设一批产品的不合格品率为0.1，从中任取3件，记X为其中的不合格品数，则下列概率计算正确的有（ ）。[2008年真题]

不合格品数X有四种可能，则其概率分别为：P(X=0) =0.9×0.9×0.9 =0.729; P(X = l) = 3×0. 1×0. 9×0. 9 = 0. 243 ； P(X = 2) =3×0. 1×0. 1×0.9 =0.027； P(X =3) =0. 1×0. 1×0. 1 =0.001。所以，P(X≤l) =P(X=0) +P(X=1) =0. 972, P(0≤X≤ 3) = P(X = 0) +P(X = 1) +P(X = 2) +P(X=3) =1。