Question

甲、乙、丙三人同时向一架飞机设计,设他们击中飞机的概率均为0.4,如果只有一人击中飞机,

Answer 1

解1P=(0.4+0.4+0.4)×0.2+C(3,2)×0.6^1×0.4^2×0.6+C(3,3)×0.6^0×0.4^3

=0.24+0.1728+0.064

=0.4768

2P=C(3,2)×0.6^1×0.4^2

=0.288