I computed the probability if an event E given an event T. Now I want to compute the probability of E given T happened twice independently? How can I go about this? So I'm basically looking for $$P(E|(T_1, T_2).$$ $$P(T_1, T_2)=P(T_1)P(T_2)$$ since they are independent. Now, $$P(E|(T_1, T_2) = \frac{P((T_1, T_2)|E)P(E)}{P(T_1)P(T_2)}$$ should be what I'm looking for, correct? I know every probability but $$P((T_1, T_2)|E).$$ How do I find that out?

Problem set My solution

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    $\begingroup$ Your basic manipulations are good. There is no way to determine $P[T_1,T_2|E]$ without more information. $\endgroup$ – Michael Oct 29 '19 at 20:58
  • $\begingroup$ I'm quite confused why that's not possible. I mean, I believe you, but then I must have understood something wrong. I think my professor is not very accurate when it comes to asking questions .. I edited my question, you can now see the problem set (this question is referring to number 2) and how I solved the first part. $\endgroup$ – Marc Nov 1 '19 at 11:56

You missed an important word. In the problem statement, we are told that the two tests are conditionally independent. That is to say: $$P(T_1 , T_2 | E) = P(T_1 | E) . P(T_2 | E)$$ and: $$P(T_1 , T_2 | \overline E) = P(T_1 | \overline E) . P(T_2 | \overline E)$$ But this is not the same as being independent. And in fact we have: $$P(T_1 , T_2) \neq P(T_1) . P(T_2)$$

Intuitively, this is because an employee who has passed the first test is probably innocent, so they will probably pass the second test too. While an employee who has failed the first test is probably guilty, and the second test is likely to confirm that. It's unusual to fail one test and pass the other - so the two events can't possibly be independent.

  • $\begingroup$ So if I understand correctly, all I have to do now is square the $P(T|E)$ in the solution provided? $\endgroup$ – Marc Nov 1 '19 at 16:31
  • $\begingroup$ And sorry for confusion. Bayesian statistics is quite new for me. $\endgroup$ – Marc Nov 1 '19 at 16:31
  • $\begingroup$ That's right. That gives you the probability that an employee failed both tests given that they are guilty. Then you use Bayes' Theorem to find the probability that they are guilty given that they failed both tests. $\endgroup$ – Christopher Wells Nov 1 '19 at 17:17

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