28. From the six digits $0,1,2, 3, 4,5$, select 2 odd numbers and 2 even numbers to form a 4-digit number without repeated digits. What is the probability of forming a 4-digit even number?
28. The difficulty of this problem lies in the digit "0", so we can use the binary method, dividing the problem into two categories: "taking 0" and "not taking 0".

First category: Taking 0, there are $C_{3}^{2} \cdot C_{2}^{1}$ different ways to choose, each way corresponds to $A_{3}^{1} \cdot A_{3}^{3}$ different 4-digit numbers without repeated digits, and corresponds to $A_{3}^{3}+A_{2}^{1} \cdot A_{2}^{2}$ different 4-digit even numbers without repeated digits. At this time, a total of $n_{1}=C_{3}^{2} \cdot C_{2}^{1} \cdot A_{3}^{1} \cdot A_{3}^{3}=108$ different 4-digit numbers without repeated digits can be formed, and $m_{1}=C_{3}^{2} \cdot C_{2}^{1} \cdot\left(A_{3}^{3}+A_{2}^{1} \cdot A_{2}^{2}\right)=60$ different 4-digit even numbers without repeated digits can be formed;

Second category: Not taking 0, there are $C_{3}^{2} \cdot C_{2}^{2}$ different ways to choose, each way corresponds to $A_{4}^{4}$ different 4-digit numbers without repeated digits, and corresponds to $A_{2}^{1} \cdot$ $A_{3}^{3}$ different 4-digit even numbers without repeated digits. At this time, a total of $n_{2}=C_{3}^{2} \cdot C_{2}^{2} \cdot A_{4}^{4}=72$ different 4-digit numbers without repeated digits can be formed, and $m_{2}=$ $C_{3}^{2} \cdot C_{2}^{2} \cdot A_{2}^{1} \cdot A_{3}^{3}=36$ different 4-digit even numbers without repeated digits can be formed; From the above, we get,
$$
P=\frac{m_{1}+m_{2}}{n_{1}+n_{2}}=\frac{60+36}{108+72}=\frac{8}{15} .
$$