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. | ||
| + | <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 <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>y</math> be the number of men with no risk factors. It now follows that | ||
| + | <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>. | ||
== See also == | == See also == | ||
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 
. The probability that a man has none of the three risk factors given that he does not have risk factor A is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
We first assume a population of 
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]](http://latex.artofproblemsolving.com/1/c/a/1cabe8198c2c6126c59a6a8a52dc675c24f950b3.png)
Let 
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 ," we can tell that 
, since there are people with all three factors and 14 with only A and B. Thus 
.
Let 
be the number of men with no risk factors. It now follows that
![\[y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21.\]](http://latex.artofproblemsolving.com/4/6/a/46a7e0dd57f5bdfc92811742cf02ad4afc159b57.png)
The number of men with risk factor A is 
(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 
is 55, so the desired conditional probability is 
. So the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
