Sylvia has a bag of 10 coins. Nine are fair coins, but the tenth has tails on both sides. Sylvia draws a coin at random from the bag and flips it without looking. If the coin comes up tails, what is the probability that the coin she drew was the 2-tailed coin?
To solve this problem, we will use Bayes' Theorem, which is useful for finding the probability of an event given that another event has occurred. 

Let:
- \( A \) be the event that Sylvia drew the 2-tailed coin.
- \( B \) be the event that the coin flip resulted in tails.

We need to find \( P(A|B) \), the probability that Sylvia drew the 2-tailed coin given that the coin flip resulted in tails.

Bayes' Theorem states:
\[
P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}
\]

1. **Calculate \( P(A) \):**
   - \( P(A) \) is the probability of drawing the 2-tailed coin.
   - There is 1 two-tailed coin out of 10 coins.
   \[
   P(A) = \frac{1}{10}
   \]

2. **Calculate \( P(B|A) \):**
   - \( P(B|A) \) is the probability of flipping tails given that the 2-tailed coin was drawn.
   - Since the 2-tailed coin always shows tails, \( P(B|A) = 1 \).

3. **Calculate \( P(B) \):**
   - \( P(B) \) is the total probability of flipping tails.
   - This can happen in two ways: either the 2-tailed coin is drawn, or a fair coin is drawn and it lands on tails.
   - The probability of drawing a fair coin is \( \frac{9}{10} \).
   - The probability of flipping tails with a fair coin is \( \frac{1}{2} \).
   - Therefore, the total probability of flipping tails is:
   \[
   P(B) = P(B|A) \cdot P(A) + P(B|\text{fair coin}) \cdot P(\text{fair coin})
   \]
   \[
   P(B) = 1 \cdot \frac{1}{10} + \frac{1}{2} \cdot \frac{9}{10}
   \]
   \[
   P(B) = \frac{1}{10} + \frac{9}{20}
   \]
   \[
   P(B) = \frac{2}{20} + \frac{9}{20}
   \]
   \[
   P(B) = \frac{11}{20}
   \]

4. **Apply Bayes' Theorem:**
   \[
   P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}
   \]
   \[
   P(A|B) = \frac{1 \cdot \frac{1}{10}}{\frac{11}{20}}
   \]
   \[
   P(A|B) = \frac{\frac{1}{10}}{\frac{11}{20}}
   \]
   \[
   P(A|B) = \frac{1}{10} \cdot \frac{20}{11}
   \]
   \[
   P(A|B) = \frac{2}{11}
   \]

The final answer is \(\boxed{\frac{2}{11}}\).