求概率论大神解答一道概率论题目,可追加悬赏,在线等,急!

如题所述

X服从二项分布,x~B(4,p)
所以得到P(x=0)=(1-p)^4
P(x=1)=C(4,1) * p * (1-p)^3 =4p * (1-p)^3
P(x=2)=C(4,2) * p^2 * (1-p)^2=6p^2 * (1-p)^2
P(x=3)=C(4,3) * p^3 * (1-p)=4p^3 * (1-p)
P(x=4)= p^4
而Y=sin(xπ/2)=0,1,-1
x=1时Y=1,x=3时Y= -1,其余为0
那么EY= 4p * (1-p)^3 - 4p^3 * (1-p)
=4p(1-p) *[(1-p)^2 -p^2]
=4p(1-p)(1-2p)
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