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?
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2 Answers
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An answer without using the inclusion exclusion principle.
answered May 8, 2018 at 15:14
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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 $P(A\cap B\cap C) = 20$.
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$\begingroup$ 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? $\endgroup$ Commented May 8, 2018 at 14:59