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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$

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
$$p=frac12cdot 1+frac12cdotfrac38 $$

Where you have gone wrong is that there are $14$ different possible results of this process, but they are not all equally likely, so you can't just count how many are successful. (An easy way to see this doesn't work is that the probability of getting exactly one head can't depend on the number of faces of the die, but if you apply your method with a $20$-sided die, say, you will get a completely different answer.)

In fact each coin-die combination has probability $frac12timesfrac16=frac112$, and each four-coin combination has probability $frac12timesfrac12timesfrac12timesfrac12=frac116$. Since your successes counted $6$ of the former and $3$ of the latter, the correct answer is $frac612+frac316$.

Alternative route:

Throw $4$ coins.

Let $H_1$ denote the first gives heads, and let $E$ denote the event that exactly one coin gives heads.

Then $H_1cup E$ can be recognized as the event that the first coin gives heads or it does not and exactly one of the other coins gives heads.

$$P(H_1cup E)=P(H_1)+P(E)-P(H_1cap E)=frac12+4timesfrac116-frac116=frac1116$$