Official solution
Given, set P = {1, 2, 3, 4, 5}
Let the two subsets be A and B
Then, n (A \[\cap\] B) = 2 (as given in question)
We can choose two elements from set P in 5C2 ways.
After choosing two common elements for set A and B, each of remaining three elements from set P have three choice (1) It can go to set A (2) It can go to set B (3) It don't go to any sets it stays at set P.
\[ \therefore \] Total ways for the three elements = 3 \[ \times \] 3 \[ \times \] 3 = 33
\[\therefore\] Required probability = \[{{{}^5{C_2} \times {3^3}} \over {\left( {{2^5}} \right)\left( {{2^5}} \right)}}\] = \[{{{}^5{C_2} \times {3^3}} \over {{4^5}}} = {{10 \times 27} \over {{2^{10}}}} = {{135} \over {{2^9}}}\]