相互独立以X表示同时需要调整的部件数试求X的概率分布数学E(x)和方差
设A,B,C是三个需要调整的部件,调整概率分别为0.1,0.2,0.3
计算概率:
X=0,概率:(1-0.1)*(1-0.2)*(1-0.3)=0.5040
X=1,概率:0.1*(1-0.2)*(1-0.3)+0.2*(1-0.1)*(1-0.3)+0.3*(1-0.1)*(1-0.2)=0.1327
X=2,概率:0.1*0.2*(1-0.3)+0.1*0.3*(1-0.2)+0.2*0.3*(1-0.1)=0.0980
X=3,概率:0.1*0.2*0.3=0.006
E(X)=1*0.1327+2*0.0980+3*0.006=0.612
E(X^2)=!^2*0.1327+2^2*0.0980+3^2*0.006=0.398
D(X)=E(X^2)-(E(X))^2=0.398-0.612^2=0.0235
计算概率:
X=0,概率:(1-0.1)*(1-0.2)*(1-0.3)=0.5040
X=1,概率:0.1*(1-0.2)*(1-0.3)+0.2*(1-0.1)*(1-0.3)+0.3*(1-0.1)*(1-0.2)=0.1327
X=2,概率:0.1*0.2*(1-0.3)+0.1*0.3*(1-0.2)+0.2*0.3*(1-0.1)=0.0980
X=3,概率:0.1*0.2*0.3=0.006
E(X)=1*0.1327+2*0.0980+3*0.006=0.612
E(X^2)=!^2*0.1327+2^2*0.0980+3^2*0.006=0.398
D(X)=E(X^2)-(E(X))^2=0.398-0.612^2=0.0235
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