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What is the probability that Sam is guilty if Tom and Devi gave conflicting testimonies?

Is it conditional probability? Or intersection simply?

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    $\begingroup$ Is this a homework question? Because if it is, it should be tagged as such. Is there any other context? $\endgroup$ – naught101 Feb 17 '15 at 6:45
  • $\begingroup$ @naught101 I found this while reading through a book online. But i cant figure out what to do. i know the probabilty of conflicting testimonies is 0.3 and P(guilty) = 0.4. $\endgroup$ – Zum Feb 17 '15 at 6:48
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It's a pretty basic probability question, so there is little room for hints. I will give an answer instead.

Try to feel the difference between these two questions:

  1. What is the probability that {Sam is guilty} AND {Tom and Devi gave conflicting testimonies}?
  2. What is the probability that {Sam is guilty} IF {Tom and Devi gave conflicting testimonies}?

You have two events
$A$: {Sam is guilty}
$B$: {Tom and Devi gave conflicting testimonies}

Question 1. asks: what is the probability that both $A$ AND $B$ happened. In other words, $P(A,B)=?$ This is what you would call intersection. No tricks here, just translating from everyday language to probability notation. This step is supposed to be intuitive.

Question 2. asks: what is the probability that $A$ happened IF we know that $B$ happened. I.e. what is the probability of $A$ given $B$. In other words, $P(A|B)=?$ This is what you would call conditional probability.

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  • $\begingroup$ This is of course, assuming independence - that is, that a conflicting testimony is equally as likely if Sam is guilty or if she is innocent. $\endgroup$ – naught101 Feb 17 '15 at 10:47
  • $\begingroup$ I am not sure if I follow your argument. I did not suggest that $P(A,B)=P(A) \cdot P(B)$ or the like. But perhaps I miss something. Could you point which statements in my answer do not hold under dependence? That would be very helpful. $\endgroup$ – Richard Hardy Feb 17 '15 at 11:00
  • $\begingroup$ You're right, I misread your answer. But in the context of the two probabilities given by Zum in the comments on the question, independence is relevant - If you can't assume independence, then you can't calculate an exact probability. $\endgroup$ – naught101 Feb 17 '15 at 11:09
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    $\begingroup$ OK, that makes sense. I did not consider those probabilities in my answer. I was trying to give all the intuition without giving the exact solution (since it is a self-study problem). $\endgroup$ – Richard Hardy Feb 17 '15 at 11:15
  • $\begingroup$ You are welcome! You can accept it (by clicking on the tick mark to the left) and upvote it if you find it useful. $\endgroup$ – Richard Hardy Feb 17 '15 at 13:16

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