Is $P(A | B, C') + P(A | B, C) = P(A | B)$? Why or Why not?

Question

I know this question might be pretty basic for most but I'm stuck. I've tried a few approaches but couldn't get to it. Intuitively the answer seems to be true as I feel that by conditioning on $C$ and $C'$ both, the dependency on $C$ is removed. But I parallelly feel that I'm missing something.

Answer 1

No it is not. If you draw a Venn diagram, things become much clearer.

$$ \begin{align} & P(A|B)=\frac{w+z}{w+z+x+b} \\ & P(A|B,C)=\frac{z}{z+x} \\ & P(A|B,C')=\frac{w}{w+b} \\ \end{align} $$

As you can see, we are comparing very different things. Adding $P(A|B,C)$ and $P(A|B,C')$ does not give us the result that we want. What you were probably in your head thinking of adding together was

$$ \frac{w}{w+z+x+b}+\frac{z}{w+z+x+b} $$

which would indeed add to $P(A|B)$. Which events correspond to these? $P(A, C|B)$ and $P(A, C'|B)$. That makes perfect sense, as these two events cover all of the ways in which $P(A|B)$ can happen, and they are mutually exclusive. Changing the 'given' part of the probability does not work because we are considering cases with more information than usual about $C$. Generally, we don't know whether or not $C$ has happened, hence why the denonimator is $w+z+x+b$ for $P(A|B)$. However, once we are told this information, the denominator changes, and so the probabilities don't add up properly.

My maths teacher had a mantra for working out probabilities that went 'what is my world'? That is, what information have I been given, and what could happened within this constrained world? Think about the $|B,C$ part of the probability for a moment. That looks at a certain probability given $B$ and $C$ have happened. In other words, we take it for granted that $B$ and $C$ have happened. We can also look at probabilities given $B$ and $C'$, but there is no reason to expect that these probabilities will add up nicely. It's like saying 'why doesn't the probability of getting heads when you flip a coin and rolling four sixes in a row not add up to $1$?' Well, we are just looking at different types of events—the games we are playing are not the same, and so there's no reason to assume that the probabilities will add up nicely. Your example is essentially just a more subtle version of the coin/die example: you are comparing two different 'games', one where $B$ and $C$ have already happened, and one where $B$ and $C'$ have already happened. Again, we are not comparing the same thing, and so things don't end up well when we try to add those probabilities together.