Question

在一次抗洪抢险中,准备用射击的方法引爆从桥上游漂流而下的一巨大汽油罐.已知只有5发子弹备用,且首次命中只能使汽油流出,再次命中才能引爆成功,每次射击命中率都是,每次命中与否互相独立.

(1)求油罐被引爆的概率.

(2)如果引爆或子弹打光则停止射击,设射击次数为ξ,求ξ的分布列及ξ的数学期望；

Answer 1

解:(1)记“油罐被引爆”的事件为事件A,其对立事件为,?

则P(A)=C¹₅·()()⁴+()⁵.                                                                                   ?

∴P(A)=1-［C¹₅·()·()⁴+()⁵］=.                                                       ?

(2)射击次数ξ的可能取值为2,3,4,5.?

P(ξ=2)=()²=,                                                                                            ?

P(ξ=3)=C¹₂···=,                                                                        ?

P(ξ=4)=C¹₃·()²·=,                                                                            ?

P(ξ=5)=C¹₄·()()³+()⁴=.                                                                       ?

故ξ的分布列为

ξ	2	3	4	5
P

?

Eξ=2×+3×+4×+5×=.