The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is :

Given, set P = {1, 2, 3, 4, 5}<br/><br/>Let the two subsets be A and B<br/><br/>Then, n (A $\cap$ B) = 2 (as given in question) <br/><br/>We can choose two elements from set P in ⁵C₂ ways. <br/><br/>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. <br/><br/>$\therefore$ Total ways for the three elements = 3 $\times$ 3 $\times$ 3 = 3³ <br/><br/>$\therefore$ Required probability = ${{{}^5{C_2} \times {3^3}} \over {\left( {{2^5}} \right)\left( {{2^5}} \right)}}$ = $${{{}^5{C_2} \times {3^3}} \over {{4^5}}} = {{10 \times 27} \over {{2^{10}}}} = {{135} \over {{2^9}}}$$