Is it true that $P[A|B]+P[A|B’]=P[A]$ when B,B’ are exhaustive?

Question

Essentially what the title says. For reference, I am working on the following problem and know that the answer above the solid line is wrong and that the answer below the line is right. However, I am not sure why the top one is wrong snd believe it’s because I made up this formula. Could someone explain why it’s wrong? thank you very much

Answer 1

We'll be looking at $B,B'$ that are both exhaustive and disjoint, which I think is implicit in your scenario.

The probabilities should be $\mathbb{P}(A)=\mathbb{P}(A\cap B)+\mathbb{P}(A\cap B')$, per the diagram you've given. The Venn diagram shows the intersection of the events $A,B,B'$, not conditioned events (which would need some sort of diagram with an order to it: we first see if $B$ or $B'$ occurs and then if $A$ occurs).

Another way of writing this is $\mathbb{P}(A)=\mathbb{P}(B)\mathbb{P}(A|B)+\mathbb{P}(B')\mathbb{P}(A|B')$.

This is the law of total probability.

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For a simple counterexample to your rule, flip a fair coin and let $B$ be the event that it's heads, $B'$ the event that it's tails and $A$ be the event that it lands (i.e. $\mathbb{P}(A)=1$). You'll get a probability of $2$ from your formula, which is impossible.