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2014 AIME II Problems/Problem 2: Difference between revisions

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Created page with "==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 th..."
 
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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 <math>\frac{1}{3}</math>. The probability that a man has none of the three risk factors given that he doest not have risk factor A is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
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 <math>\frac{1}{3}</math>. The probability that a man has none of the three risk factors given that he doest not have risk factor A is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
==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 <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>.

Revision as of 10:22, 28 March 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:

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

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