$\frac{\left(\begin{array}{l}3 \\ 1\end{array}\right)\left(\begin{array}{c}10-3 \\ 5-1\end{array}\right)}{\left(\begin{array}{c}10 \\ 5\end{array}\right)}$

Step 1 ：First, we need to calculate the combinations separately. The formula for combination is \(\frac{n!}{r!(n-r)!}\), where \(n\) is the total number of items, and \(r\) is the number of items to choose.

Step 2 ：Calculate the first combination \(\left(\begin{array}{l}3 \\ 1\end{array}\right)\), which equals to 3.

Step 3 ：Calculate the second combination \(\left(\begin{array}{c}10-3 \\ 5-1\end{array}\right)\), which equals to 35.

Step 4 ：Calculate the third combination \(\left(\begin{array}{c}10 \\ 5\end{array}\right)\), which equals to 252.

Step 5 ：Substitute these values into the original equation: \(\frac{3 \times 35}{252}\), which simplifies to approximately 0.4166666666666667.

Step 6 ：Final Answer: \(\boxed{0.4166666666666667}\)