某球员投球命中率为0.7,他连投三个球,求他投中球的概率颁布和数学期望.这是一道数学题

由题可知,投中球的个数可为0,1,2,3,
P(X=0)=（1-0.7)^3=0.027,
P(X=1)=3×(1-0.7)^2×0.7=0.189,
P(X=2)=3×(1-0.7)×0.7^2=0.441,
P(X=3)=0.7^3=0.343.
所以他投中球的概率分布为：
X 0 1 2 3
P 0.027 0.189 0.441 0.343
数学期望EX=0×P(X=0)+1×P(X=1)+2×P(X=2)+3×P(X=0)
=0×0.027+1×0.189+2×0.441+3×0.343=2.1