p(其中至少有2片安慰剂的概率) = 1 - p(都不是安慰剂) - p(只有一片安慰剂)
= 1 - C(5,5)/C(10,5) - C(5,1)C(5,4)/C(10,5)
= 1 - 1/252 - 25/252
=113/126
≈0.8969
p(其中至少有2片安慰剂的概率) = 1 - p(都不是安慰剂) - p(只有一片安慰剂)
= 1 - C(5,5)/C(10,5) - C(5,1)C(5,4)/C(10,5)
= 1 - 1/252 - 25/252
=113/126
≈0.8969