# Sum of conditional probabilities equals 1?

Assuming that sum of probabilities for all possible events that can occur should sum to 1, how does one denote this for a conditional probability? Is it $P(A|E_1) + P(A|E_2) + ... = 1$, where $E_i$ is a specific event to be conditioned on? Or is the answer something else entirely?

• The first sentence assumption is incorrect, even before you get to the conditional probability question. Take $\Omega = \{heads, tails\}$ with all outcomes equally likely, with all possible events being $\mathcal{F}=\{\phi, \{heads\}, \{tails\}, \{heads, tails\}\}$ and $$P[\phi] + P[\{heads\}]+P[\{tails\}]+P[\{heads,tails\}]=0 + 1/2 + 1/2 + 1 = 2$$ Commented Feb 17 at 20:48

This is a good example where intuition on conditional probability can be a good check of the formula. Suppose $P(A)=0$ (which is clearly possible); then we know intuitively that $P(A|E_i)=0$ for any $i$ (because event $A$ never happens, so how can it happen conditioned on something?), so the sum you gave will be $0$, not $1$.

But if the $E_i$ are a partition of the sample space, we do have the formula $$P(E_1|A) + P(E_2|A) + \dots + P(E_n|A) = 1$$ with an intuitive explanation. Given that $A$ occurred, it's still true that one of the $E_i$ must occur, so the total probability of the $E_i$ occurring given $A$ must still be $1$.

• @InterstellarProbe Thank you for your comment. I believe the formula is correct as I have stated it. If it were $P(E_i \cap A)$ instead of conditional probabilities, you would be correct. For a simple example, $A$ and $A^c$ (complement of $A$) partition the sample space; $P(A|A)+P(A^c|A)=1+0$ sums to $1$, not $P(A)$. Commented Jun 28, 2018 at 18:14
• @YForman You are correct. My mistake. I deleted my comment. Commented Jun 28, 2018 at 18:20

If $E_1, E_2, \ldots$ is a collection of disjoint events whose union equals the entire sample space (exhaustive), then $P(A\cap E_1)+P(A\cap E_2)+\ldots = P(A|E_1)P(E_1)+P(A|E_2)P(E_2)+\ldots = P(A)$.

• @InterstellarProbe: Oh, my bad! It should've been intersection instead. I'll edit the answer. Commented Jun 28, 2018 at 18:13
• Thank you, that looks much better. Commented Jun 28, 2018 at 18:24

The previous answers are more than enough to understand what is going on.

I just want to state the general proposition (implicit in the answers) with a formal proof.

Proposition: Let $$E_1,...,E_n,...$$ be a countable collection of sets such that $$E_i\cap E_j=\emptyset$$ for all $$i\neq j$$ whose total union $$\bigcup_{i=1}^\infty E_i$$ is the whole sample space. If $$p(A)>0$$, then $$\sum_{i=1}^\infty p(E_i|A)=1.$$ *Remark: If $$E_i\neq\emptyset$$ for finitely many $$i$$'s, then the sum has finitely many positive terms.

Proof. Since $$\{E_i\}_i$$ is a countable collection of pairwise disjoint sets, then so is $$\{E_i\cap A\}_i$$. By the axiom of countable additivity, $$\sum_{i=1}^\infty p(E_i\cap A)=p\left(\bigcup_{i=1}^\infty (E_i\cap A)\right)=p(A).$$

By definition of conditional probability, $$p(E_i\cap A)=p(E_i|A)\cdot p(A)$$ for each $$i$$. Hence \begin{align*} \sum_{i=1}^\infty p(E_i|A)&=\sum_{i=1}^\infty \frac{p(E_i\cap A)}{p(A)}\\ &=\frac{1}{p(A)}\sum_{i=1}^\infty p(E_i\cap A)\\ &=1.\,_\blacksquare \end{align*}

One extra proof to complement the rest (the same but with a slightly different notation):

$$\sum_{i=0}^n P(E_{i}|A) = \sum_{i=0}^n(\frac{P(A|E_{i})P(E_{i})}{P(A)}) = \frac{1}{P(A)}\sum_{i=0}^n(P(A|E_{i})P(E_{i})) = \frac{P(A)}{P(A)} = 1$$

The third step holds thanks to the product rule in probability.