Understanding Conditional Probability with Dependent Variables

  • Thread starter hodor
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In summary: However, it looks like that is not the case. In summary, I'm not sure how to proceed when I'm asked to restrict an example problem.
  • #1
hodor
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Hi,

I've run into a snag trying to read a textbook problem. Here is the original example, it's pretty straightforward. The problem I have is when I get to the exercise and it asks me to place a restriction on this example. This restriction seems to break the independence of two variables and renders all the probabilities in the original example useless to me. For example:

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So, ignoring that it's asking for a code update here, I seem to have P(Butler = murderer, Maid = murderer) = 0 and so on. But it appears I can't recalculate P(K) (knife used) since B and M are no longer independent. So I really don't understand how to proceed. This leads me to believe I'm misinterpreting things so I thought I'd ask here. Thanks.
 
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  • #2
I don't see why you would need that independence. You have P(B, not M) and P(M, not B) and there are no other cases to consider.
 
  • #3
Ok, I had thought since those probabilities had changed, the P(k|B,M) probabilities would also have to change, and without ndependence I wouldn't be able to recalculate. It seems I can set P(K|B=murderer,M=murderer)=0 and reuse the others unchanged, but that still makes me a bit uncomfortable?
 
  • #4
hodor said:
It seems I can set P(K|B=murderer,M=murderer)=0 and reuse the others unchanged, but that still makes me a bit uncomfortable?

No, P(K|B=murderer,M=murderer) is not changed - it is P(B=murderer,M=murderer) that is 0, and P(B=murderer,~M=murderer) etc. are different.
 
  • #5
ok, thanks. I had assumed that since P(B=murderer,M=murderer) had changed, P(K|B=murderer,M=murderer) would change since P(K|B=murderer,M=murderer) = P(K,B=murderer,M=Murderer)/P(B=murderer,M=murderer), and similarly for the other conditionals.
 

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