Conditional Probability $P(A \cap B \cap C | ( A\cap B) \cup (A \cap C) \cup (B \cap C))$

Question

If $P(A|B \cap C)=0.6 \ \ P(B|A \cap C)=.3 \ \ P(C | A \cap B)=.5$.

Find $P(A \cap B \cap C | ( A\cap B) \cup (A \cap C) \cup (B \cap C))$

My first idea is the definition of conditional probability:

$ P(A \cap B \cap C | ( A\cap B) \cup (A \cap C) \cup (B \cap C)) = \frac{ P ((A \cap B \cap C) \cap \left( ( A\cap B) \cup (A \cap C) \cup (B \cap C) \right) ) }{ P( ( A\cap B) \cup (A \cap C) \cup (B \cap C) ) } $

We have that $(A \cap B \cap C) \cap \left( ( A\cap B) \cup (A \cap C) \cup (B \cap C) \right) ) = A \cap B \cap C$

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

Now we need to use the hypothesis:

$P(A|B \cap C)=\frac{ P(A \cap B \cap C) }{P(B \cap C) } = .6 \implies P(A \cap B \cap C)=.6 \ P(B \cap C)$

$P(B|A \cap C)=\frac{ P(A \cap B \cap C) }{P(A \cap C) } = .3 \implies P(A \cap B \cap C)=.3 \ P(A \cap C)$

$P(C|A \cap B)=\frac{ P(A \cap B \cap C) }{P(A \cap B) } = .9 \implies P(A \cap B \cap C)=.9 \ P(A \cap B)$

And we can compute $P( ( A\cap B) \cup (A \cap C) \cup (B \cap C) $

$P( ( A\cap B) \cup (A \cap C) \cup (B \cap C) = P(A \cap B)+ P(A \cap C)+P(B \cap C)-P((A \cap B) \cap (A \cap C) )-P((A \cap B) \cap (B \cap C ) )-P( (A \cap C) \cap (B \cap C) )+P((A \cap B) \cap (A \cap C ) \cap (B \cap C))$

And now I don't know how to compute the conditional probability.

Answer 1

I find this sort of thing easiest to think about using a Venn diagram:

You’re given the proportions of $A\cap B\cap C$ within $A\cap B$, within $B\cap C$ and within $A\cap C$, and you want to know its proportion within the union of all of these. With $x=\mathsf P(A\cap B\cap C)$, that proportion is

$$\frac x{\frac x{0.6}+\frac x{0.3}+\frac x{0.5}-2x}=\frac 1{\frac 1{0.6}+\frac 1{0.3}+\frac 1{0.5}-2}=0.2\;.$$