When flipped, coin A shows heads $\textstyle\frac{1}{3}$ of the time, coin B shows heads $\textstyle\frac{1}{2}$ of the time, and coin C shows heads $\textstyle\frac{2}{3}$ of the time. Anna selects one of the coins at random and flips it four times, yielding three heads and one tail. The probability that Anna flipped coin A can be expressed as $\textstyle\frac{p}{q}$ for relatively prime positive integers $p$ and $q$. Compute $p + q$.

[i]Proposed by Eugene Chen[/i]
1. **Define the problem and given probabilities:**
   - Coin A shows heads with probability \( \frac{1}{3} \).
   - Coin B shows heads with probability \( \frac{1}{2} \).
   - Coin C shows heads with probability \( \frac{2}{3} \).
   - Anna selects one of the coins at random, so the probability of selecting any one coin is \( \frac{1}{3} \).

2. **Calculate the probability of getting exactly 3 heads and 1 tail for each coin:**
   - For coin A:
     \[
     P(\text{3 heads, 1 tail} \mid A) = \binom{4}{3} \left( \frac{1}{3} \right)^3 \left( \frac{2}{3} \right)^1 = 4 \cdot \frac{1}{27} \cdot \frac{2}{3} = 4 \cdot \frac{2}{81} = \frac{8}{81}
     \]
   - For coin B:
     \[
     P(\text{3 heads, 1 tail} \mid B) = \binom{4}{3} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^1 = 4 \cdot \frac{1}{8} \cdot \frac{1}{2} = 4 \cdot \frac{1}{16} = \frac{1}{4}
     \]
   - For coin C:
     \[
     P(\text{3 heads, 1 tail} \mid C) = \binom{4}{3} \left( \frac{2}{3} \right)^3 \left( \frac{1}{3} \right)^1 = 4 \cdot \frac{8}{27} \cdot \frac{1}{3} = 4 \cdot \frac{8}{81} = \frac{32}{81}
     \]

3. **Calculate the total probability of getting 3 heads and 1 tail:**
   \[
   P(\text{3 heads, 1 tail}) = P(\text{3 heads, 1 tail} \mid A) \cdot P(A) + P(\text{3 heads, 1 tail} \mid B) \cdot P(B) + P(\text{3 heads, 1 tail} \mid C) \cdot P(C)
   \]
   \[
   = \frac{8}{81} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{1}{3} + \frac{32}{81} \cdot \frac{1}{3}
   \]
   \[
   = \frac{8}{243} + \frac{1}{12} + \frac{32}{243}
   \]
   \[
   = \frac{8}{243} + \frac{20.25}{243} + \frac{32}{243}
   \]
   \[
   = \frac{8 + 20.25 + 32}{243} = \frac{60.25}{243}
   \]

4. **Calculate the probability that Anna flipped coin A given that she got 3 heads and 1 tail:**
   \[
   P(A \mid \text{3 heads, 1 tail}) = \frac{P(\text{3 heads, 1 tail} \mid A) \cdot P(A)}{P(\text{3 heads, 1 tail})}
   \]
   \[
   = \frac{\frac{8}{81} \cdot \frac{1}{3}}{\frac{60.25}{243}}
   \]
   \[
   = \frac{\frac{8}{243}}{\frac{60.25}{243}}
   \]
   \[
   = \frac{8}{60.25}
   \]
   \[
   = \frac{8}{60.25} = \frac{8}{\frac{241}{4}} = \frac{8 \cdot 4}{241} = \frac{32}{241}
   \]

5. **Simplify the fraction \(\frac{32}{241}\):**
   - The fraction \(\frac{32}{241}\) is already in simplest form since 32 and 241 are relatively prime.

6. **Compute \( p + q \):**
   - Here, \( p = 32 \) and \( q = 241 \).
   - Therefore, \( p + q = 32 + 241 = 273 \).

The final answer is \( \boxed{273} \)