Question

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 $p + q.$

Answer

Solution:

Take a population of $100$ men and fill in a Venn diagram. Each of the three exactly-one regions contains $10$ men, and each of the three exactly-two regions contains $14.$ If $x$ men have all three factors, then the men with both $A$ and $B$ number $x + 14,$ so the given conditional probability says $\frac{x}{x + 14} = \frac{1}{3},$ giving $x = 7.$

The union of the three sets therefore contains $3 \cdot 10 + 3 \cdot 14 + 7 = 79$ men, leaving $21$ with no risk factor. The men with risk factor $A$ number $10 + 14 + 14 + 7 = 45,$ so $55$ men do not have $A.$

The desired probability is $\frac{21}{55},$ which is in lowest terms, so $p + q = 21 + 55 = 76.$

2014 AIME II Problem 2

Solution:

Problem 2 in Other Years