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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

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    I don't believe there is enough information to answer the question. We need to know more about the possible outcomes and about the independence. – 2017-01-26
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    okay, assume that the tester reads correctly 80% of the times. Also assume any other missing info that you might need. – 2017-01-26
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    What do you mean "reads correctly"? How can a read of $40$ be "correct" or not? I would want to know all the possible test outcomes and the probability distribution for how a bad machine would score plus the distribution for how a good machine would score. Too much to simply guess at. – 2017-01-26
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    To make clear what I am talking about, suppose the presence of some gene is a weak indicator of a specific defect. That is, a person with the gene has about a $25\%$ chance of having the defect. Your test is the test that looks for this gene. Your scores indicate rather high confidence that the gene is present. Then...based on your test results I can logically conclude that the gene is present and hence that the defect has about a $25\%$ probability. If I run the tests a lot more often I might expect to see the same result and reach the same conclusion. – 2017-01-26

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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.

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    But of course the tests are not independent (as they both depend on the actual health of the machine). To phrase the problem another way, the first score of $40$ by itself means there is a $.7$ chance the machine is good, and the score of $20$ by itself means there is a $.8$ chance the machine is good. Your reasoning would give a $.7\times .8=.56$ chance it was good. But $.56+.06 \neq 1$. – 2017-01-26
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    the trials are independent of each other...i mean each trial is independent of the previous trial.. – 2017-01-26
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    and i mean the way lulu described, if this way we keep on doing trials...the probablities of the machine being good or faulty would both drop to zero. – 2017-01-26
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    @MuhammadAbbas which is clearly absurd, so this analysis is wrong. Look at it differently: suppose $.3$ of the units were defective and my test constantly reported $.3$ without even looking at the machine being tested. Then my test is, in fact, statistically accurate (though purely by stupid accident). running the test a trillion times would yield $.3$ a trillion times and the answer would still just be $.3$. – 2017-01-26
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    We're given $\mathsf P(F\mid T_1)=0.3, \mathsf P(F\mid T_2)=0.2$ and wish to find $\mathsf P(F\mid T_1,T_2) $ . We cannot just multiply the conditionals to get that, and using Bayes' Rule indicates we don't have enough to obtain a solution. – 2017-01-26
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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.