Difference between revisions of "2014 AIME II Problems/Problem 2"
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(→Solution: restoring old Venn diagram) |
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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>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> 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21. </cmath> | + | <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>. | ||
Revision as of 01:16, 8 June 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 . 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.
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 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.