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

m (Problem)
(Solution: edited for easier reading (hopefully))
Line 7: Line 7:
 
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.
  
 +
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>.
  
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>.
+
It now follows that the number of men with no risk factors is
 
+
<cmath> 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21. </cmath>
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>.
+
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 ==

Revision as of 01:03, 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 $\frac{1}{3}$. 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.

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 $x=7$.

It now follows that the number of men with no risk factors is \[100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21.\] 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png