I need some help to calculate the following joint probability:
Let's consider a binary event $A=\{0,1\}$ with $P(A=1)=0.75$ and a set of finite events $B_i$, $i=1,\dots,n$, mutually independent. Let's also consider known the probabilities $P(A|B_i)=\alpha_i$ and $P(B_i)=\beta_i$, $i=1,\dots,n$. How can I derive the calculation of $P\big[A|\bigcap_{i=1}^n B_i\big]$?
Thank you in advance!
Well, the problem is that while you are promised that the set $(B_i)_{i\in[[1..n]]}$ is of mutually independent events, that does not guarantee their conditional independence given $A$.
So Bayes' Theorem takes us to:
$$\begin{align}\textstyle\mathsf P(A=a\mid\bigcap_{i=1}^n B_i)~&=~\dfrac{\mathsf P(\bigcap_{i=1}^n B_i\mid A)~\mathsf P(A=a)}{\mathsf P(\bigcap_{i=1}^n B_i)}&&\text{Bayes' Theorem}\\[2ex]&=~\dfrac{\mathsf P(\bigcap_{i=1}^n B_i\mid A)~\mathsf P(A=a)}{\prod_{i=1}^n\mathsf P(B_i)}&&\text{Mutual Independence}\\[2ex]&=\dfrac{\mathsf P(\bigcap_{i=1}^n B_i\mid A)~0.75^a0.25^{1-a}}{\prod_{i=1}^n \beta_i}\,\mathbf 1_{a\in\{0,1\}}\end{align}$$
However, because you cannot automatically claim $\mathsf P(\bigcap_{i=1}^n B_i\mid A)$ equals $\prod_{i=1}^n\mathsf P(B_i\mid A)$, you may not go any further. So we maynot express this in terms of $\alpha_i$ .