Probability Dilemma The teacher gave us a question: 

We flip a coin. If it was Heads, we roll a die and if it was Tails, we flip three other coins. What's the probability of exactly having one coin as Heads?

I first calculated $n(S)= 1 
\cdot 6 + 1 \cdot 2 \cdot 2 \cdot 2$, then $n(a) = 1 \cdot 6 + 1 \cdot 1 \cdot 1 \cdot 3$. $P(a)=n(a)/n(S) = 9/14$.
For calculating $n(S)$ I first considered the case of the (first) coin being Heads and multiplied by $6$ (for the die), then calculated the case of the (first) coin being Tails and a $2 \cdot 2 \cdot 2$ for the coins. You'll get a sense of what I did for calculating $n(a)$. The teacher told that my answer was incorrect and then wrote the answer as $6 \cdot (1/12) + 3 \cdot (1/16)=11/16$ and told me to find the problem of my answer for the next session. I don't think there's anything wrong with my answer and my teacher's wrong.

It looks like you are trying to calculate the number of states and divide the number of states fitting some criteria by the total number of states. One thing to note with your approach is that not all of the states you sum up are equally likely. In particular if I asked what is the probability of the first coin flip being heads you would get 6/14, but we know it to be 1/2=7/14
A more systematic approach is to look at the probability of the first coin being heads. If it isn't heads then we flip the three coins so you would need the probability of exactly one of those coins being heads:
$P($H$)+P($1 H out of 3$|$T$)P($T$) = (1/2)+(3/8)(1/2)=11/16$
