Does independence of events depend upon underlying probability model? Consider a sample space of two coin tosses = $\{HH, HT, TH, TT\}$. 
Suppose that the coin is fair and therefore every outcome has probability $\frac{1}{4}$.
Now, consider another probability model where the coin is biased and head occurs with probability $\frac{3}{4}$.
In this model the corresponding probabilities = $\{\frac{9}{16}, \frac{3}{16}, \frac{3}{16}, \frac{1}{16}\}$.
Definition of independence $P(A \cup B) = P(A)P(B)$ 
Event $A$ = first coin toss results in head        = $\{HT, HH\}$ 
Event $B$ = both coin toss results in same outcome = $\{HH, TT\}$ 
First model: 
$P(A) = P(\{HT,HH\}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
$P(B) = P(\{HH,TT\}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
$P(A \cup B) = P(\{HH\}) = \frac{1}{4} = \frac{1}{2}\times\frac{1}{2} = P(A)P(B)$
Therefore, in this probability model events $A$ and $B$ are independent.
Second model : 
$P(A) = P(\{HT,HH\}) = \frac{3}{16} + \frac{9}{16} = \frac{12}{16}$
$P(B) = P(\{HH,TT\}) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16}$
$P(A \cup B) = P(\{HH\}) = \frac{9}{16}$ which is not equal to $P(A)P(B) = \frac{15}{32}$
Therefore, in this probability model events $A$ and $B$ are not independent
Thus, independence of the events depend upon the underlying probability model.
However, I am not getting the intuition. Can anyone explain?
Also Bayes' rule and law of total probability is applicable irrespective of underlying probability model so why the independence of events differs from one probability model to another? 
 A: The individual coin tosses are still independent, but now the outcome of the first toss is informative of your odds on the second. Imagine a bet based on a slightly modified version of your experiment: You toss a coin, see the results, then determine how much you want to bet that the second toss matches the first toss.
Under Model 1, the amount you would bet wouldn't change based on the outcome of the first toss, since it is equally likely that you could get a macth or not. This is not so under Model 2, where you would be willing to bet less money if the first toss were Tails. So, under Model 2, the first toss tells you something.
You seem to have the mathematical details down, so I'm not sure if you need some sort of alternative derivation. Independence depends on the underlying model, as you indicated. In particular, if the conditional distribution is simply proportional to the unconditional distribution, then you have independence. 
A: Both events have $HH$ in them. Thus, if prob of heads is higher than $0.5$, there is a stronger chance they will both happen. Another way to think of independence is:
Originally,
$P(B \cup A) = P(A)P(B|A)$
$A$ and $B$ are independent iff $P(B|A) = P(B)$
If $A$ occurs, there is a chance because it is due to the $HH$ in it. If $A$ occurs due to the $HH$, then $B$ occurs. In second model, the possibility of heads is higher and thus the possibility of $HH$ is higher.
Possibly related: http://en.wikipedia.org/wiki/Bertrand's_box_paradox
