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Let the three independent events $ A, B,$ and $ C$ be such that $P(A)=P(B)=P(C)= \frac14.$ find $P[(A^*\cap B^*) \cup C].$

My solution starts from using the probability of their complements which is $\frac34$, I do not know how to answer this question. Please help.

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  • $\begingroup$ Hi Jonarie - It's considered polite on this site to share what you've thought about and tried, and to formulate your question as a question rather than seeming like a textbook exercise. $\endgroup$ – Ben Blum-Smith Oct 4 '14 at 3:24
  • $\begingroup$ Hi.. thanks for the reminder $\endgroup$ – Jonarie Ramos Vergara Oct 4 '14 at 3:48
  • $\begingroup$ Also asked at stats.stackexchange.com/q/117821/10259 $\endgroup$ – Joel Reyes Noche Oct 8 '14 at 4:00
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It might help your understanding to break down the problem based upon a few simple rules:

\begin{align} P[(A^*\cap B^*) \cup C] &= P(A^* \cap B^*) + P(C) - P[(A^*\cap B^*) \cap C] \\ &= P(A^*)P(B^*) + P(C) - P(A^* \cap B^*)P(C) \\ &= P(A^*)P(B^*) + P(C) - P(A^*)P(B^*)P(C) \\ &= P(A^*)P(B^*)(1-P(C)) + P(C) \\ &= P(A^*)P(B^*)P(C^*) + P(C) \end{align}

The general idea here being that, if $A$ and $B$ are independent, then:

  1. $P(A \cap B) = P(A)P(B)$
  2. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A \cup B) = P(A) + P(B) - P(A)P(B)$

And of course $P(A^*) = 1-P(A)$.

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