The numbers $1,2,\ldots,9$ are arranged so that the $1$st term is not $1$ and the $9$th term is not $9$. What is the probability that the third term is $3$?

$\text{(A) }\frac{17}{75}\qquad\text{(B) }\frac{43}{399}\qquad\text{(C) }\frac{127}{401}\qquad\text{(D) }\frac{16}{19}\qquad\text{(E) }\frac{6}{7}$
1. **Calculate the total number of valid arrangements:**
   - We need to count the number of permutations of the numbers \(1, 2, \ldots, 9\) such that the 1st term is not 1 and the 9th term is not 9.
   - The total number of permutations of 9 distinct numbers is \(9!\).
   - We use the principle of inclusion-exclusion (PIE) to exclude the invalid cases:
     - Let \(A\) be the set of permutations where the 1st term is 1.
     - Let \(B\) be the set of permutations where the 9th term is 9.
     - We need to find \(|A \cup B|\), the number of permutations where either the 1st term is 1 or the 9th term is 9.
     - By PIE, \(|A \cup B| = |A| + |B| - |A \cap B|\).
     - \(|A| = 8!\) (fixing 1 at the 1st position and permuting the remaining 8 numbers).
     - \(|B| = 8!\) (fixing 9 at the 9th position and permuting the remaining 8 numbers).
     - \(|A \cap B| = 7!\) (fixing 1 at the 1st position and 9 at the 9th position, and permuting the remaining 7 numbers).
     - Therefore, the number of valid permutations is:
       \[
       9! - (8! + 8! - 7!) = 9! - 2 \cdot 8! + 7!
       \]
       \[
       = 362880 - 2 \cdot 40320 + 5040 = 362880 - 80640 + 5040 = 287280
       \]

2. **Calculate the number of valid arrangements with the 3rd term being 3:**
   - We fix the 3rd term as 3 and count the number of valid permutations of the remaining 8 numbers such that the 1st term is not 1 and the 9th term is not 9.
   - The total number of permutations of the remaining 8 numbers is \(8!\).
   - Using PIE again:
     - Let \(A'\) be the set of permutations where the 1st term is 1.
     - Let \(B'\) be the set of permutations where the 9th term is 9.
     - We need to find \(|A' \cup B'|\), the number of permutations where either the 1st term is 1 or the 9th term is 9.
     - By PIE, \(|A' \cup B'| = |A'| + |B'| - |A' \cap B'|\).
     - \(|A'| = 7!\) (fixing 1 at the 1st position and permuting the remaining 7 numbers).
     - \(|B'| = 7!\) (fixing 9 at the 9th position and permuting the remaining 7 numbers).
     - \(|A' \cap B'| = 6!\) (fixing 1 at the 1st position and 9 at the 9th position, and permuting the remaining 6 numbers).
     - Therefore, the number of valid permutations is:
       \[
       8! - (7! + 7! - 6!) = 8! - 2 \cdot 7! + 6!
       \]
       \[
       = 40320 - 2 \cdot 5040 + 720 = 40320 - 10080 + 720 = 30960
       \]

3. **Calculate the probability:**
   - The probability that the 3rd term is 3 is the ratio of the number of valid arrangements with the 3rd term being 3 to the total number of valid arrangements.
   \[
   \text{Probability} = \frac{30960}{287280} = \frac{43}{399}
   \]

The final answer is \( \boxed{ \frac{43}{399} } \).