0
$\begingroup$

I ran across the following set identity in a probability problem: $$(E_1 E_2 E_3)^c = E_1 \cup (E_1 E_2^c) \cup (E_1 E_2 E_3^c)$$ and can't make heads or tails of it. I tried applying De Morgan's law as follows: $$ (E_1 E_2 E_3)^c = E_1^c \cup E_2^c \cup E_3^c$$ but don't see how this helps. I am not necessarily looking for a proof of the identity, but rather a way to understand heuristically why it is true, although a proof would probably help with that.

$\endgroup$
1
$\begingroup$

A consequcne of your identity is $$(E_1\cap E_2 \cap E_3)^c \subseteq E_1;$$ do you see a problem?

$\endgroup$
  • 1
    $\begingroup$ Yes, I see the contradiction here. Thank you, the solution I was looking at came from a less than completely reputable source and I won't be as trusting the next time. $\endgroup$ – Is12Prime Mar 14 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.