(1)记:“该射手恰好命中一次”为事件A,“该射手射击甲靶命中”为事件B,“该射手第一次射击乙靶命中”为事件C,“该射手第二次射击乙靶命中”为事件D,由题意知 P(B)= ,P(C)=P(D)= , 由于A=(B )∪( C )∪( D), 根据事件的独立性和互斥性得 P(A)=P((B )∪( C )∪( D))=P(B )+P( C )+P( D) =P(B)P( )P( )+P( )P(C)P( )+P( )P( )P(D) = ×(1- )×(1- )+(1- )× ×(1- )+(1- )×(1- )× = . (2)根据题意,X的所有可能取值为0,1,2,3,4,5. 根据事件的独立性和互斥性得 P(X=0)=P( ) =[1-P(B)][1-P(C)][1-P(D)] =(1- )×(1- )×(1- )= , P(X=1)=P(B )=P(B)P( )P( ) = ×(1- )×(1- ) = , P(X=2)=P( C ∪ D)=P( C )+P( D) =(1- )× ×(1- )+(1- )×(1- )×![](http://img.shitiku.com.cn/uploads/allimg/20191026/20191026182450-28799.png) = , P(X=3)=P(BC ∪B D)=P(BC )+P(B D) = × ×(1- )+ ×(1- )×![](http://img.shitiku.com.cn/uploads/allimg/20191026/20191026182450-28799.png) = , P(X=4)=P( CD) =(1- )× ×![](http://img.shitiku.com.cn/uploads/allimg/20191026/20191026182450-28799.png) = , P(X=5)=P(BCD) = × ×![](http://img.shitiku.com.cn/uploads/allimg/20191026/20191026182450-28799.png) = . 故X的分布列为 |