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

[CAH]

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?

 

[mathman]

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.

 

[da_nang]

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.

 

[Stephen Tashi]

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.