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Set theory and baye's theorem problem

  • Thread starter bdh2991
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  • #1
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A computer consulting firm presently has bids out on three projects. Let Ai = {awarded project i}, for i = 1, 2, 3, and suppose that P(A1) = 0.22, P(A2) = 0.25, P(A3) = 0.29, P(A1 ∩ A2) = 0.07, P(A1 ∩ A3) = 0.09, P(A2 ∩ A3) = 0.05, P(A1 ∩ A2 ∩ A3) = 0.02.

The question is to find the probability of P(A2 U A3 | A1)

I found that A2 U A3 is equal to 0.4675 but i'm confused on which equation to use to get the rest of the answer...

I originally thought you would just do P((A2 U A3) [itex]\bigcap[/itex] A1) / P(A1) but i'm not getting the correct answer...any help?
 

Answers and Replies

  • #2
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There are 8 cases, you can calculate the probabilities of all of them (or at least the 4 with A1 in it) and then just calculate P(A2 U A3 | A1) based on all cases with A1.
 
  • #3
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Try this:

P(A2 U A3|A1) = P(A2|A1) + P(A3|A1) - P(A2,A3|A1)

and you can switch around each term using Bayes theorem to get

= [ P(A1,A2) + P(A1,A3) - P(A1,A2,A3) ] / P(A1)

if I am not mistaken

edit: Oh, I just realised that is what you said you tried. Looks like it should work to me...
 
  • #4
haruspex
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I originally thought you would just do P((A2 U A3) [itex]\bigcap[/itex] A1) / P(A1) but i'm not getting the correct answer...any help?
What answer did you get?
 

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