Probability Theory Question on Expected Value and Variance of Random Variable I've been trying to solve this question, any help would be much appreciated!

There are three coins. 2 coins are fair, the other has two tails. A
  random coin from this pile is chosen and tossed, it lands heads. This
  is done two more times without replacement.
What is the conditional probability that the third toss will land
  heads?
If X is a random variable which counts the number of heads
  total over the three tosses (assuming each coin is tossed exactly
  once) what is the expected value E(X)?
what is the variance of X?

I believe each coin is tossed only once.
 A: The two-tailed coin can't be the one that was chosen.  So one of the two fair coins was chosen, and we know it landed heads.  The fair coin that was not chosen could land either heads or tails, with equal probabilities.
So the possible values of $X$ are $1$ and $2$, each with probability $1/2$.
As for the conditional probability of the "third toss", the question is a bit
ambiguous.  Is the coin that was chosen the first toss, or could the coins have been
tossed in any order?
A: The random variable $X$ has precisely the same distribution as the number of heads when two fair coins are tossed. So $E(X)=1$ and $\operatorname{Var}(X)=2(1/2)(1/2)=1/2$. 
As to the conditional probability question, with probability $1/2$, the third coin is twwo-tailed, in which case we cannot get a head, and with probability $1/2$ the third coin is fair, in which case the probability of head is $1/2$. So the conditional probability is $1/4$. 
A: The probability that the first and third coins are heads is $1/3\times 1/4=1/12$.  The probability that the first coin is heads is $2/3\times 1/2=1/3$.  So the conditional probability that the third coin is heads, given that the first coin is heads, is $(1/12) \div(1/3)=1/4$.
As for the second part, the total number of heads is $0$ with probability $1/4$, $1$ with probability $1/2$, and $2$ with probability $1/4$.  From this, you can find
$$
E[X]=\frac{1}{4}\cdot 0+\frac{1}{2}\cdot 1 + \frac{1}{4}\cdot 2=1;
$$
and
$$
E[X^2]=\frac{1}{4}\cdot 0^2+\frac{1}{2}\cdot 1^2 + \frac{1}{4}\cdot 2^2=\frac{3}{2},
$$
so $\text{Var}[X]=E[X^2]-E[X]^2=3/2-1=1/2.$
