一道概率论的问题具体问题是：甲乙二人轮流射击,直到某人集中目标为止.已知甲击中目标的概率为0.6,乙击中目标的概率为0.

设X、Y、Z分别代表目标被击中时总的射击次数、甲射击次数、乙射击次数.
X=1,2,3,4.
Y=1,2,3,4,.
Z=0,1,2,3,4.
P(X=1)=0.6,P(X=2)=0.4*0.5,
P(X=3)=0.4*0.5*0.6,P(X=4)=0.4*0.5*0.4*0.5,
P(X=5)=0.4*0.5*0.4*0.5*0.6,P(X=6)=0.4*0.5*0.4*0.5*0.4*0.5
.
可以总结出X的分布律为
X为奇数时,P(X=2k-1)=0.6*(0.4*0.5)^(k-1)=0.6*0.2^(k-1)
X为偶数时,P(X=2k)=(0.4*0.5)^k=0.2^k
k=1,2,3,4,5,.
Y的分布律为：
P(Y=k)=P(X=2k-1)+P(X=2k)=0.6*0.2^(k-1)+0.2^k
k=1,2,3,4,.
Z的分布律为
P(Z=k)=P(X=2k)+P(X=2k+1)=0.2^k+0.6*0.2^k=1.6*0.2^k
P(Z=0)=0.6
k=1,2,3,4,.
解毕.