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The question was to prove that $P(E_1 \cup E_2^C |E_3 \cap E_4) = P(E_1 \cup E_2^C)$ when $E_1,E_2,E_3,E_4$ are independent events. What I know and can prove is that $P(E_1 \cap E_2) = P(E_1) P(E_2)$, $P(E_1 \cap E_2^c) = P(E_1) P(E_2^c)$ and $P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{P(E_1)P(E_2)}{P(E_2)} = P(E_1)$

From here, my attempt was that at first I used De Morgan's law to transform $P(E_1 \cup E_2^C) = P(E_1^c \cap E_2)^c = (P(E_1^c)P(E_2))^c$

From here I am kind of lost how to proceed.

Alternatively,

$P(E_1 \cup E_2^C |E_3 \cap E_4) = \frac{P(E_1 \cup E_2^C \cap E_3 \cap E_4)}{P(E_3 \cap E_4)} = \frac{P(E_1 \cup E_2^C)P(E_3)P(E_4)}{P(E_3)P(E_4)} = P(E_1 \cup E_2^C)$

But I thought this would be too simple.. and also even though it is independent, I don't know whether $P(E_1 \cup E_2^C)$ are independent with $P(E_3 \cap E_4)$. I would appreciate some help on this.

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Hint: $$P((E_1\cup E_2^{c}) \cap E_3\cap E_4)$$ $$=P((E_1\cap E_3 \cap E_4) \cup ( E_2^{c} \cap E_3 \cap E_4))$$ $$ =P(E_1\cap E_3 \cap E_4) + P( E_2^{c} \cap E_3 \cap E_4)-P(E_1\cap E_3 \cap E_4 \cap E_2^{c}).$$ Note that $E_1, E_3,E_4 , E_2^{c}$ are independent too. So all these terms can be computed by applying definition of independence. I hope you can finish the proof using this computation.

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