2014 AIME II Problem 2

Attempt Problem 2 of the 2014 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AIME II solutions, or check the answer key.

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2.

Arnold is studying the prevalence of three health risk factors, denoted by A,A, B,B, and C,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 0.1.0.1. 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 0.14.0.14. The probability that a randomly selected man has all three risk factors, given that he has AA and B,B, is 13.\frac{1}{3}. The probability that a man has none of the three risk factors given that he does not have risk factor AA is pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q.

Answer: 76
Concepts:Venn Diagramconditional probability
Difficulty rating: 2110
Solution:

Take a population of 100100 men and fill in a Venn diagram. Each of the three exactly-one regions contains 1010 men, and each of the three exactly-two regions contains 14.14. If xx men have all three factors, then the men with both AA and BB number x+14,x + 14, so the given conditional probability says xx+14=13,\frac{x}{x + 14} = \frac{1}{3}, giving x=7.x = 7.

The union of the three sets therefore contains 310+314+7=793 \cdot 10 + 3 \cdot 14 + 7 = 79 men, leaving 2121 with no risk factor. The men with risk factor AA number 10+14+14+7=45,10 + 14 + 14 + 7 = 45, so 5555 men do not have A.A.

The desired probability is 2155,\frac{21}{55}, which is in lowest terms, so p+q=21+55=76.p + q = 21 + 55 = 76.

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