The set $$A=\{1,2,3\}$$ has $$|A|=3$$ elements, so the Cartesian product $$A\times A$$ contains $$3\times3=9$$ ordered pairs.
Any relation on $$A$$ is a subset of these nine ordered pairs.
Total number of relations $$=2^{9}=512.$$
Step 1: Count the symmetric relations.
For symmetry, whenever $$(a,b)$$ is included, $$(b,a)$$ must also be included.
• Diagonal pairs $$(1,1),(2,2),(3,3)$$ are independent; each can be either present or absent.
Number of choices for the diagonal $$=2^{3}$$.
• Off-diagonal ordered pairs come in three unordered pairs: $$\{(1,2),(2,1)\}, \{(1,3),(3,1)\}, \{(2,3),(3,2)\}$$.
For each unordered pair we either include both ordered pairs or exclude both, giving $$2$$ choices per unordered pair.
Number of choices for the off-diagonal $$=2^{3}$$.
Hence, total symmetric relations $$=2^{3}\times2^{3}=2^{6}=64.$$
Step 2: Count the relations that are both symmetric and reflexive.
Reflexivity requires all three diagonal pairs to be included, leaving no choice there.
• Diagonal: exactly $$1$$ way (all present).
• Off-diagonal: still $$2^{3}=8$$ ways.
So, symmetric & reflexive relations $$=1\times8=8.$$
Step 3: Symmetric but not reflexive relations.
$$\text{Required count}=64-8=56.$$
Step 4: Probability.
$$\text{Probability}= \frac{56}{512}= \frac{7}{64}.$$
Option B which is: $$\dfrac{7}{64}$$
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