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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?

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

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  • $\begingroup$ @InterstellarProbe: Oh, my bad! It should've been intersection instead. I'll edit the answer. $\endgroup$ – Shirish Kulhari Jun 28 '18 at 18:13
  • $\begingroup$ Thank you, that looks much better. $\endgroup$ – InterstellarProbe Jun 28 '18 at 18:24
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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 $1$.

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  • $\begingroup$ @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 $P(A)$. $\endgroup$ – BallBoy Jun 28 '18 at 18:14
  • $\begingroup$ @YForman You are correct. My mistake. I deleted my comment. $\endgroup$ – InterstellarProbe Jun 28 '18 at 18:20

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