Independence of coin flips 
A fair coin is tossed three times in succession. If at least one of
  the tosses has resulted in Heads, what is the probability that at
  least one of the tosses resulted in Tails?

My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is $ \frac 12 + \frac 12 - \frac 14 = \frac 34 $. 
The given answer: $ \frac 67 $. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is $ \frac 67 $.
Why is my answer wrong? What am I missing?


The coin flips are all independent of each other, 

Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.  

Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1\cup H_2\cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.  
$$\mathsf P(T_1\mid E)~\mathsf P(T_2\mid E)~\mathsf P(T_3\mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$\mathsf P(T_1\cap T_2\cap T_3\mid E)=0$$
