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 has 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 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 $\frac{1}{3}$. The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

We are given the following information about three health risk factors $ A $, $ B $, and $ C $:

1. The probability that a randomly selected man has only one of the risk factors $ A $, $ B $, or $ C $ (and none of the others) is 0.1.
2. The probability that a randomly selected man has exactly two of the three risk factors is 0.14.
3. The probability that a randomly selected man has all three risk factors given that he has $ A $ and $ B $ is $ \frac{1}{3} $.

We are asked to find $ p+q $ where the probability that a man has none of the three risk factors given that he does not have risk factor $ A $ is $ \frac{p}{q} $, and $ p $ and $ q $ are relatively prime positive integers.

### Step 1: Define the relevant probabilities
Let:
- $ P(A) $, $ P(B) $, $ P(C) $ be the probabilities of having risk factors $ A $, $ B $, and $ C $, respectively.
- $ P(A \cap B) $, $ P(A \cap C) $, and $ P(B \cap C) $ be the probabilities of having exactly two of the three risk factors.
- $ P(A \cap B \cap C) $ be the probability of having all three risk factors.

### Step 2: Express the given probabilities
Given:
- $ P(A \cap \overline{B} \cap \overline{C}) = P(\overline{A} \cap B \cap \overline{C}) = P(\overline{A} \cap \overline{B} \cap C) = 0.1 $.
- $ P(A \cap B \cap \overline{C}) = P(A \cap \overline{B} \cap C) = P(\overline{A} \cap B \cap C) = 0.14 $.
- $ P(A \cap B \cap C \mid A \cap B) = \frac{1}{3} $.

### Step 3: Calculate $ P(A \cap B \cap C) $
First, we need to find $ P(A \cap B \cap C) $.

From the conditional probability formula:
\[
P(A \cap B \cap C \mid A \cap B) = \frac{P(A \cap B \cap C)}{P(A \cap B)}
\]
We have:
\[
\frac{P(A \cap B \cap C)}{0.14} = \frac{1}{3}
\]
Thus:
\[
P(A \cap B \cap C) = \frac{1}{3} \times 0.14 = \frac{14}{300} = \frac{7}{150}
\]

### Step 4: Use the inclusion-exclusion principle
The probability of having at least one risk factor is given by:
\[
P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)
\]
Substituting the known values:
\[
P(A) + P(B) + P(C) = 3 \times 0.1 = 0.3
\]
\[
P(A \cap B) + P(A \cap C) + P(B \cap C) = 3 \times 0.14 = 0.42
\]
Thus:
\[
P(A \cup B \cup C) = 0.3 - 0.42 + \frac{7}{150} = -0.12 + \frac{7}{150}
\]
Converting $ -0.12 $ to a fraction:
\[
-0.12 = -\frac{12}{100} = -\frac{18}{150}
\]
Therefore:
\[
P(A \cup B \cup C) = -\frac{18}{150} + \frac{7}{150} = -\frac{11}{150}
\]

### Step 5: Probability that none of the risk factors are present
The probability that a man has none of the risk factors $ A $, $ B $, or $ C $ is:
\[
P(\overline{A} \cap \overline{B} \cap \overline{C}) = 1 - P(A \cup B \cup C) = 1 - P(A \cup B \cup C) = 1 + \frac{11}{150} = \frac{161}{150}
\]

### Step 6: Calculate the conditional probability
We now need to find $ P(\overline{A} \cap \overline{B} \cap \overline{C} \mid \overline{A}) $:
\[
P(\overline{A} \cap \overline{B} \cap \overline{C} \mid \overline{A}) = \frac{P(\overline{A} \cap \overline{B} \cap \overline{C})}{P(\overline{A})}
\]
We already know $ P(\overline{A} \cap \overline{B} \cap \overline{C}) = \frac{161}{150} $.

$ P(\overline{A}) = 1 - P(A) = 1 - 0.1 = 0.9 $.

Therefore:
\[
P(\overline{A} \cap \overline{B} \cap \overline{C} \mid \overline{A}) = \frac{\frac{161}{150}}{0.9} = \frac{161}{135}
\]
Simplifying the fraction $ \frac{161}{135} $ gives us $ p = 161 $ and $ q = 135 $. Therefore, $ p + q = 161 + 135 = 296 $. The final answer is $ \boxed{296} $.

Explore the solution to a probability problem involving the prevalence of health risk factors A, B, and C in a population of men. Learn about conditional probabilities, the inclusion-exclusion principle, and calculate the probability that a man has none of these risk factors given that he does not have risk factor A. Suitable for advanced high school students.

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