------------------(1)3件中恰有1件次品的概率
C(4,1)*C(36,2)/C(40,3)
=4*36*35*3/(40*39*38)
=9*7*3/(39*19)
=189/741
------------------(2)3件全是正品的概率
C(36,3)/C(40,3)
=36*35*34/(40*39*38)
=3*7*17/(13*38)
=357/494
------------------(3)3件中至少有1件次品的概率
1-C(36,3)/C(40,3)
=1-36*35*34/(40*39*38)
=1-357/494
=137/494
------------------(4)3件中至少有2件次品的概率
C(4,2)*C(36,1)/C(40,3)+C(4,3)/C(40,3)
=4*3*36*3/(40*39*38)+4*3*2/(40*39*38)
=9*3*2/(5*13*38)+1/(5*13*38)
=55/(5*13*38)
=11/494