I couldn't fit all of the problem in the title, but here it is in full:

Events A and B are independent, events A and C are mutually exclusive, and events B and C are independent.

If P(A) = $\frac{1}{2}$; P(B) = $\frac{1}{4}$; P(C) = $\frac{1}{8}$; what is P($A \cup B \cup C$)?

I'm a little lost on how to approach this - I understand the difference between mutually exclusive and independent events (being that mutually exclusive events both cannot happen at once, and independent events do not affect one another) but I don't entirely understand what formula I should be using to figure out how to solve the problem.


Note that the event $A\cup B\cup C$ can happen in the following two disjoint ways: (i) $B$ holds or (ii) $B$ fails and one of $A$ or $C$ holds.

The probability of (i) is $1/4$.

For probability of (ii) we need to work somewhat harder. Since $A$ and $C$ are mutually exclusive, we want $\Pr(B'\cap A)+\Pr(B'\cap C)$. Here $B'$ denotes the complement of $B$. The required probabilities can be found by independence.

It remains to put the pieces together.

  • $\begingroup$ Thanks! This made things a bit easier, I was attempting to draw venn diagrams of the problem and it wasn't helping $\endgroup$ – secondubly Sep 22 '15 at 0:00
  • $\begingroup$ You are welcome. A Venn diagram that breaks things up into all the basic ways the event can happen, like $A\cap B\cap C'$ and so on can be used, I tried to do the calculation reasonably efficiently. But the analysis came from a Venn Diagram visualization, done in the head because I can't draw and type at the same time. $\endgroup$ – André Nicolas Sep 22 '15 at 0:08
  • $\begingroup$ Here is the visualization. Two ovals, nicely separated, representing $A$ and $C$. A nice fat circle in the middle, representing $B$ and overlapping with $A$ and with $C$. I want $B$ plus the two outer ears. $\endgroup$ – André Nicolas Sep 22 '15 at 0:16

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.