I'm trying to find the conditional probability of with 4 separate events. For example, I want to find:

P(A| B,C,D) and also all of the different variations of this i.e [P(B|A,C,D) etc.].

I've started out by finding the union of B, C, D (U) through p(B) + p(C) + p(D) - p(BC) - p(BD) - p(CD) + p(BC*D).

But I'm confused on what I should do next. Could someone please explain to me how I find:

P(A|U) = P(A) * P(U|A) / P(U) ==> I think I'm confused on the P(U|A) part.

  • 1
    $\begingroup$ What is the meaning of $P(A|B,C,D)$? $\endgroup$
    – Xi'an
    Commented Jan 5, 2021 at 17:37

1 Answer 1


Depends what you mean by "B,C,D"; if you mean B AND C AND D, that's an intersection (union corresponds to OR).

Supposing you mean B AND C AND D, the basic Bayesian formula applies as follows:

$$ P(A|(B\cap C\cap D))=\frac{P(A\cap B\cap C\cap D)}{P(B\cap C\cap D)} $$

It is true that this also equals

$$ \frac{P(A)P((B\cap C\cap D)|A)}{P(B\cap C\cap D)} $$

Whether this is a useful equality depends on which of the various terms you have values for.


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