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Difference between revisions of "2014 AIME II Problems/Problem 2"

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(→‎Solution)
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==Solution==
 
==Solution==
  
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We first assume a population of 100 to facilitate solving. Then we simply organize the statistics given into a Venn diagram:
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We first assume a population of 100 to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
  
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<asy>
 
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pair A,B,C,D,E,F,G;
 
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A=(0,55);
 
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B=(60,55);
 
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C=(60,0);
 
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D=(0,0);
 
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draw(A--B--C--D--A);
 
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E=(30,35);
 
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F=(20,20);
 
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G=(40,20);
 
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draw(circle(E,15));
 
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draw(circle(F,15));
 
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draw(circle(G,15));
 
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draw("$A$",(30,52));
 
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draw("$B$",(7,7));
 
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draw("$C$",(53,7));
 
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draw("100",(5,60));
 
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draw("10",(30,40));
 
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draw("10",(15,15));
 
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draw("10",(45,15));
 
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draw("14",(30,16));
 
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draw("14",(38,29));
 
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draw("14",(22,29));
 
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draw("$x$",(30,25));
 
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draw("$y$",(10,45));
 
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</asy>
 
  
 
Now from "The probability that a randomly selected man has all three risk factors, given that he has A and B is <math>\frac{1}{3}</math>." we can tell that <math>\frac{x}{\frac{1}{3}}=14+x</math>, so <math>x=7</math>. Thus <math>y=21</math>.
 
Now from "The probability that a randomly selected man has all three risk factors, given that he has A and B is <math>\frac{1}{3}</math>." we can tell that <math>\frac{x}{\frac{1}{3}}=14+x</math>, so <math>x=7</math>. Thus <math>y=21</math>.
  
 
So our desired probability is <math>\frac{y}{y+10+14+10}</math> which simplifies into <math>\frac{21}{55}</math>. So the answer is <math>21+55=\boxed{076}</math>.
 
So our desired probability is <math>\frac{y}{y+10+14+10}</math> which simplifies into <math>\frac{21}{55}</math>. So the answer is <math>21+55=\boxed{076}</math>.

Revision as of 21:33, 1 April 2014

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.


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 $21+55=\boxed{076}$.

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