Need help with conditional probability with multiple trials.

Question

So lets say, a if a tester gives me value of 40, then there is a 30% chance the machine is faulty. If the testers gives me a value of 20, there is a 20% chance the machine is faulty.

I perform two trials with the tester. The first time, it gives me 40, the second time it gives me 20. What is the probability that the machine is faulty...

So intuitively, without involving any complexities. I'd say just take the average of the two probabilities and say there is a 25% chance...if that's right, please let me know the working principles behind my intuition

Answer 1

With two independent events where one event happens $given$ another, you will need to multiply the probabilities of the two. Therefore you need to multiply $.2$ * $.3$ which gives you a result of $.06$, or a $6$% percent chance that the machine is faulty.

Answer 2

We are given $\mathsf P(F\mid T_1)=0.3, \mathsf P(F\mid T_2)=0.2$, the probability of the machine being faulty when given the results of two independent trials (we presume independence comes from testing independent samples).

We wish to ascertain $\mathsf P(F\mid T_1,T_2)$ the probability of the machine being faulty when given the results of both trials.

We can not.

$$\begin{align}\mathsf P(F\mid T_1, T_2)~&=~\dfrac{\mathsf P(T_1,T_2\mid F)~\mathsf P(F)}{\mathsf P(T_1,T_2)}&&\text{By Bayes' Rule }\\[1ex] &=~\dfrac{\mathsf P(T_1,T_2\mid F)~\mathsf P(F)}{\mathsf P(T_1)~\mathsf P(T_2)}&&\text{By independence}\end{align}$$

And... we can not go any further.   We don't have any of the three marginal probability, and independence of the tests does not imply independence of the test results when given a faulty device.