$\frac{\left(\begin{array}{l}3 \\ 0\end{array}\right)\left(\frac{10-3}{5-0}\right)}{\left(\begin{array}{c}10 \\ 5\end{array}\right)}$

Step 1 ：First, we need to calculate the combinations. The combination \(\begin{array}{l}n \\ k\end{array}\) can be calculated as \(\frac{n!}{k!(n-k)!}\), where \(n!\) is the factorial of \(n\), which is the product of all positive integers up to \(n\).

Step 2 ：For the combination \(\begin{array}{l}3 \\ 0\end{array}\), we have \(n=3\) and \(k=0\). So, \(\begin{array}{l}3 \\ 0\end{array}\) = \(\frac{3!}{0!(3-0)!}\) = 1.

Step 3 ：For the combination \(\begin{array}{c}10 \\ 5\end{array}\), we have \(n=10\) and \(k=5\). So, \(\begin{array}{c}10 \\ 5\end{array}\) = \(\frac{10!}{5!(10-5)!}\) = 252.

Step 4 ：Next, we need to perform the operations in the fraction \(\frac{10-3}{5-0}\). This simplifies to \(\frac{7}{5}\) = 1.4.

Step 5 ：Finally, we divide the results of the combination and the fraction. So, \(\frac{\left(\begin{array}{l}3 \\ 0\end{array}\right)\left(\frac{10-3}{5-0}\right)}{\left(\begin{array}{c}10 \\ 5\end{array}\right)}\) = \(\frac{1*1.4}{252}\) = 0.005555555555555555.

Step 6 ：So, the final answer is \(\boxed{0.005555555555555555}\).