如题所述
解:(Ⅰ)求该运动员两次都命中7环的概率为P(7)=0.2×0.2=0.04;(Ⅱ)ξ的可能取值为7、8、9、10P(ξ=7)=0.04P(ξ=8)=2×0.2×0.3+0.32=0.21P(ξ=9)=2×0.2×0.3+2×0.3×0.3+0.32=0.39P(ξ=10)2×0.2×0.2+2×0.3×0.2+2×0.3×0.2+0.22=0.36ξ分布列为(Ⅲ)ξ的数学希望是Eξ=7×0.04+8×0.21+9×0.39+10×0.36=9.0719.