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. | + | 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>. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2014|n=II|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 20:33, 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 
. The probability that a man has none of the three risk factors given that he doest 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.
Now from "The probability that a randomly selected man has all three risk factors, given that he has A and B is ." we can tell that 
, so 
. Thus 
.
So our desired probability is 
which simplifies into 
. 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 | ||
