The idea is same , just a different combinatoric approach (as it a counting problem only) :
# : no of (shorthand)
#reflexive relations = #supersets of identity relation on set A = {1,2,3} (all reflexive relations are just supersets of identity relation only on any set)
identity relation, I = {(1,1),(2,2),(3,3)} (they are fixed to appear in every reflexive relation , therefore these 3 pairs will create only 1 choice )
maximum #pairs in relation R = |AxA| = 3x3 = 9 (this is the biggest relation AxA only)
#binary relations on A = #subsets of AxA = $2^{9}$ = 512 (these are all possible binary relations on set A )
now the choices are created by non-reflexive pairs : (x,y) such that x, y are distinct (every pair appear in reflexive relation or not – 2 choices)
#non-reflexive pairs = 3 x 2 = 6 (#permutations with non-repetitions)
#reflexive relations on A = 1x$2^{6}$ (1 is fixed for reflexive pairs and 2^6 for all subsets of non-reflexive pairs )
therefore,
P(R is reflexive) = #reflexive rel / #binary relations on A = $\frac{2^{6}}{2^{9}}$ = 1/8 = 0.125