The question:
Assume you have 10 coins labeled 1, 2, 3, ..., 10. The probability of flipping a heads on the $i^{th}$ coin is $\frac{i}{10}$. You flip a coin, and it comes up heads. What is the probability the coin is a 2 headed coin.
If the question is just $\frac{1}{10}$ for each coin, I think it's just a straightforward application of Bayes Rule
$P(2H|H) = \frac{P(2H)P(H|2H)}{P(H)}$
But, I'm really not sure how to do this with with the varying probability.
On
true blue anil's answer is correct and a fuller explanation is that if $C_i$ is the event that the $i$th coin is selected and $H$ that a head shows after the flip then $$\mathbb P(C_i \mid H) = \dfrac{\mathbb P(C_i \cap H)}{\mathbb P(H)}= \dfrac{\mathbb P(H \mid C_i )\mathbb P(C_i)}{\sum_{j=1}^{10} \mathbb P(H \mid C_j )\mathbb P(C_j)} = \dfrac{\frac{i}{10}\times \frac1{10}}{\sum_{j=1}^{10} \frac{j}{10}\times \frac1{10}}= \dfrac{i}{\sum_{j=1}^{10} j} =\dfrac{i}{55}$$
and so for the double-headed coin you have $\mathbb P(C_{10} \mid H)=\dfrac{10}{55}=\dfrac{2}{11}$
A simple way is to realize that each coin has an equal probability of being chosen,
thus in $100$ trials, each coin is likely to be chosen $10$ times,
and total number of heads expected will be $(1+2+3+...+10) = 55$
Thus $P(C_{10}|H) = \frac{10}{55}$