AIME 2014 II · 第 2 题

AIME 2014 II — Problem 2

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

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 $p+q$ .

解析

Solution

We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.

$AIME diagram$

Let $x$ be the number of men with all three risk factors. Since "the probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$ ," we can tell that $x = \frac{1}{3}(x+14)$ , since there are $x$ people with all three factors and 14 with only A and B. Thus $x=7$ .

Let $y$ be the number of men with no risk factors. It now follows that

$y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21.$ The number of men with risk factor A is $10+2 \cdot 14+7 = 45$ (10 with only A, 28 with A and one of the others, and 7 with all three). Thus the number of men without risk factor $A$ is 55, so the desired conditional probability is $21/55$ . So the answer is $21+55=\boxed{076}$ .