2015 AIME I Problems/Problem 5

Revision as of 17:49, 31 July 2020 by Rj5303707 (talk | contribs) (Solution 1)

Problem

In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, Find $m+n$.

Solution 1

Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\dfrac{1}{9}$.

Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\dfrac{6}{7}.$

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 $\dfrac{4}{5}$ (5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus \[\frac{1}{9} \cdot \frac{6}{7} \cdot \left(\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{4}{5}\right) = \frac{26}{315}.\] The sum is therefore $26+315=\boxed{341}.$

Solution 2

The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ 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 $4$ pairs, the number of ways to do this is $\dbinom{8}{2}-4$. Then, there are two pairs and two nonmatching socks for you to pick from on Monday, a total of $6$ socks. Since you don't want to pick a pair, the number of ways to do this is $\dbinom{6}{2}-2$. Thus the answer is \[\dfrac{\left(5\right)\left(\dbinom{8}{2}-4\right)\left(\dbinom{6}{2}-2\right)}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}}=\dfrac{26}{315}.\] $26 + 315 = \boxed{341}$.

See also

2015 AIME I (ProblemsAnswer KeyResources)
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

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