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.
Best Answer
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.