Values from (incorrectly used?) Bayes' Theorem Given a bag of 15 coins:
 5 of which are   fair coins with P(Heads) = 0.5  
10 of which are unfair coins with P(Heads) = 0.8  

We select a coin at random, flip it, recorded the result of the flip, then returned the coin to the bag.
We do this a total of 6 times, and recorded 4 heads.  
I've worked out that for any individual trial, P(Heads) is 0.7  
So, out of curiosity, I applied Bayes' theorem:
P(Fair|4/6 Heads) = P(4/6 Heads|Fair) * P(Fair) / P(4/6 Heads)

P(4/6 Heads|Fair) = $\binom{6}{4}*0.5^{6}$
P(Fair) = $\frac{1}{3}$
P(4/6 Heads) = $\binom{6}{4}*0.7^{4}*0.3^{2}$  
P(Fair|4/6 Heads) = 0.241026 = 24.1026%
I'm puzzled as to what that value of 24.1026% represents.
Initially, I thought that it must be the ratio of fair coins within the selected six coins (i.e. on average, given multiple trials which results in an outcome of 4/6 heads, 24.1026% of the 6 selected coins were fair).
So I wrote a program to test this:

I simulated multiple trials that picks 6 coins from the bag, with the original 1/3 fair coin chance.
  I kept a count for each trial which resulted in 4/6 Heads, and summed up the number of fair coins in each of them.
  Then I tried to find the ratio: $\frac{numberOfFairCoins}{6*numberOf4HeadTrials}$
  But I got a value of 34.38% instead.

So, any ideas?
As a side note, how does one calculate that value of 34.38%?

$P(\text{Fair}\mid \text{4/6 Heads})= P(\text{4/6 Heads}|\text{Fair}) \cdot P(\text{Fair}) / P(\text{4/6 Heads})$ is meaningless.  
From the values you used, that's the probability that 4 of 6 coins are heads if they are all fair times the probability that one coin is fair divided by the probability that 4 of 6 coins are heads when you are returning the coins to the bag after each flip.   That makes no sense at all.
This has more meaning.
$$P(\text{All Fair}\mid \text{4/6 Heads})= P(\text{4/6 Heads}|\text{All Fair}) \cdot P(\text{Fair})^\color{red}{6} / P(\text{4/6 Heads})$$
