta$ and they perform some test with two possible issues: the first possible is that an individual falls under the above condition $A$, the second possible issue is another condition $B$ (obviously $B = \neg A$). Among these individuals, $x=2$ individuals fall under condition $A$.
Now consider an individual under condition $\Delta$ who perform experiment 2) and then experiment 1) in case when he falls under condition $A$. We are interested in the probability for this individual to succeed in the second test.
Of course this probability is naturally estimated by $x/m \times y/n$. But how to assess the uncertainty about this estimate ?
I have in mind a valuable Bayesian solution: using a (possibly noninformative) prior distribution on the first probability of success $\theta_1$ and another (possibly noninformative) prior distribution on the second probability of success $\theta_2$ then $(x,m)$ yields a posterior distribution for $\theta_1$ and $(y,n)$ yields a posterior distribution on $\theta_2$, and finally we get a posterior distribution on the probability of interest $\theta_1\theta_2$ (assuming independent posterior distributions of $\theta_1$ and $\theta_2$).
Do you know/imagine another solution ?
EDIT : Maybe the following frequentist solution is valuable : estimating the asymptotic variance of $\hat\theta_1$ and the asymptotic variance of $\hat\theta_2$ then we get an asymptotic variance of $\hat\theta_1 \hat\theta_2$ by multiplying the two variances. But, in fact, my real problem is a bit more complicated : there is a third experiment and I am interested in the estimation of the probability of two possible issues at the last stage.
EDIT2: I was tired when writing the previous "EDIT". More correctly I was having in mind the Delta-method to derive the asymptotic behaviour of $\hat\theta_1 \hat\theta_2 - \theta_1\theta_2$ and then an asymptotic confidence interval.