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I'm trying to express conditional probability of $p(C|A \cup B)$ (probability of $C$, given $A$ or $B$) using $p(C|A)$, $p(C|B)$, and $p(C|A \cap B)$.

Using Venn's diagram I came up with: $$p(C|A \cup B) = p(C|A) + p(C|B) - p(C|A \cap B).$$

Is this correct? How can I prove this more formally?

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  • $\begingroup$ I suspect you want to replace the plus sign with a union $\cup$ symbol. Thr former is an arithmetic operator. The latter is a set operator. $\endgroup$ – Taylor May 23 '17 at 7:10
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    $\begingroup$ @Taylor The textbook I was using (Jaynes, Probability Theory) was using +, but I can go with $\vee$ if that is more common. $\endgroup$ – genericname May 23 '17 at 7:15
  • $\begingroup$ 1. Hint: define $A$ and $B$ as independent coin flips and let $C=AB$. What happens? 2. Can you also show your work with Venn diagrams? $\endgroup$ – Juho Kokkala May 23 '17 at 15:01
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    $\begingroup$ Another textbook using $\plus$ for union (of events) and concatenation for intersection of events is A. Papoulis (1984) Probability, Random Variables and Stochastic Processes. But using the more common/modern(?) notation $A\cup B,~A\cap B$ is more likely to be understood. (@user135519 please 9 note also the difference between $\vee$ and $\cup$) $\endgroup$ – Juho Kokkala May 23 '17 at 15:06
  • $\begingroup$ @JuhoKokkala Thanks for heads up about unions. Spending some more time with this formula I realised that it is definitely wrong, but I am not sure how to proof that $p(C|A\cup B)$ cannot be expressed using only $p(C|A)$, $p(C|B)$, and $p(C|A \cap B)$. $\endgroup$ – genericname May 24 '17 at 10:32
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Sorry, but your formula is wrong. Simplest similar formula is below:

$$p(C|A \cup B) =\frac{p(C(A \cup B))}{p(A \cup B)}=\frac{p(CA) + p(CB) - p(CAB)}{p(A) + p(B) - p(AB)}=$$

$$=\frac{p(C|A)p(A) + p(C|B)p(B) - p(C|AB)p(AB)}{p(A) + p(B) - p(AB)}$$

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