# Conditional probability sum not equal to one?

If I have $P(A|E_1) = 0.3$ and $P(A|E_2) = 0.7$, where $E_1$ and $E_2$ are just two events upon which $A$ is conditioned, is it coincidental that their sum is $1$, or must this always be the case, if $E_1$ and $E_2$ are the only possible events that can occur?

Thanks.

• This is purely coincidental. For a trivial counterexample, suppose $A$ is the entire sample space. In that case we have $P(A\mid E_1)=P(A\mid E_2)=1$ and so $P(A\mid E_1)+P(A\mid E_2)=2$ – JMoravitz Jun 28 '18 at 18:57

It can happens, but it is not a rule. On the other hand,

$$P(E_1|A) + P(E_2|A) = 1,$$

provided that $E_1$ and $E_2$ are disjoint events (i.e. $E_1 \cap E_2 = \emptyset$) covering the whole event space (i.e. $E_1 \cup E_2 = \Omega$).

Proof $$P(E_1|A) + P(E_2|A) = \frac{P(E_1 \cap A)}{P(A)} + \frac{P(E_2 \cap A)}{P(A)} =\\ = \frac{1}{P(A)}(P(E_1 \cap A) + P(E_2 \cap A)).$$

Since $E_1 \cap E_2 = \emptyset$, then also $(E_1 \cap A) \cap (E_2 \cap A) = \emptyset$ are disjoint events. Therefore: $$P(E_1 \cap A) + P(E_2 \cap A) = P((E_1 \cap A) \cup (E_2 \cap A)).$$

Notice that:

$$(E_1 \cap A) \cup (E_2 \cap A) = (E_1 \cup E_2) \cap A = A,$$

since $E_1 \cup E_2 = \Omega$.

We conclude that:

$$P(E_1 \cap A) + P(E_2 \cap A) = P(A),$$

and hence:

$$P(E_1|A) + P(E_2|A)=1.$$