Finding the probability that the message will find it's way through the network, why this approach doesn't work. 
The three switches in the figure above operate independently of one another. Switch 1 allows a message to go through with probability of .88, switch 2 allows a message to go through with probability of .92 and switch 3 allows a message to go through with a probability of .90. What is the probability that a message will find it's way through the network. 
This is what I've got, a message can either go the route with switches 1 and 2 or the route with switch 3. If it goes the route of switches 1 and 2 then both must be working for the message to go through. This gives me the probability for the message to go through in the top route as $P(1 \cap 2) = .88 * .92 = .8096$ Then, the probability that a message will find it's way through the network will  be: 
$$P(3\cup(1 \cap 2)) =$$
$$P(3\cup 1)\cap P(3\cup2) =$$
$$(P(3)+P(1)-P(3\cap1)) * (P(3)+P(2)-P(3\cap2))=$$
So my book has the answer as .98096, while the approach shown above gives me .980096. I'm guessing this makes it incorrect? Could someone explain why it doesn't work? 
 A: You mistakenly assumed that the events $3\cup1$ and $3\cup2$ are independent; but they are not. Therefore you cannot multiply their probabilities in order to obtain $P\bigl((3\cup1)\cap(3\cup2)\bigr)$.
A: The message will get through if it successfully passes through the top route, or the bottom route or both. There is a handy trick to remember in probability problems for evaluating these "either" problems:
\begin{align}
P(\text{Either top route or bottom route}) &= 1 - P(\text{Neither route})\\
  &= 1 - P(\text{Top route fails})\times P(\text{Bottom route fails})
\end{align}
Now, we also have that $P(\text{Top route fails}) = 1 - P(\text{Top route succeeds})$ and $P(\text{Bottom route fails}) = 1 - P(\text{Bottom route succeeds})$. You've correctly calculated the top route success probability, and the bottom route success probability is given. Can you proceed from here?
A: The method to be solving this problem assumes that you consider the following


*

*1 and 2 are independent events

*1 and 3, as well as 2 and 3 are dependent events.

*$\mathbb{p(\mathsf{(1 \cap 2)\complement}\cap 3\complement)=p(1 \cap 2)\complement\space * p(3)\complement}$
The last fact can only be learned by looking at the diagram. Since $( 1\space \cap \space  2)\complement$  or the probability does not take the upper path does not effect the probability of $C$ or $C\complement$ (whether or not a message takes a lower path), which means the message may not go through then we can learn the probability the message will go through by taking the intersection of the complements which are independent events and applying the complement rule to it or
$$ 1-{p(1 \cap 2)\complement\space * p(3)\complement} $$
This would only not apply if the probability of message being delivered  was said to be 1.0 and a message must either take the upper path or the lower path.
Since however the question asked what the probability of a message being sent is, then we can assume the probability is not 1.0 and a third option exist in which the message does not take either path.
