Each unit square of a 3 by 3 unit grid is to be cloured either green or blue for each square, either colour is equally likely. If the probability of obtaining a grid that does not have any 2 by 2 green square is  $\frac{\text{p}}{\text{q}}$ , where  $\text{p,\hspace{0.17em}q\hspace{0.17em}}\in \text{\hspace{0.17em}N}$  and  $\text{p,\hspace{0.17em}q\hspace{0.17em}}$  are coprime, then the value of  $|\text{p\hspace{0.17em}}-\text{\hspace{0.17em}q\hspace{0.17em}}|$

Let  $A_1,A_2,A_3,A_4$  be events that left top corner, right top corner, left bottom corner, right bottom corner square is coloured green and is part of  $2\times 2$  square with all faces green, respectively. By principle of inclusion and exclusion $\Rightarrow \text{\hspace{0.17em}P}({\text{A}}_{\text{1}}\cup {\text{\hspace{0.17em}A}}_{\text{2}}\cup {\text{A}}_{\text{3}}\cup {\text{A}}_{\text{4}})=\frac{95}{512}\Rightarrow \text{\hspace{0.17em}P}({\text{A}}_{\text{1}}^{\text{c}}\cap {\text{\hspace{0.17em}A}}_{\text{2}}^{\text{c}}\cap {\text{\hspace{0.17em}A}}_{\text{3}}^{\text{c}}\cap {\text{\hspace{0.17em}A}}_{\text{4}}^{\text{c}})=\frac{417}{512}$
  so   $|p-q|=95$