# How to find Pr(A,B,C)? (Probability of Intersection of three circles)

Can someone please help to find the intersection of three circles given following conditions: $$n|A\cup B\cup C|=750$$ $$|A|=412$$ $$|B|=321$$ $$|C|=338$$ $$|A\cap B|=121$$ $$|A\cap C|=115$$ $$|B\cap C|=105$$ All three events are basically three intersecting circles. So far I got this: $P(A∩B∩C)=P(A)P(B│A)P(C│A,B)$
Calculating it one at a time I come to:
$$P(A)=219/750=0.2920$$ $$P(B│A)=P(A∩B)/P(A) =121/412=0.2937$$ $$P(C│A,B)=P(C│A)∩P(C│B)=P(C∩A)/P(A) ∩P(C∩B)/(P(B))$$ And this is where I'm stuck.
Using Chain Rule I have this expression: $P(A∩B∩C)=P(C│A∩B)*P(B│A)*P(A)$
But I can't see how to calculate $P(C│A,B)$. Can some one please explain?

An answer without using the inclusion exclusion principle.

Assuming that $n=|P(A\cup B\cup C)|$ (since you didn't specify), apply the Inclusion-Exclusion Principle:

$$750=(412+321+338)-(121+115+105)+P(A\cap B\cap C)$$ It follows that $P(A\cap B\cap C) = 20$.

• Oh... this makes sense. Just checked the wording of the question, and this seems to be the case. Thank you. I've updated my question. But then when would I use the chain rule, like I tried? Commented May 8, 2018 at 14:59