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.

$p1'$ is the chance of escaping through road A, so $1/8$ in your example (because its random). $p2'$ would be the chance of escaping though road B (so $1/6$), et cetera.
Anyhow, $P(E)$ consists of the summation of two factors: $p_n$ being the chance to pick the road (randomly, so $1/4$) and $p_n'$ the chance to escape through that road.
