某种产品共5件，其中有2件次品，3件正品，从中任取3件，设X表示取出的3件产品中次品的个数，求(1)X的分布律;(2)X的分布函数F(x);(3)期望E(X);(4)方差D(X).

(1) **分布律** $X$表示取出的次品数，可能取值为0, 1, 2。 \[ P(X=0) = \frac{C_2^0 \cdot C_3^3}{C_5^3} = \frac{1}{10}, \quad P(X=1) = \frac{C_2^1 \cdot C_3^2}{C_5^3} = \frac{3}{5}, \quad P(X=2) = \frac{C_2^2 \cdot C_3^1}{C_5^3} = \frac{3}{10} \] 分布律为： \[ \boxed{ \begin{array}{c|c} X & 0 & 1 & 2 \\ \hline P & \frac{1}{10} & \frac{3}{5} & \frac{3}{10} \\ \end{array} } \] (2) **分布函数** \[ F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{10} & 0 \leq x < 1 \\ \frac{7}{10} & 1 \leq x < 2 \\ 1 & x \geq 2 \end{cases} \] (3) **期望** \[ E(X) = 0 \times \frac{1}{10} + 1 \times \frac{3}{5} + 2 \times \frac{3}{10} = \frac{6}{5} \] \[ \boxed{\frac{6}{5}} \] (4) **方差** \[ E(X^2) = 0^2 \times \frac{1}{10} + 1^2 \times \frac{3}{5} + 2^2 \times \frac{3}{10} = \frac{9}{5} \] \[ D(X) = E(X^2) - [E(X)]^2 = \frac{9}{5} - \left(\frac{6}{5}\right)^2 = \frac{9}{25} \] \[ \boxed{\frac{9}{25}} \]