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

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==Solution==
 
==Solution==
  
We first assume a population of 100 to facilitate solving. Then we simply organize the statistics given into a 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.
  
  
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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>.
 
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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== See also ==
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{{AIME box|year=2014|n=II|num-b=1|num-a=3}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 21:33, 20 May 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 doest 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.


Now from "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 $\frac{x}{\frac{1}{3}}=14+x$, so $x=7$. Thus $y=21$.

So our desired probability is $\frac{y}{y+10+14+10}$ which simplifies into $\frac{21}{55}$. 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