2015 AIME I Problem 5

Below is the professionally curated solution for Problem 5 of the 2015 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AIME I solutions, or check the answer key.

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Concepts:basic probabilitysampling without replacementsymmetry

Difficulty rating: 2510

5.

In a drawer Sandy has 55 pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 1010 socks in the drawer. On Tuesday Sandy selects 22 of the remaining 88 socks at random and on Wednesday two of the remaining 66 socks at random. The probability that Wednesday is the first day Sandy selects matching socks is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

Imagine dealing all ten socks out two per day for five days; every assignment of unordered pairs to days is equally likely, and permuting the days does not change this distribution. Swapping Monday and Wednesday therefore shows that the desired probability (mismatch, mismatch, match) equals the probability of a match on Monday followed by mismatches on Tuesday and Wednesday.

That pattern is easy to compute in order. Monday matches with probability 19\frac{1}{9} (the second sock must be the first sock's mate). The remaining 88 socks then form 44 complete pairs, so Tuesday mismatches with probability 14(82)=67.1 - \frac{4}{\binom{8}{2}} = \frac{6}{7}. Tuesday's mismatch breaks two pairs, leaving 22 complete pairs among the 66 remaining socks, so Wednesday mismatches with probability 12(62)=1315.1 - \frac{2}{\binom{6}{2}} = \frac{13}{15}.

The probability is 19671315=26315,\frac{1}{9} \cdot \frac{6}{7} \cdot \frac{13}{15} = \frac{26}{315}, so m+n=26+315=341.m + n = 26 + 315 = 341.

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