Let the matrix be $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where each entry is chosen from the set of first $10$ primes: $S=\{2,3,5,7,11,13,17,19,23,29\}$

Since each of the $4$ entries can be chosen independently in $10$ ways, the total number of matrices is $10^4=10000.$

We need the probability that the matrix is singular.

1. Condition for singularity

A $2\times 2$ matrix is singular iff its determinant is zero: $ad-bc=0$ that is, $ad=bc.$

So we must count the number of ordered quadruples $(a,b,c,d)\in S^4$ such that $ad=bc.$

2. Use prime factorization structure

All entries are prime numbers.

If $ad=bc,$ then the product of the two primes on the left equals the product of the two primes on the right.

By unique factorization, the multiset of primes in $(a,d)$ must be the same as the multiset of primes in $(b,c)$.

So the only possibilities are:

1. $a=b$ and $d=c$, or
2. $a=c$ and $d=b$.

Equivalently, once $(a,d)$ is chosen, $(b,c)$ must be either $(a,d)$ or $(d,a)$.

3. Count singular matrices carefully

We count ordered pairs $(a,d)$ first.

There are $10\times 10=100$ choices for $(a,d)$.

Case 1: $a\neq d$

Number of such pairs: $10\cdot 9=90.$

For each such pair, $(b,c)$ can be chosen in exactly $2$ ways: $(b,c)=(a,d) \quad \text{or} \quad (d,a).$ So contribution is $90\cdot 2=180.$

Case 2: $a=d$

Number of such pairs: $10.$

Then the two possibilities for $(b,c)$ coincide, since $(a,d)=(d,a).$ So for each such pair there is only $1$ valid choice: $(b,c)=(a,a).$ Thus contribution is $10\cdot 1=10.$

Hence total number of singular matrices is $180+10=190.$

4. Compute probability

Therefore, $P(\text{singular})=\frac{190}{10^4}=\frac{19}{1000}.$

5. Compare with options

$\frac{19}{1000}$ corresponds to Option C.

Final Answer

$\boxed{\frac{19}{10^3}}$