Conditional probability rules with set $P(A \cup B\mid C) = P(A\mid C) + P(B\mid C) - P(A\cap B\mid C)$

Question

I am learning about conditional properties for probability. And I am wondering about one particular rule for three events A, B, C where the probability of event C Is greater than 0.

Can someone explain this rule perhaps via a Venn diagram?

I tried to draw it but the LHS does not seem to match RHS.

Thanks

Answer 1

$P(A\cup B\mid C) = \dfrac{P((A\cup B)\cap C)}{P(C)} = \dfrac{P((A\cap C)\cup (B\cap C))}{P(C)}$

Now, apply inclusion-exclusion to the numerator using the events $A\cap C$ and $B\cap C$ (instead of how they normally appear in inclusion exclusion as $A$ and $B$)

$=\dfrac{P(A\cap C)+P(B\cap C)-P(A\cap C\cap B\cap C)}{P(C)}$

This all separates and simplifies as $=P(A\mid C)+P(B\mid C)-P(A\cap B\mid C)$

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If you insist on using venn diagrams, then due to the fact that we are using conditional probabilities conditioned on $C$, you should ignore everything outside of $C$.

The ratio of area that is blue compared to the size of the circle for $C$ here, $P(A\cup B\mid C)$:

Is equal to the sum of the blue areas here, $P(A\mid C)$:

and here, $P(B\mid C)$:

Except that by adding those areas we added the bit in the middle, $P(A\cap B\mid C)$, too much so to correct that we subtract by that amount again after the fact to fix.

Answer 2

$A \cap C$ includes a copy of $A \cap B \cap C$. $B \cap C$ also includes a copy of $A \cap B \cap C$. To not double count the probability contribution from $A \cap B \cap C$, you must subtract one of the two copies of contributions from $A \cap B \cap C$ from $P(A \mid C) + P(B \mid C)$.