一道关于概率的问题,请能人帮帮忙

2个回答

  • 楼主你好

    1. pX(1)+pX(2)+……+pX(n)=a×[(1/3)^1+(1/3)^2+……+(1/3)^n]=a×(1/3)×(1-(1/3)^n)÷(1-(1/3))=a×1/2×(1-(1/3)^n).当n趋近于∞的时候,(1-(1/3)^n)趋近于1,所以pX(1)+pX(2)+……+pX(n)+……=a×1/2=1,所以a=2

    2. x是奇数的概率=pX(1)+pX(3)+……+pX(2n-1)=2×((1/3)^1+(1/3)^3+……+(1/3)^(2n-1))=2×(1/3)×(1-(1/9)^n)÷(1-(1/9))=(3/4)×(1-(1/9)^n).当n趋近于∞的时候,(1-(1/9)^n)趋近于1,所以X是奇数的概率=pX(1)+pX(3)+……+pX(2n-1)+……=(3/4)×1=3/4,那么X是偶数的概率=1-(3/4)=1/4,所以X多为奇数

    3. E(X)=2×(1×(1/3)+2×(1/3)^2+……+n×(1/3)^n),n趋近于∞,那么(1/3)×E(X)=2×(1×(1/3)^2+2×(1/3)^3+……+(n-1)×(1/3)^n+n×(1/3)^(n+1))=2×(0+1×(1/3)^2+2×(1/3)^3+……+(n-1)×(1/3)^n+n×(1/3)^(n+1)),所以E(X)-[(1/3)×E(X)]=2×(1×(1/3)+(2-1)×(1/3)^2+(3-2)×(1/3)^3+……+(n-(n-1))×(1/3)^n-n×(1/3)^(n+1))=2×[(1/3)+(1/3)^2+……+(1/3)^n-n×(1/3)^(n+1)]=[2×(1/3)×(1-(1/3)^n)÷(1-(1/3))]-2×n×(1/3)^(n+1)=(1-(1/3)^n)-2n×(1/3)^(n+1)=(1-(1/3)^n)-(2/3)×n×(1/3)^n.当n趋近于∞的时候,E(X)-[(1/3)×E(X)]=(2/3)×E(X)=1-0=1,所以E(X)=3/2

    至于Var(X),我们可以用Var(X)=E(X^2)-[E(X)]^2来做.观察一下儿,E(X^2)=2×((1^2)×(1/3)+(2^2)×(1/3)^2+……+(n^2)×(1/3)^n),n趋近于∞.再看2E(X)=2×(2×1×(1/3)+2×2×(1/3)^2+……+2×n×(1/3)^n),n趋近于∞.再设Sn=pX(1)+pX(2)+……+pX(n)=2×((1/3)+(1/3)^2+……+(1/3)^n),n趋近于∞.好,现在n趋近于∞的时候,你发现E(X^2)+2E(X)+Sn=2×([(1^2)+2×1+1]×(1/3)+[(2^2)+2×2+1]×(1/3)^2+……+[(n^2)+2×n+1]×(1/3)^n+……)=2×((2^2)×(1/3)+(3^2)+(1/3)^2+……+((n+1)^2)×(1/3)^n+……)=3×2×((2^2)×(1/3)^2+(3^2)+(1/3)^3+……+((n+1)^2)×(1/3)^(n+1)+……)=3×2×(-(1^2)×(1/3)+(1^2)×(1/3)+(2^2)×(1/3)^2+(3^2)+(1/3)^3+……+((n+1)^2)×(1/3)^(n+1)+……)=3×2×-(1^2)×(1/3)+3×2×((1^2)×(1/3)+(2^2)×(1/3)^2+(3^2)+(1/3)^3+……+((n+1)^2)×(1/3)^(n+1)+……)=-2+3E(X^2)=3E(X^2)-2.你已经知道2E(X)=3,且当n趋近于∞时,Sn=1,所以E(X^2)+3+1=3E(X^2)-2,所以E(X^2)=3.所以Var(X)=E(X^2)-[E(X)]^2=3^2-(3/2)^2=27/4

    希望你满意