2014 AIME II 第 2 题

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2.

Arnold 正在研究某男性群体中三种健康风险因素的流行情况,分别记为 AABBCC。对这三种因素中的每一种,随机选中一名男子只具有这一种风险因素(且不具有另外两种)的概率都是 0.10.1。对任意两种风险因素,随机选中一名男子恰好具有这两种风险因素(但不具有第三种)的概率都是 0.140.14。已知一名男子具有 AABB 的条件下,他同时具有三种风险因素的概率为 13\frac{1}{3}。在一名男子不具有风险因素 AA 的条件下,他三种风险因素都不具有的概率为 pq\frac{p}{q},其中 ppqq 是互质正整数。求 p+qp + q

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.

答案:76
知识点:韦恩图条件概率
难度评级:2110
解答:

假设群体中有 100100 名男子并填写维恩图。三个“恰好一种”的区域各有 1010 人,三个“恰好两种”的区域各有 1414 人。若 xx 人具有三种风险因素,则同时具有 AABB 的人数为 x+14x + 14,所以给定的条件概率给出 xx+14=13\frac{x}{x + 14} = \frac{1}{3},从而 x=7x = 7

三个集合的并集中共有 310+314+7=793 \cdot 10 + 3 \cdot 14 + 7 = 79 人,所以有 2121 人没有任何风险因素。 具有风险因素 AA 的人数为 10+14+14+7=4510 + 14 + 14 + 7 = 45,所以不具有 AA 的人数为 5555

所求概率为 2155\frac{21}{55},已经是最简分数,因此 p+q=21+55=76p + q = 21 + 55 = 76

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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