Problem:
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
Number the squares as shown. For i=1,2,4, and 5 , let Qi​ be the event that i is the upper left corner of a 2-by-2 red square, and let p(E) be the probability that event E will occur. By the Inclusion-Exclusion Principle, the probability that the grid does have at least one 2 -by-2 red square is
147​258​369​​
−+−​p(Q1​)+p(Q2​)+p(Q4​)+p(Q5​)p(Q1​∩Q2​)−p(Q1​∩Q4​)−p(Q2​∩Q5​)−p(Q4​∩Q5​)−p(Q1​∩Q5​)−p(Q2​∩Q4​)p(Q1​∩Q2​∩Q5​)+p(Q1​∩Q2​∩Q4​)+p(Q1​∩Q4​∩Q5​)+p(Q2​∩Q4​∩Q5​)p(Q1​∩Q2​∩Q4​∩Q5​)​
or
4(21​)4−[4(21​)6+2(21​)7]+4(21​)8−(21​)9=51295​
The probability that the grid does not have at least one 2-by-2 red square is therefore 1−95/512=417/512, so m+n=929​.
The problems on this page are the property of the MAA's American Mathematics Competitions