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$\newcommand{\P}{\mathbb{P}}$Trying to figure out the probability of the following $\P(A|B,C)$ while knowing that $B$ and $C$ are independent.

I know $\P(A|B) = \P(A \cap B)/\P(B)$.

I saw this link: How can I calculate the conditional probability of several events?

But I can't figure out how to do $\P(A \cap B \cap C)$ because $B$ and $C$ are independent. Any help is greatly appreciated.

Context (simplified): Trying to find the probability that Bob will buy product $X$ (event $A$) given a few independent factors, for example: Probability that Bob wants it enough to buy it (event $B$) and probability that product $X$ works (event $C$).

So if there is a high probability that he wants it enough to buy it, $\P(A|B)$ aka the probability Bob buys product $X$ given he wants it bad enough is high. Same goes with $\P(A|C)$. Now I want to see how both independent events $B$ and $C$ affect the probability that Bob buys product $X$.

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user3073431
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    if this is HW please consider adding the self-study tag – Antoine Oct 10 '15 at 10:07
  • Have you tried to use the definition $$P(A|B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}$$ – Xi'an Oct 10 '15 at 12:32
  • It's not hw :/ But I dont think that definition would work because B and C are independent.. right? or would the denominator just be P(B)*P(C)? – user3073431 Oct 10 '15 at 14:39
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    In general, the independence of $B$ and $C$ does not help in finding the conditional probability of an arbitrary event $A$ given that $B\cap C$ occurred because it may well be that $P(A\cap B\cap C) = 0$ even though $P(A\cap B) > 0$ and $P(A\cap C) > 0$, that is, $P(A\mid B)>0$ and $P(A\mid C)>0$ but $P(A\mid B\cap C)=0$. For example, suppose that $A$ is a subset of $B^c\cap C^c = (B\cap C)^c$, say $A=(B\cap C^c)\cup(B^c\cap C)$. Then, $P(A\cap B). P(A\cap C) >0$ but $P(A\cap B\cap C) = P(\emptyset)=0$. – Dilip Sarwate Oct 10 '15 at 16:03
  • For your proof, you'd like to perform steps like $\mathbb{P}(B,C|A)=\mathbb{P}(B|A)\mathbb{P}(C|A)$ ... but sadly that would be wrong because it is not true in general that $B\perp!!!\perp C \implies B\perp!!!\perp C | A$ (where $\perp!!!\perp$ means "is independent of"). See wikipedia's rules of conditional independence. – Creosote Oct 10 '15 at 17:35

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