2001 AIME II Problems/Problem 9
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 , where and are relatively prime positive integers. Find .
We can use complementary counting, counting all of the colorings that have at least one red square.
- For at least one red square:
- There are four squares to choose which one will be red. Then there are ways to color the rest of the squares.
- For at least two squares:
- There are two cases: those with two red squares on one side and those without red squares on one side.
- The first case is easy: 4 ways to choose which the side the squares will be on, and ways to color the rest of the squares, so 32 ways to do that. For the second case, there will by only two ways to pick two squares, and ways to color the other squares.
- For at least three squares:
- Choosing three such squares leaves only one square left, with four places to place it. This is ways.
- For at least four squares, we clearly only have one way.
By the Principle of Inclusion-Exclusion, there are (alternatively subtracting and adding) ways to have at least one red square.
There are ways to paint the square with no restrictions, so there are ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a red square is , and .
We consider how many ways we can have 2*2 grid
: All the girds are red-- case
: One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid cases
: Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, cases. Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. cases
Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are
Four unit squares are blue, no overcomunted case will be considered. there are
Five unit squares are blue, cases in all
Sum up those cases, there are cases that a 2*2 grid can be formed.
In all, there are possible ways to paint the big square, so the answer is leads to
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