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Given this joint probability distribution table

     |   A   |   B   |   C  
X    | 0.15  | 0.03  | 0.07
notX | 0.05  | 0.25  | 0.45

I've to calc the following probabilities

P(B or X) = ?
P(C or notX) = ?

So far I got this solution:

P(B or X) = 0.15 + 0.03 + 0.07 + 0.25 
P(C or notX) = 0.07 + 0.45 + 0.05 + 0.25 

but I'm very unsure...

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    $\begingroup$ This looks fine to me, though this isn't so much a question as it is procedural confirmation. $\endgroup$ – Jonathan Thiele Jun 13 '12 at 17:31
  • $\begingroup$ ok, sorry for that! $\endgroup$ – alepfu Jun 13 '12 at 17:41
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    $\begingroup$ I agree with Jonathan. What you did to compute P(B or X) was to take P(X and A)+P(X and B) + P(X and C) to get P(X). Then you need to count all the probabilities associated with B occurring but not X ( you do this to avoid double counting). That is the term with the value 0.25. You use the same idea for the second probability. $\endgroup$ – Michael Chernick Jun 13 '12 at 18:00
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By addition theorem of probability,

P(A U B) = P(A) + P(B) – P(A ⋂ B)

U denotes union of events, and ⋂ denotes intersection of events.

In other words,

P(A or B) = P(A) + P(B) – P(A and B)

So, your calculations are correct. Here's the proof

P(B or X) = P(B) + P(X) - P(B and X)

          = (0.03 + 0.25) + (0.15 + 0.03 + 0.07) - 0.03

          = 0.15 + 0.03 + 0.07 + 0.25

Similarly,

P(C or not X) = 0.07 + 0.45 + 0.05 + 0.25 
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