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

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==Problem==
 
==Problem==
  
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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>.
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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 does 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==
 
==Solution==
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We first assume a population of <math>100</math> to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
 
We first assume a population of <math>100</math> to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
  
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<asy>
  
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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>.
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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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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>.
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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>
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 +
Let <math>x</math> 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 <math>\frac{1}{3}</math>," we can tell that <math>x = \frac{1}{3}(x+14)</math>, since there are <math>x</math> people with all three factors and 14 with only A and B.  Thus <math>x=7</math>.
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Let <math>y</math> be the number of men with no risk factors.  It now follows that
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<cmath> y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21. </cmath>
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The number of men with risk factor A is <math>10+2 \cdot 14+7 = 45</math> (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 <math>A</math> is 55, so the desired conditional probability is <math>21/55</math>. So the answer is <math>21+55=\boxed{076}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 14:19, 28 February 2022

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 $p+q$.

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]

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

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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