Question

A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

Solution:

The combinations match exactly when both draw two reds, both draw two blues, or both draw one candy of each color. The probability that Terry draws two reds is $\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38},$ after which $8$ reds and $10$ blues remain, so Mary draws two reds with probability $\frac{\binom{8}{2}}{\binom{18}{2}} = \frac{28}{153}.$ That case has probability $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323},$ and by symmetry two blues each is also $\frac{14}{323}.$

For mixed draws, Terry succeeds with probability $\frac{10 \cdot 10}{\binom{20}{2}} = \frac{10}{19},$ leaving $9$ of each color, and Mary with probability $\frac{9 \cdot 9}{\binom{18}{2}} = \frac{9}{17},$ for $\frac{10}{19} \cdot \frac{9}{17} = \frac{90}{323}.$

The total is $\frac{14 + 14 + 90}{323} = \frac{118}{323}.$ Since $118 = 2 \cdot 59$ and $323 = 17 \cdot 19,$ the fraction is in lowest terms, and $m + n = 118 + 323 = 441.$

2004 AIME II Problem 2

Solution:

Problem 2 in Other Years