What is the probability of escaping from jail if....? Disclaimer: The story given below is purely fictional and does not, in any way, relate to a prison break. :P  

Four roads lead away from a jail. A prisoner is trying to escape from
  the jail and selects one road at random. If road A is selected, the
  probability of escaping is (1/8), or road B it is (1/6), for road C it
  is (1/4) and for road D it is (9/10). What is the probability that the
  prisoner will succeed in escaping from the jail.  

Now, the book gives the formula as follows:  
P(E) = $p1p1' + p2p2' + p3p3'+p4p4'$ where
P(E) is probability of escape,
$p1$ is the probability of choosing road A.
What I would like to ask is.. what is $p1'$ ?  
On the previous page it explains something about the 'theorem on total probability' in really shoddy terms. It is something related but  cannot understand what.

Set up a tree of possibilities, where each branching point is a decision (take road 1, 2, 3, or 4, then escape or not for each), each edge is labeled with the probability of taking that option . Mark the ends (leaves) that interest you (escape successful), for each of them compute the probability by multiplying probabilities along the path from the root, add all up. Rigorous justification of this is quite simple.
This is overkill in your case, the tree is just two steps deep; but having an orderly, uniform technique to handle more complex situations that helps avoiding forgetting or duplicating one of a myriad of cases is valuable.
