Difference between revisions of "2004 AIME II Problems/Problem 2"
(→Solution) |
|||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | A jar has 10 red candies and 10 blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the [[probability]] that they get the same color combination, irrespective of order, is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> | + | A jar has <math>10</math> red candies and <math>10</math> blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the [[probability]] that they get the same color combination, irrespective of order, is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> |
== Solution == | == Solution == | ||
| Line 6: | Line 6: | ||
The probability that Terry picks two different candies is <math>\frac{20\cdot10}{20\cdot19} = \frac{10}{19}</math>, and the probability that Mary picks two different candies after Terry picks two different candies is <math>\frac{18\cdot 9}{18\cdot 17} = \frac{9}{17}</math>. Thus, the probability that they both choose two different candies is <math>\frac{10}{19}\cdot\frac{9}{17} = \frac{90}{323}</math>. Then the total probability is | The probability that Terry picks two different candies is <math>\frac{20\cdot10}{20\cdot19} = \frac{10}{19}</math>, and the probability that Mary picks two different candies after Terry picks two different candies is <math>\frac{18\cdot 9}{18\cdot 17} = \frac{9}{17}</math>. Thus, the probability that they both choose two different candies is <math>\frac{10}{19}\cdot\frac{9}{17} = \frac{90}{323}</math>. Then the total probability is | ||
| − | |||
| − | |||
| + | <cmath>2 \cdot \frac{14}{323} + \frac{90}{323} = \frac{118}{323}</cmath> | ||
| − | In the above calculations, we treated the choices as ordered; that is, Terry chose first one candy, then a second, and so on. We could also solve the problem using unordered choices. The probabilities calculated will all be the same, but the calculations will appear somewhat different. For instance, the probability that Mary chooses two red candies after Terry chose two red candies will have the form <math>\frac{{8\choose 2}}{{18 \choose 2}}</math>, and the probability that Terry chooses two different candies will have the form <math>\frac{{10\choose 1}\cdot{10\choose 1}}{{20\choose2}}</math>. It is not difficult that these yield the same results as our calculations above, as we would expect. | + | and so the answer is <math>118 + 323 = \boxed{441}</math>. |
| + | |||
| + | |||
| + | In the above calculations, we treated the choices as ordered; that is, Terry chose first one candy, then a second, and so on. We could also solve the problem using unordered choices. The probabilities calculated will all be the same, but the calculations will appear somewhat different. For instance, the probability that Mary chooses two red candies after Terry chose two red candies will have the form <math>\frac{{8\choose 2}}{{18 \choose 2}}</math>, and the probability that Terry chooses two different candies will have the form <math>\frac{{10\choose 1}\cdot{10\choose 1}}{{20\choose2}}</math>. It is not difficult to see that these yield the same results as our calculations above, as we would expect. | ||
== See also == | == See also == | ||
| Line 16: | Line 18: | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:42, 1 January 2015
Problem
A jar has 
red candies and blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is 
where 
and 
are relatively prime positive integers, find 
Solution
The probability that Terry picks two red candies is 
, and the probability that Mary picks two red candies after Terry chooses two red candies is 
. So the probability that they both pick two red candies is 
. The same calculation works for the blue candies.
The probability that Terry picks two different candies is 
, and the probability that Mary picks two different candies after Terry picks two different candies is 
. Thus, the probability that they both choose two different candies is 
. Then the total probability is
![\[2 \cdot \frac{14}{323} + \frac{90}{323} = \frac{118}{323}\]](http://latex.artofproblemsolving.com/9/e/2/9e281ac310860beb6bc8499faab958586f45fcad.png)
and so the answer is 
.
In the above calculations, we treated the choices as ordered; that is, Terry chose first one candy, then a second, and so on. We could also solve the problem using unordered choices. The probabilities calculated will all be the same, but the calculations will appear somewhat different. For instance, the probability that Mary chooses two red candies after Terry chose two red candies will have the form 
, and the probability that Terry chooses two different candies will have the form 
. It is not difficult to see that these yield the same results as our calculations above, as we would expect.
See also
| 2004 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. 
