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.

If the first flip is head, then the die won't produce coins showing head, hence we will have exactly one head. (You could just leave the die alone).
If the first flip is tails, then there are exactly three out of eight  outcomes from the three extra coins that result in exactly one head.
Hence
$$p=\frac12\cdot 1+\frac12\cdot\frac38 $$
