The probability that a randomly chosen $$2 \times 2$$ matrix with all the entries from the set of first 10 primes, is singular, is equal to

The first 10 primes are $$\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$$.

A $$2 \times 2$$ matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ is singular when $$ad = bc$$.

Total number of matrices = $$10^4$$.

Step 1: Count the number of ordered pairs $$(a, d)$$ and $$(b, c)$$ from the set such that $$ad = bc$$.

For each product value $$p$$, let $$f(p)$$ be the number of ordered pairs from the set with that product. The number of favorable outcomes is $$\sum f(p)^2$$.

Same element pairs (e.g., $$(2,2), (3,3), \ldots$$): There are 10 products ($$4, 9, 25, \ldots$$), each with $$f = 1$$.

Contribution: $$10 \times 1^2 = 10$$.

Different element pairs (e.g., $$(2,3)$$ and $$(3,2)$$): Since all 10 numbers are distinct primes, each product of two different primes is unique. There are $$\binom{10}{2} = 45$$ such products, each with $$f = 2$$.

Contribution: $$45 \times 2^2 = 180$$.

Note: No product of two distinct primes can equal a perfect square of a prime, and no two different pairs of distinct primes can give the same product (by unique factorization). So all products are distinct.

Total favorable = $$10 + 180 = 190$$.

Probability = $$\frac{190}{10^4} = \frac{19}{10^3}$$.

The answer is Option B: $$\dfrac{19}{10^3}$$.