Question

概率论，求解 要过程

Answer 1

P(0发炮弹命中) = (0.3)^10

P(1发炮弹命中) = C(10,1) (0.7)(0.3)^9 =10(0.7)(0.3)^9

P(2发炮弹命中) = C(10,2) (0.7)^2.(0.3)^8 =45(0.7)^2.(0.3)^8

P(3发炮弹命中) = C(10,3) (0.7)^3.(0.3)^7 =120(0.7)^3.(0.3)^7

P(目标被击毁)

=P(不少于3发炮弹命中)

=1-P(0发炮弹命中)-P(1发炮弹命中)-P(2发炮弹命中)-P(3发炮弹命中)

=1-(0.3)^10 -10(0.7)(0.3)^9 -45(0.7)^2.(0.3)^8 -120(0.7)^3.(0.3)^7

=0.9894