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I am trying to understand law of total probability in the context of conditional probability.

Lets say we are trying to calculate $P(A)$, but it may be easier to calculate $P(A \cap B_i)$ given $B_i$ are all disjointed. Law of total probability gives us, $P(A) = \sum_i P(A \cap B_i) = \sum_i P(A | B_i) P(B_i)$

Now, lets say we are trying to calculate $P(A|C)$, but it may be easier to calculate $P(A|C \cap B_i)$ given $B_i$ are all disjointed, so we start similarly, $P(A|C) = \sum_i P(A \cap C \cap B_i)$

Am I correct so far?

Now, I am having trouble breaking down the term $\sum_i P(A \cap C \cap B_i)$. Assuming I have to use Bayes theorem, I am trying to figure out what some event X and Y are,

$P(X) = \frac{\sum_i P(A \cap C \cap B_i)}{P(Y)} ... 0$

I understand that there are many X and Y event that makes the above equation true.

From the Law of total probability wiki page, it says, $P(A|C) = \sum_i P(A|C \cap B_i) P(B_i | C)$, which means their X and Y events are,

$P(A|C \cap B_i) = \frac{\sum_i P(A \cap C \cap B_i)}{P(B_i | C)} ... 1$

I am under the assumption that in Bayes theorem the event we are conditioning on goes to the denominator, so $P(A|{\color{red}B}) = \frac{P(A\cap B)}{P({\color{red}B})}$, which means the above express should be,

$P(A|{\color{red}C} \mathbin{\color{red}\cap} {\color{red}B_i}) = \frac{\sum_i P(A \cap C \cap B_i)}{P({\color{red}C} \mathbin{\color{red}\cap} {\color{red}B_i})} ... 2$

As you can see the denominator in equation 1 and 2 are very different. I am trying to reconcile $\sum_i P(A \cap C \cap B_i) = \sum_i P(A|C \cap B_i) P(B_i | C)$ to what I know.

Is both 1 and 2 are true? As I've mentioned there are many X and Y that makes equation 0 true. Trying to understand the rational behind it.

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Disjoint $B_i$ is not enough. They also need to form up the sample space. So, I'm assuming that's the case. Also, $P(A|C) = \sum_i P(A \cap C \cap B_i)$ is not correct. It's actually the following:

$$P(A|C) = \sum_i P(A \cap B_i|C)=\sum_i P(A|B_i,C)P(B_i|C)$$

The following is not correct as well.

$$P(A|C \cap B_i) = \frac{\sum_i P(A \cap C \cap B_i)}{P(B_i | C)} ... 1$$

It should be $$P(A|C\cap B_i)=\frac{P(A\cap C\cap B_i)}{P(B_i\cap C)}$$

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  • $\begingroup$ Thank you. I have a basic understand of Bayes theorem where the numerator is both events happening together divided by the probability of the event we are conditioning on. Can you please explain in plain English why this follows the same logic: $P(A|B_i \cap C) = \frac{P(A \cap B_i|C)}{P(B_i|C)}$ I don't understand why we have $P(A \cap B_i|C)$ as numerator. Also why $P(B_i|C)$ is the denominator when the event we are conditioning on it $B_i \cap C$. Can you please relate all these back to $P(A|B) = \frac{P(A \cap B)}{P(B)}$ $\endgroup$
    – Quazi Irfan
    Commented Feb 9, 2021 at 21:08
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    $\begingroup$ You can derive that using the last formula I wrote. Numerator is $P(A\cap B_i|C)P(C)$ and denominator is $P(B_i|C)P(C)$, and $P(C)$'s cancel. Intuitively, you can think of it as $C$ is always given, so it should appear at the given side of the Bayes formula at all times. $\endgroup$
    – gunes
    Commented Feb 9, 2021 at 21:11
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    $\begingroup$ It's again the same logic. The simpler equation is $P(A)=\sum_i P(A\cap B_i)$, and in the conditional version of this, you just add $C$ to each expression's given side. Or, mathematically, $$P(A|C)=\frac{P(A\cap C)}{P(C)}=\frac{\sum P(A\cap C\cap B_i)}{P(C)}=\frac{\sum P(A\cap B_i|C)P(C)}{P(C)}=\sum P(A\cap B_i|C)$$ $\endgroup$
    – gunes
    Commented Feb 9, 2021 at 21:36
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    $\begingroup$ Disjoint B_i is not enough. They also need to form up the sample space. To clarify, $B_i$ needs to form up the space of A only. Am I correct? $\endgroup$
    – Quazi Irfan
    Commented Sep 19, 2021 at 21:45
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    $\begingroup$ It might be misleading to say it that way, e.g. A can be result of a coin toss, B can be result of a dice throw. But, when you define the sample space as pairs of (coin toss, dice thrown), e.g. (T, 1), (T, 2) ..., (H, 5), (H, 6), B1, B2, ... B6 can be dice thrown = 1 (i.e. (T,1), (H,1)), dice thrown = 2 (i.e. (T,2), (H,2)) etc. So, B covers the whole sample space. Of course, if any of the subsets (e.g. Bi) doesn't intersect with A, then it's enough to have cover the part of the sample space covered by A since the other terms will be 0 because of the intersection term, P(A ∩ Bi). $\endgroup$
    – gunes
    Commented Sep 22, 2021 at 19:30

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