至少有两张A的概率为 1-C(48,4)/C(52,4)-C(48,3)*C(4,1)/C(52,4) =1-48*47*46*45/(52*51*50*49)-48*47*46*4*4/(52*51*50*49) ≈1-0.25555-0.71874 ≈0.0257
