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Inclusion/Exclusion Principle for 3 events
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[QUOTE="fishturtle1, post: 6634744, member: 606256"] I have a little trouble following it, but it looks like you've made a mistake in distribution e.g., ##(A \cup B) \cup Z \neq (A \cup Z) \cap (B \cup Z)## in general. We have ##P(Y \cup Z) = P(Y) + P(Z) - P(Y \cap Z)##. We can then apply I/E to ##P(Y)## (i'll leave that to you). We can also use distribution (as you suggested) to get $$Y \cap Z = (A \cup B) \cap Z = (A \cap Z) \cup (B \cap Z)$$ And now we can use I/E on ##P((A \cap Z) \cup (B \cap Z))## (what are the two events?). Putting it all together should give us the identity! [/QUOTE]
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Inclusion/Exclusion Principle for 3 events
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