Question

1．设事件A和B的概率为p（A) =1\2，P(B)=2/3 则 P(AB)可能为（）

Answer 1

1.P(A)=1/4,P(B|A)=1/3,P(A|B)=1/2,可得P(AB)=1/12,P(B)=1/6
P(X=0,Y=0)=P(A非B非)=1-P(A+B)=1-[P(A)+P(B)-P(AB)]=2/3
P(X=1,Y=0)=P(AB非)=P(A)-P(AB)=1/6
P(X=0,Y=1)=P(BA非)=P(B)-P(AB)=1/12
P(X=1,Y=1)=P(AB)=1/12 联合分布就可得出.
2.E(X)=1/4,E(Y)= 1/6,E(XY)=1/12,COV(X,Y)=E(XY)-E(X)E(Y)=1/24
D(X)=E(X2)-[E(X)]2=3/16,D(Y)=E(Y2)-[E(Y)]2=5/36
ρ=COV(X,Y)=COV(X,Y)/[√D(X)√D(Y)]=1/√15

Answer 2

①(sinx-xcosx)/{(e^x-1)[(1+x^2)^(1/3)-1]}(x->0)
=(sinx-xcosx)/{x[(1+x^2)^(1/3)-1]}(x->0)（根据当x->0时e^x-1等价于x)
=(sinx-xcosx)*[(1+x^2)^(2/3)+(1+x^2)^(1/3)+1]/x*x^2(x->0)
=3(sinx-xcosx)/x^3(x->0)
=3(cosx+xsinx-cosx)/3x^2(x->0)(罗必塔法则）
=sinx/x^2(x->0)
=cosx/2x(x->0)(罗必塔法则)
=∞
②tan3x/tan5x=sin3x*cos5x/[cos3x*sin5x](x->π/2）
=sin(3π/2)*cos5x/[sin(5π/2)*cos3x](x->π/2)
=-cos5x/cos3x(x->π/2)
=-(cos3x*cos2x-sin3x*sin2x)/cos3x(x->π/2)
=-cos2x-(sin2x/cos3x)(x->π/2)
=1-2sinx*cosx/[4(cosx)^3-3cosx](x->π/2)
=1-2sinx/[4(cosx)^2-3](x->π/2)
=1-2/(0-3)
=5/3

Answer 3

1/3

追问

答案是6分之1