Why P(A|B') not P(A)-P(A n B)?

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SUMMARY

The discussion clarifies the distinction between conditional probability P(A|B') and the expression P(A) - P(A ∩ B). It establishes that P(A|B') = P(A ∩ B) / P(B') is valid only when B' is considered, excluding events in A ∩ B from the sample space. The participants emphasize that the probability space for P(A|B') does not include events from A ∩ B, which is crucial for accurate probability calculations. Misinterpretation arises when applying Venn diagrams without recognizing the different probability spaces involved.

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CAH
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I know the equations that P(A|B') = P(AnB) / P(B')

But why isn't it P(A) - P(A n B)

See photo attachment

We know b didn't happen so isn't it just A minus the middle?
 

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P(A)=P(A ∩ B )+P(A ∩ B')
P(A|B')=P(A ∩ B')/P(B')

Your version will hold only of P(B')=1.
 
Specifically, P(A | B^{C}) := P(A) - P(A \cap B) would still include B in the sample space. Given that B hasn't happen, you don't want B in the sample space.
 
CAH said:
I know the equations that P(A|B') = P(AnB) / P(B')

But why isn't it P(A) - P(A n B)

In an expression for probability P(S), there are more things involved that the set S. The expression for a probability involves (perhaps implicitly) a particular "probability space". The expressions P(A|B') and P(A \cap B') both refer to the same set. However, they refer to different probability spaces. In the probability space for P(A|B') no events in A \cap B exist. In the probability space for A \cap B' , events in A \cap B may exist and may be assigned nonzero probabilities.

Your reasoning with the Venn diagram doesn't include the information about what sets are in the two different probability spaces.
 

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