The probability that a relation $$R$$ from $$\{x, y\}$$ to $$\{x, y\}$$ is both symmetric and transitive, is equal to:

We need to find the probability that a relation $$R$$ from $$\{x, y\}$$ to $$\{x, y\}$$ is both symmetric and transitive. Such a relation is a subset of $$\{(x,x), (x,y), (y,x), (y,y)\}$$, so there are $$2^4 = 16$$ possible relations.

Symmetry requires that if $$(a,b)\in R$$ then $$(b,a)\in R$$, and transitivity requires that if $$(a,b)\in R$$ and $$(b,c)\in R$$ then $$(a,c)\in R$$. The symmetric subsets are those where $$(x,y)$$ and $$(y,x)$$ either both appear or both don’t. There are 2 choices for including or excluding $$(x,x)$$, 2 choices for $$(y,y)$$, and 2 choices for the pair $$\{(x,y),(y,x)\}$$, giving $$2^3 = 8$$ symmetric relations.

Checking these for transitivity shows that the empty set, $$\{(x,x)\}$$, $$\{(y,y)\}$$, $$\{(x,x),(y,y)\}$$, and the full set $$\{(x,x),(y,y),(x,y),(y,x)\}$$ are transitive, while the other three fail because having $$(x,y)$$ and $$(y,x)$$ demands inclusion of $$(x,x)$$ or $$(y,y)$$, which they lack. Thus there are 5 relations that are both symmetric and transitive.

Therefore, the probability is $$\f\frac{5}{16}$$, and the answer is $$\boldsymbol{\f\frac{5}{16}}$$.