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

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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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 +
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>
  
 
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>.
 
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>.
  
It now follows that the number of men with no risk factors is
+
Let <math>y</math> be the number of men with no risk factors.  It now follows that
<cmath> 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21. </cmath>
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<cmath> y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21. </cmath>
 
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>.
 
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>.
  

Latest revision as of 15:19, 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 (ProblemsAnswer KeyResources)
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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