1. Key Concepts and Formulas

• Relation: A relation $R$ from set $A$ to set $B$ is any subset of the Cartesian product $A \times B$. If $A=B$, it's a relation on set $A$.
• Total Number of Relations: If $|A|=m$ and $|B|=n$, then $|A \times B|=mn$. The total number of possible relations from $A$ to $B$ is $2^{mn}$. For a relation on set $A$, it is $2^{|A|^2}$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for all $a, b \in A$, whenever $(a, b) \in R$, then $(b, a) \in R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if for all $a, b, c \in A$, whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.
• Probability Formula: The probability of an event is given by: $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

2. Step-by-Step Solution

Step 1: Determine the total number of possible relations.

• Identify the sets: The relation $R$ is from the set $A = \{x, y\}$ to itself. So, $R \subseteq A \times A$.
• Calculate the Cartesian product $A \times A$: This set consists of all possible ordered pairs $(a, b)$ where $a, b \in A$. $A \times A = \{(x, x), (x, y), (y, x), (y, y)\}$ The number of elements in $A \times A$ is $|A \times A| = 2 \times 2 = 4$.
• Calculate the total number of possible relations: A relation is any subset of $A \times A$. If a set has $n$ elements, it has $2^n$ subsets. $\text{Total number of relations} = 2^{|A \times A|} = 2^4 = 16$ This forms the denominator of our probability.

Step 2: Identify and count relations that are both symmetric and transitive.

We need to systematically enumerate all relations on $A = \{x, y\}$ and check for both properties. Let the elements of $A \times A$ be $e_1=(x,x)$, $e_2=(x,y)$, $e_3=(y,x)$, $e_4=(y,y)$.

For a relation to be symmetric, if it contains $(a,b)$, it must also contain $(b,a)$. Specifically, if $(x,y) \in R$, then $(y,x)$ must also be in $R$. If $(y,x) \in R$, then $(x,y)$ must also be in $R$. The pairs $(x,x)$ and $(y,y)$ are self-symmetric.

We can categorize relations based on the presence of the pair $(x,y)$:

• Case A: Relations that do NOT contain the pair $(x, y)$. If $(x, y) \notin R$, then for symmetry, $(y, x)$ must also not be in $R$ (or if it were, it would imply $(x,y)$ must be in $R$, leading to a contradiction). So, in this case, neither $(x,y)$ nor $(y,x)$ are present in $R$. The relations can only be formed using $e_1=(x,x)$ and $e_4=(y,y)$. There are $2^2 = 4$ such relations:

  1. $R_1 = \emptyset$ (the empty relation)

     • Symmetric? Yes (vacuously true, as there are no pairs $(a,b)$ to contradict the condition).
     • Transitive? Yes (vacuously true, as there are no pairs $(a,b)$ and $(b,c)$ to check).
     • Result: Symmetric and Transitive.

  2. $R_2 = \{(x, x)\}$

     • Symmetric? Yes (only $(x,x)$ is present, and its symmetric pair is $(x,x)$ itself, which is in $R_2$).
     • Transitive? Yes (the only sequence to check is $(x,x) \in R_2$ and $(x,x) \in R_2$, which implies $(x,x) \in R_2$. This holds).
     • Result: Symmetric and Transitive.

  3. $R_3 = \{(y, y)\}$

     • Symmetric? Yes (similar to $R_2$).
     • Transitive? Yes (similar to $R_2$).
     • Result: Symmetric and Transitive.

  4. $R_4 = \{(x, x), (y, y)\}$

     • Symmetric? Yes (both $(x,x)$ and $(y,y)$ are self-symmetric).
     • Transitive? Yes (no sequences like $(a,b), (b,c)$ exist where $a \ne b$ or $b \ne c$. The self-loops $(x,x)$ and $(y,y)$ satisfy transitivity on their own).
     • Result: Symmetric and Transitive.

  So far, we have found 4 relations that are both symmetric and transitive.

• Case B: Relations that DO contain the pair $(x, y)$. For such a relation to be symmetric, it must also contain the pair $(y, x)$. So, $R$ must include the set $\{(x, y), (y, x)\}$. Now, we consider the possible inclusions of the remaining elements: $(x, x)$ and $(y, y)$. There are $2^2 = 4$ such relations:

  1. $R_5 = \{(x, y), (y, x)\}$

     • Symmetric? Yes (by construction, $(x,y)$'s symmetric pair $(y,x)$ is present, and vice-versa).
     • Transitive? No.
       • Consider $(x, y) \in R_5$ and $(y, x) \in R_5$. For transitivity, $(x, x)$ must be in $R_5$. It is not.
       • Consider $(y, x) \in R_5$ and $(x, y) \in R_5$. For transitivity, $(y, y)$ must be in $R_5$. It is not.
     • Result: Symmetric but NOT Transitive.

  2. $R_6 = \{(x, y), (y, x), (x, x)\}$

     • Symmetric? Yes (by construction).
     • Transitive? No.
       • Consider $(y, x) \in R_6$ and $(x, y) \in R_6$. For transitivity, $(y, y)$ must be in $R_6$. It is not.
       • (Note: $(x,y)$ and $(y,x)$ implies $(x,x)$ which is present, but the other implication $(y,x)$ and $(x,y)$ implying $(y,y)$ fails).
     • Result: Symmetric but NOT Transitive.

  3. $R_7 = \{(x, y), (y, x), (y, y)\}$

     • Symmetric? Yes (by construction).
     • Transitive? No.
       • Consider $(x, y) \in R_7$ and $(y, x) \in R_7$. For transitivity, $(x, x)$ must be in $R_7$. It is not.
     • Result: Symmetric but NOT Transitive.

  4. $R_8 = \{(x, y), (y, x), (x, x), (y, y)\}$ (This is $A \times A$, the universal relation)

     • Symmetric? Yes (as it contains all pairs, it is trivially symmetric).
     • Transitive? Yes (If $R$ contains all possible pairs, then for any $(a, b) \in R$ and $(b, c) \in R$, $(a, c)$ will always be in $R$ because all pairs are in $R$).
     • Result: Symmetric and Transitive.

Combining the results from Case A and Case B: Number of symmetric and transitive relations = (4 from Case A) + (1 from Case B) = 5. These 5 relations are: $\emptyset$, $\{(x,x)\}$, $\{(y,y)\}$, $\{(x,x), (y,y)\}$, and $\{(x,x), (x,y), (y,x), (y,y)\}$.

Step 3: Calculate the probability.

Using the probability formula: $P(\text{R is symmetric and transitive}) = \frac{\text{Number of favorable relations}}{\text{Total number of relations}} = \frac{5}{16}$

3. Common Mistakes & Tips

• Vacuously True Conditions: Remember that properties like symmetry and transitivity are "vacuously true" for relations that don't have the prerequisite conditions. For example, the empty set is always symmetric and transitive.
• Transitivity with Inverse Pairs: A common error is overlooking the implication of $(a,b) \in R$ and $(b,a) \in R$. For transitivity to hold in this scenario, both $(a,a) \in R$ and $(b,b) \in R$ must be true. This was critical in identifying non-transitive relations in Case B.
• Systematic Enumeration: For small sets, a systematic listing and checking of relations, perhaps by grouping them based on key properties (like the presence of specific pairs), helps avoid missing cases or making errors.
• Universal Relation: The universal relation (containing all possible pairs) is always reflexive, symmetric, and transitive.

4. Summary

To find the probability, we first determined the total number of relations on the set $\{x,y\}$ by considering all subsets of its Cartesian product, which yielded 16 relations. Next, we systematically identified relations that satisfy both symmetry and transitivity. This involved two main cases: relations not containing $(x,y)$ (and thus $(y,x)$), and relations containing both $(x,y)$ and $(y,x)$. We found 4 relations in the first case and 1 in the second, totaling 5 favorable relations. Finally, dividing the number of favorable relations by the total number of relations gave the probability.

5. Final Answer

The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive is $\frac{5}{16}$.

The final answer is $\boxed{\frac{5}{16}}$, which corresponds to option (A).