例22.4某射手在相同条件下独立地进行5次射击每次击中目标的 概率是0.6,求击中目标次数X的概率分布 解X的可能取值为0,1,2,3,4,5,设事件A表示第次射中,(i=1,2,,5 则A相互独立, P(X=0)=P(工)=10.65=04=C5×06×(1-06 P(X=1)=P(AA2A3A443+A42A2A,45+A42A3A,45+AA243A1A 类推得 +444)=5×06×(10.6)=C3×06×(1-06 P(X=)=C3×062×(1-06)P(X=3)=C3×0.63×(1-06 即: P(X=4)=C×0.6×(1-06 P(x=1)=0k×0.6×(1-0.6)6 P()-0×0.63×/1-0 i=0,1,2,3,4,5例2.2.4 某射手在相同条件下独立地进行5次射击,每次击中目标的 概率是0.6,求击中目标次数X的概率分布. 解:X的可能取值为0,1,2,3,4,5,设事件Ai表示第i次射中,(i=1,2,...,5), 则Ai相互独立, P(X=0)= ( ) P A1 A2 A3 A4 A5 =(1-0.6)5 =0.45 P(X=1)= A A A A A ) P( A A A A A A A A A A A A A A A A A A A A 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 + + + + =5×0.6×(1-0.6)4 0 0 5 5 = C 0.6 (1− 0.6 ) 1 1 4 5 类推得 = C 0.6 (1− 0.6 ) : P(X=3) 3 3 2 5 = C 0.6 (1− 0.6 ) P(X=4) 4 4 1 5 = C 0.6 (1− 0.6 ) P(X=5) 5 5 0 5 = C 0.6 (1− 0.6 ) 即: i i 5 i 5 P(X i ) C 0.6 ( 1 0.6) − = =   − i=0,1,2,3,4,5 P(X=2) 2 2 3 5 = C 0.6 (1− 0.6 )