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Thread: Need help with conditional probability with multiple trials.

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    Need help with conditional probability with multiple trials.




    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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    Re: Need help with conditional probability with multiple trials.

    Are there only two values that the tester can give? If so, I believe you are correct. What I'm thinking of is the law of total probability.

    Actually, I might be wrong. Didn't take into account the two trials.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: Need help with conditional probability with multiple trials.


    I suspect that the question is not well defined, but, because the title mentioned conditional probability here is a conditional probability approach which may satisfy.
    To calculate conditional probability you select all the situations which are possible and add them. Then you take the probability of the thing you want and divide it by the total. If it the machine is bad there is a 30%x20% chance of getting 40 and 20 (= 6%), and if it is good, a 70%x80% ( =56%) chance of it being good. So the conditional probability of being faulty should be 6%/(6%+56%).
    However, like Buckeye, I am by no means convinced that this answer is right, or even that the question makes perfect sense. It is the sort of question I might have given to a student to illustrate conditional probability without really thinking through the details.
    kat

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