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.

  • $\begingroup$ 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

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 $P(A^*) = 1-P(A)$.


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.