2014 AIME II Problems/Problem 2

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 doest not have risk factor A is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

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

$[asy] pair A,B,C,D,E,F,G; A=(0,55); B=(60,55); C=(60,0); D=(0,0); draw(A--B--C--D--A); E=(30,35); F=(20,20); G=(40,20); draw(circle(E,15)); draw(circle(F,15)); draw(circle(G,15)); draw("$A$",(30,52)); draw("$B$",(7,7)); draw("$C$",(53,7)); draw("100",(5,60)); draw("10",(30,40)); draw("10",(15,15)); draw("10",(45,15)); draw("14",(30,16)); draw("14",(38,29)); draw("14",(22,29)); draw("$x$",(30,25)); draw("$y$",(10,45)); [/asy]$

Now from "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 $\frac{x}{\frac{1}{3}}=14+x$ , so $x=7$ . Thus $y=21$ .

So our desired probability is $\frac{y}{y+10+14+10}$ which simplifies into $\frac{21}{55}$ . So the answer is $21+55=\boxed{076}$ .

Page

Toolbox

Search

2014 AIME II Problems/Problem 2

Problem

Solution