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?

Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$  Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$\frac{1-P(T=0)-P(X=0)}{1-P(X=0)}=\frac{1-\frac{1}{8}-\frac{1}{8}}{1-\frac{1}{8}}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
