Are the first flip and second flip of this problem independent events? 

Problem
    Given 5 coins:




*

*2 double-headed coins 

*2 fair coins 

*1 double-tailed coin




A coin is chosen at random and flipped. The coin is then flipped in
    a second time. What's the probability that face showing is a head?


My attempt was using conditional probability on each type of coin:
P( 1st = whatever, 2nd = head | DH ) * P( DH )
+ P( 1st = whatever, 2nd = head | DT ) * P( DT )
+ P( 1st = whatever, 2nd = head | FC ) * P( FC )   
Apparently, P( 1st = whatever, 2nd = head | DT ) = 0.
So I ended up with:
$$1 \cdot \dfrac{2}{5} + \dfrac{1}{2} \cdot \dfrac{2}{5} = \dfrac{3}{5} $$
But my teacher said the answer is $\dfrac{5}{6}$ which I couldn't understand how did he come up with that solution! So my question is, are the 1st flip and 2nd flip two independent events? I don't see how should we take the 1st flip into account?  
 A: Converting my comment into an answer, the question asked in the title is 

"Are the first flip and the second flip independent events?" 

to which the answer is 

"No, they are not independent events, but they are conditionally independent events conditioned on which coin is drawn."

Let $H_1$ and $H_2$ denote the events that the first flip and the second flip respectively resulted in Heads, and let $H$, $F$, and $T$ denote the events that that the coin drawn is double-Headed, Fair, and double-Tailed respectively, where $P(H) = P(F) = \frac{2}{5}$ and $P(T) = \frac{1}{5}$.  Then, given that a colin has been chosen, the two tosses are (conditionally) independent events, that is,
$$\begin{align*}
P(H_1 \mid H) = P(H_2 \mid H) = 1; 
~~ &P(H_1 \cap H_2 \mid H) = P(H_1 \mid H)P(H_2 \mid H) = 1,\\
P(H_1 \mid F) = P(H_2 \mid F) = \frac{1}{2}; 
~~ &P(H_1 \cap H_2 \mid F) = P(H_1 \mid F)P(H_2 \mid F) = \frac{1}{2}\times \frac{1}{2} = \frac{1}{4},\\
P(H_1 \mid T) = P(H_2 \mid T) = 0; 
~~ &P(H_1 \cap H_2 \mid T) = P(H_1 \mid T)P(H_2 \mid T) = 0.
\end{align*}
$$
The law of total probability then gives that for $i = 1, 2$,
$$
\begin{align*}
P(H_i) &= P(H_i \mid H)P(H) + P(H_i \mid F)P(F) + P(H_i \mid T)P(T)\\ 
&= 1 \times \frac{2}{5} + \frac{1}{2}\times \frac{2}{5} + 0 \times \frac{1}{5}
= \frac{3}{5},\\
P(H_1 \cap H_2) &= P(H_1 \cap H_2 \mid H)P(H) + P(H_1 \cap H_2 \mid F)P(F) + P(H_1 \cap H_2 \mid T)P(T)\\ 
&= 1 \times \frac{2}{5} + \frac{1}{4}\times \frac{2}{5} + 0 \times \frac{1}{5}
= \frac{1}{2} \neq \left ( \frac{3}{5}\right )^2 = P(H_1)P(H_2),
\end{align*}
$$
showing that $H_1$ and $H_2$ are not independent events.  Finally,
$$
P(H_2 \mid H_1) = \frac{P(H_1 \cap H_2)}{P(H_1)}
= \frac{1/2}{3/5} = \frac{5}{6}
$$
as calculated by the OP's teacher.
A: As the comments say, this looks like a matter of checking the question. 
If the first flip is ignored, then $\frac{3}{5}$ is correct, for the reasons you give.  It is also the probability that the first flip is heads.
If you condition on the first flip being heads, then you would have 
P( 2nd = head | DH, 1st = head ) * P( DH| 1st = head )
+ P( 2nd = head | FC, 1st = head ) * P( FC| 1st = head )
+ P( 2nd = head | DT, 1st = head ) * P( DT| 1st = head )   
which gives
$$1 \cdot \dfrac{2}{3} + \dfrac{1}{2} \cdot \dfrac{1}{3} + ? \cdot 0 = \dfrac{5}{6}.$$
Similarly if you conditioned on the first flip being tails, the probabilitiy of the second being heads would be 
$$? \cdot 0 + \dfrac{1}{2} \cdot \dfrac{1}{2} + 0 \cdot \dfrac{1}{2} = \dfrac{1}{4}.$$
Combining these two results for the overall marginal probability that the second flip is heads would give 
$$ \dfrac{5}{6} \cdot \dfrac{3}{5} + \dfrac{1}{4} \cdot \dfrac{2}{5} = \dfrac{3}{5},$$
confirming your earlier result.
