概率题。6瓶牛奶。三瓶好的三瓶坏的。完全检测出坏牛奶的次数为ξ，求ξ的分布列

回答：

计算P(ξ=k)的基本思路是，前k-1次检测中包含2瓶坏奶的概率乘以第k次检测到坏奶的概率。于是

P(ξ=3)＝[C(3, 2)C(3, 0)/C(6, 2)] x (1/4) = 1/20;
P(ξ=4) = [C(3, 2)C(3, 1)/C(6, 3)] x (1/3) = 3/20;
P(ξ=5) = [C(3, 2)C(3, 2)/C(6, 4)] x (1/2) = 3/10;
P(ξ=6) = [C(3, 2)C(3, 3)/C(6, 5)] x 1 = 1/2.

〔作为验证，∑{k=3, 6}P(ξ=k) = 1。〕