Difference between revisions of "2015 AIME I Problems/Problem 5"
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In a drawer Sandy has <math>5</math> pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the <math>10</math> socks in the drawer. On Tuesday Sandy selects <math>2</math> of the remaining <math>8</math> socks at random and on Wednesday two of the remaining <math>6</math> socks at random. The probability that Wednesday is the first day Sandy selects matching socks is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, Find <math>m+n</math>. | In a drawer Sandy has <math>5</math> pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the <math>10</math> socks in the drawer. On Tuesday Sandy selects <math>2</math> of the remaining <math>8</math> socks at random and on Wednesday two of the remaining <math>6</math> socks at random. The probability that Wednesday is the first day Sandy selects matching socks is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, Find <math>m+n</math>. | ||
| + | |||
| + | ==Hint== | ||
| + | Notice that we can allow the sample space of the problem to be the <math>\dfrac{10!}{(2!)^5}</math> possible permutations of socks, and that the desired outcomes are those for which the fifth and sixth items are the same color, the first and second are different colors, and the third and fourth are different colors. | ||
| + | |||
| + | ==Solution== | ||
| + | But this probability is simple to count. Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is <math>\dfrac{1}{9}</math>. Let the first sock be arbitrary; the probability that the second sock does not match is <math>\dfrac{6}{7}.</math> | ||
| + | |||
| + | The only "hard" part is the third and fourth sock. But that is simple casework. If the third sock's color matches the color of one of the first two socks (which occurs with probability <math>\dfrac{2}{6} = \dfrac{1}{3}</math>), then the fourth sock can be arbitrary. Otherwise (with probability <math>\dfrac{2}{3}</math>), the fourth sock can be chosen with probability <math>\dfrac{4}{5}</math> (5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus | ||
| + | <cmath>\frac{1}{9} \cdot \frac{6}{7} \cdot (\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{4}{5}) = \frac{26}{315}.</cmath> | ||
| + | The answer is <math>\dfrac{341}.</math> | ||
== See also == | == See also == | ||
Revision as of 20:49, 20 March 2015
Contents
Problem
In a drawer Sandy has 
pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 
socks in the drawer. On Tuesday Sandy selects 
of the remaining 
socks at random and on Wednesday two of the remaining 
socks at random. The probability that Wednesday is the first day Sandy selects matching socks is 
, where 
and 
are relatively prime positive integers, Find 
.
Hint
Notice that we can allow the sample space of the problem to be the 
possible permutations of socks, and that the desired outcomes are those for which the fifth and sixth items are the same color, the first and second are different colors, and the third and fourth are different colors.
Solution
But this probability is simple to count. Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is 
. Let the first sock be arbitrary; the probability that the second sock does not match is 
The only "hard" part is the third and fourth sock. But that is simple casework. If the third sock's color matches the color of one of the first two socks (which occurs with probability 
), then the fourth sock can be arbitrary. Otherwise (with probability 
), the fourth sock can be chosen with probability 
(5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus
![\[\frac{1}{9} \cdot \frac{6}{7} \cdot (\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{4}{5}) = \frac{26}{315}.\]](http://latex.artofproblemsolving.com/b/9/0/b90a6d0291d23f9fa4b506e86524458bbd43b911.png)
The answer is 
See also
| 2015 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
