2.设10件产品中有7件正品、3件次品，随机地抽取产品，每次抽取1到正品为止.(1)若有放回地抽取，求抽取次数X的分布律.(2)若不放回地抽取，求抽取次数X的分布律.(3)针对以上两种情形，分别求“至少抽取3次才能拿到正品”的概率

(1) **有放回抽取** 每次抽到次品概率为 $0.3$，正品为 $0.7$。 分布律： \[ P(X = k) = (0.3)^{k-1} \times 0.7, \quad k = 1, 2, 3, \ldots \] (2) **不放回抽取** 总共3次品，7正品。 分布律： \[ P(X = k) = \begin{cases} \frac{7}{10} & k = 1 \\ \frac{3}{10} \times \frac{7}{9} = \frac{7}{30} & k = 2 \\ \frac{3}{10} \times \frac{2}{9} \times \frac{7}{8} = \frac{7}{120} & k = 3 \\ \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} \times \frac{7}{7} = \frac{1}{120} & k = 4 \end{cases} \] (3) **至少3次才拿到正品** **有放回**： \[ P(X \geq 3) = (0.3)^2 \times 0.7 + (0.3)^3 \times 0.7 + \cdots = \frac{(0.3)^2 \times 0.7}{0.7} = 0.09 \] **不放回**： \[ P(X \geq 3) = P(X = 3) + P(X = 4) = \frac{7}{120} + \frac{1}{120} = \frac{1}{15} \] \[ \boxed{ \begin{array}{ll} \text{(1)} & P(X = k) = (0.3)^{k-1} \times 0.7, \quad k = 1, 2, 3, \ldots \\ \text{(2)} & P(X = k) = \begin{cases} \frac{7}{10} & k = 1 \\ \frac{7}{30} & k = 2 \\ \frac{7}{120} & k = 3 \\ \frac{1}{120} & k = 4 \end{cases} \\ \text{(3)} & \text{有放回：} 0.09, \quad \text{不放回：} \frac{1}{15} \end{array} } \]