Denote by $D$ the event that the coin is the one with two heads.

(a) $\mathrm{P}(D)=1 / 10$.

(b) By Bayes theorem, we have

$$

\mathrm{P}(D \mid H)=\frac{\mathrm{P}(H \mid D) \mathrm{P}(D)}{\mathrm{P}(H)}

$$

We have $\mathrm{P}(H \mid D)=1$ and $\mathrm{P}(D)=\frac{1}{10}$. Now, we need to think about $H$, getting a head, in terms of getting a head with either a double headed or single headed coin. Using the idea of a partition,

$$

\begin{aligned}

\mathrm{P}(H) &=\mathrm{P}(H \cap D)+\mathrm{P}\left(H \cap D^{c}\right) \\

&=\mathrm{P}(H \mid D) \mathrm{P}(D)+\mathrm{P}\left(H \mid D^{c}\right) \mathrm{P}\left(D^{c}\right) \\

&=(1)\left(\frac{1}{10}\right)+\left(\frac{1}{2}\right)\left(\frac{9}{10}\right) \\

&=\frac{11}{20}

\end{aligned}

$$

Finally, here is another way of calculating $\mathrm{P}(H):$ think of the bag as containing the possible tosses. As the bag contains 9 fair coins and one double-headed coin, it must contains 11 heads and 9 tails, so that the probability of choosing a head is $11 /(11+9)=11 / 20$. To return to the original question, we now obtain the answer

$$

\mathrm{P}(D \mid H)=\frac{\frac{1}{10}}{\frac{11}{20}}=\frac{2}{11}

$$

(c) 1. If it comes up tails, it can't be the coin with two heads. Therefore it must be one of the other nine.