i番目のコインが表の時、Xi=1、裏の時Xi=0とする。
(1)
EXi=1*(\(\frac{1}{2}\))+0*(\(\frac{1}{2}\))=\(\frac{1}{2}\)
E(X1+X2+...+X8)=8*(\(\frac{1}{2}\))=4 答４枚

(2)
V(Xi)=E[Xi-(\(\frac{1}{2}\))]^2=(1-(\(\frac{1}{2}\))\()^{2}\)*(\(\frac{1}{2}\))+(0-(\(\frac{1}{2}\))\()^{2}\)*(\(\frac{1}{2}\))=\(\frac{1}{8}\)+\(\frac{1}{8}\)=\(\frac{1}{4}\)
V(X1+X2+...+X8)=8*(\(\frac{1}{4}\))=2

(3)
S=X1+X2+...+X8とする。
E[S(8-S)]=E[8S-\(S^{2}\)]=8ES-E[X1+X2+...+X8]^2
ここで、8ES=8*4=32
E[X1+X2+...+X8]^2=E[ΣX\(i^{2}\)+2ΣXiXj](i＜j)
EX\(i^{2}\)=\(1^{2}\)*(\(\frac{1}{2}\))+\(0^{2}\)*(\(\frac{1}{2}\))=\(\frac{1}{2}\)
EXiXj=EXiEXj=(\(\frac{1}{2}\))(\(\frac{1}{2}\))=\(\frac{1}{4}\)
だから、E[S(8-S)]=8*(\(\frac{1}{2}\))+[2*8(8-1)/2]*(\(\frac{1}{4}\))
=4+\(\frac{56}{4}\)=4+14=18