BasicProbability InclusionExclusion AIME Intermediate
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 $\frac{1}{3}$. Find the probability that a man has none of the three risk factors given that he does not have risk factor $A$.
For the sake of convenience, let's assume there are $100$ men being studied. Most given conditions are represented in the following Venn diagram. Let the number of people having all the three risk factors be $x$.
Then, "the probability of people having all three risk factors given that he as $A$ and $B$ is $1/3$ can be translated to $\frac{1}{3}\times(x + 14)$. Therefore we have $$x=\frac{1}{3}\times(x+14)\implies x =7$$
Therefore, people with no risk factor is $$100- 3\times 10-3\times 14 - 7=21$$
People with risk factor $A$ is $$10+2\times 14 + 7 = 45$$
It follows that the final answer is $$\frac{21}{100-45}=\boxed{\frac{21}{55}}$$