2015 AIME I Problems/Problem 5
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 $\left($ (Error compiling LaTeX. Unknown error_msg)which occurs with probability $\dfrac{2}{6} = \dfrac{1}{3}\right)$ (Error compiling LaTeX. Unknown error_msg), then the fourth sock can be arbitrary. Otherwise $\left($ (Error compiling LaTeX. Unknown error_msg)with probability $\dfrac{2}{3}\right)$ (Error compiling LaTeX. Unknown error_msg), 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
The sum is therefore
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
.
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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