Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C. within a population of men. For each of the three factors, the probability that a randomly selected man in the population as only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly two of these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\tfrac{1}{3}$. The probability that a man has none of the three risk factors given that he does not have risk factor A is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
1. **Define the problem and given probabilities:**
   - Let \( P(A) \) be the probability that a randomly selected man has risk factor \( A \).
   - Similarly, define \( P(B) \) and \( P(C) \) for risk factors \( B \) and \( C \).
   - Given:
     - \( P(\text{only } A) = 0.1 \)
     - \( P(\text{only } B) = 0.1 \)
     - \( P(\text{only } C) = 0.1 \)
     - \( P(A \cap B \cap \overline{C}) = 0.14 \)
     - \( P(A \cap \overline{B} \cap C) = 0.14 \)
     - \( P(\overline{A} \cap B \cap C) = 0.14 \)
     - \( P(A \cap B \mid A \cap B) = \frac{1}{3} \)

2. **Calculate the number of people with each combination of risk factors:**
   - Assume there are 1000 people in the population.
   - Number of people with only \( A \): \( 1000 \times 0.1 = 100 \)
   - Number of people with only \( B \): \( 1000 \times 0.1 = 100 \)
   - Number of people with only \( C \): \( 1000 \times 0.1 = 100 \)
   - Number of people with \( A \) and \( B \) but not \( C \): \( 1000 \times 0.14 = 140 \)
   - Number of people with \( B \) and \( C \) but not \( A \): \( 1000 \times 0.14 = 140 \)
   - Number of people with \( A \) and \( C \) but not \( B \): \( 1000 \times 0.14 = 140 \)

3. **Calculate the number of people with all three risk factors:**
   - Given \( P(A \cap B \mid A \cap B) = \frac{1}{3} \), we can find the number of people with all three risk factors.
   - Let \( x \) be the number of people with all three risk factors.
   - \( P(A \cap B \cap C \mid A \cap B) = \frac{x}{140 + x} = \frac{1}{3} \)
   - Solving for \( x \):
     \[
     \frac{x}{140 + x} = \frac{1}{3} \implies 3x = 140 + x \implies 2x = 140 \implies x = 70
     \]
   - Therefore, 70 people have all three risk factors.

4. **Calculate the number of people with no risk factors:**
   - Total number of people with at least one risk factor:
     \[
     100 (\text{only } A) + 100 (\text{only } B) + 100 (\text{only } C) + 140 (A \cap B \cap \overline{C}) + 140 (B \cap C \cap \overline{A}) + 140 (A \cap C \cap \overline{B}) + 70 (A \cap B \cap C) = 790
     \]
   - Number of people with no risk factors:
     \[
     1000 - 790 = 210
     \]

5. **Calculate the number of people without risk factor \( A \):**
   - Number of people without risk factor \( A \):
     \[
     100 (\text{only } B) + 100 (\text{only } C) + 140 (B \cap C \cap \overline{A}) + 210 (\text{no risk factors}) = 550
     \]

6. **Calculate the probability that a man has no risk factors given that he does not have risk factor \( A \):**
   - \( P(\text{no risk factors} \mid \overline{A}) = \frac{P(\text{no risk factors} \cap \overline{A})}{P(\overline{A})} \)
   - \( P(\text{no risk factors} \cap \overline{A}) = P(\text{no risk factors}) = \frac{210}{1000} = 0.21 \)
   - \( P(\overline{A}) = \frac{550}{1000} = 0.55 \)
   - Therefore:
     \[
     P(\text{no risk factors} \mid \overline{A}) = \frac{0.21}{0.55} = \frac{21}{55}
     \]

7. **Find \( p + q \) where \( \frac{p}{q} = \frac{21}{55} \):**
   - Since 21 and 55 are relatively prime, \( p = 21 \) and \( q = 55 \).
   - Therefore, \( p + q = 21 + 55 = 76 \).

The final answer is \( \boxed{76} \).