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>. | ||
| − | + | ==Solution 1== | |
| − | |||
Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is <math>\dfrac{1}{9}</math>. | Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is <math>\dfrac{1}{9}</math>. | ||
Revision as of 19:23, 2 August 2018
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 
.
Solution 1
Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is 
.
Assuming this, then 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 
Solution 2
The key is to count backwards. First, choose the pair which you pick on Wednesday in ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are 
pairs, the number of ways to do this is 
. Then, there are two pairs and two nonmatching socks for you to pick from on Monday, a total of socks. Since you don't want to pick a pair, the number of ways to do this is 
. Thus the answer is
![\[\dfrac{\left(5\right)\left(\dbinom{8}{2}-4\right)\left(\dbinom{6}{2}-2\right)}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}}=\dfrac{26}{315}.\]](http://latex.artofproblemsolving.com/9/5/c/95cd0ea855b67859a8b01759503cd7963ca12ba6.png)
.
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. 
