urn type problem with bayes theorem (and M&Ms) don't understand how probability of seeing 'evidence' was calculated. I was trying to follow the solution to the 'urn type' 
probability problem with M&M's from this page ->
http://allendowney.blogspot.com/2011/10/all-your-bayes-are-belong-to-us.html
I have reproduced the problem below, but what I don't get is the 
skipped steps after he says 

Plugging the likelihoods and the priors into Bayes's theorem, we get P(A|E) = 40 / 54 ~ 0.74 "

I understand the formula:  
P(E) P(A|E) =  P(A) P(E|A)  

    =>

P(A|E) = P(A) P(E|A)
         -----------
            P(E)

And I got this far:
P(A) = .5
P(E|A) = .2 * .2

P(A|E) =  (.5)  (.2) (.2) 
         -----------
            P(E)

But I am stuck on how the author of the post calculated P(E)
(the probability of the evidence).  Any guidance much appreciated !
M&M Problem
The blue M&M was introduced in 1995.  Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% Red, 10% Green, 10% Orange, 10% Tan).  Afterward it was (24% Blue , 20% Green, 16% Orange, 14% Yellow, 13% Red, 13% Brown). 
A friend of mine has two bags of M&Ms, and he tells me that one is from 1994 and one from 1996.  He won't tell me which is which, but he gives me one M&M from each bag.  One is yellow and one is green.  What is the probability that the yellow M&M came from the 1994 bag?
Hypotheses: A: Bag #1 from 1994 and Bag #2 from 1996
B: Bag #2 from 1994 and Bag #1 from 1996
Again, P(A) = P(B) = 1/2.
The evidence is:
E: yellow from Bag #1, green from Bag #2
We get the likelihoods by multiplying the probabilities for the two M&M:
P(E|A) = (0.2)(0.2)
P(E|B) = (0.1)(0.14)
For example, P(E|B) is the probability of a yellow M&M in 1996 (0.14) times the probability of a green M&M in 1994 (0.1).
Plugging the likelihoods and the priors into Bayes's theorem, we get P(A|E) = 40 / 54 ~ 0.74
A: A = given yellow M & M from 1994 bag
B = given yellow and green M & M
P(A) = .2
P(A and B) = .2 x .2 = .04
the above is to say, that a Yellow comes from A, with probability .2 AND the other (now necessarily from B) is Blue, with probability 0.2
for B to be true, Green was taken from either 1994 or 1996
P(B) = P(yellow from 1994)P(green from 1996) + P(green from 1994)P(yellow from 1996)
= .2 x .2 + .1 x .14 = .04 + .014 = .054
P(A | B) = P(A and B) / P(B) 
 = .04 / .054 = 0.74074
..................................................
I don't really agree with his hypothesis about bag A and bag B with probability 1/2, that isn't necessary, it is very confusing
A: Notice that:
\begin{align*}
\Pr[E]
&= \Pr[E \text{ and } (A \text{ or } B)] &\text{since $A$ and $B$ are the only two possibilities}\\ 
&= \Pr[(E \text{ and } A) \text{ or } (E \text{ and }B)] \\ 
&= \Pr[E \text{ and } A] + \Pr[E\text{ and }B] &\text{since $A$ and $B$ are mutually exclusive}\\ 
&= \Pr[A]\Pr[E \mid A] + \Pr[B]\Pr[E \mid B] \\ 
&= \frac{1}{2}(0.2)(0.2) + \frac{1}{2}(0.14)(0.1) \\
&= 0.027
\end{align*}
