<s>[INST] Probability that no $2 \times 2$ red square is formed Each square of a $3 × 3$ board is coloured either red or blue at random (each having probability $1/2$). Then find
the probability that there is no $2 × 2$ red square.
I was trying to calculate favorable cases using inclusion-exclusion principle.
Total ways=$2^9$
Now there can be $4-$  $2 \times 2$ square so I first subtracted $C(4,1) 2^5$ but once 2 of the 4 square colored then number of remaining square starts depending on position of two squares. How to tackle it? Is inclusion-exclusion correct way to proceed or is there a better method? 
Given answer is $\frac{417}{512}$
 [/INST] there is a $\frac {1}{16}$ chance a that any $2\times 2$ block is all red.
There are four $2\times 2$ blocks. 
Now we must exclude the cases where we have double, triple or quadruple counted.
two $2\times 2$ blocks -- adjacent blocks $4\frac {1}{2^6}$ + opposite blocks $ 2\frac {1}{2^7}$
three $2\times 2$ blocks, $4\frac {1}{2^8}$
four $2\times 2$ blocks (all red) $\frac {1}{2^9}$
Chance that there is at least one  $2\times 2$ block $= \frac 1{4}  - \frac 1{16} - \frac 1{2^6} + \frac {1}{2^6} - \frac {1}{2^9} = \frac {95}{512}$
Chance that there no  $2\times 2$ block $1 - \frac {95}{512} = \frac {417}{512}$
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