①它们的积不大于2/9的概率为
P1=(2/9)×1+∫(2/9到1)(2/9x)dx=(2/9)×1+(2/9)×[ln1-ln(2/9)]
=2/9+(2/9)ln(9/2).
②它们的积大于1的概率为0.
③当还须满足条件x+y>1时,求它们的积不大于2/9的概率
x+y=1与xy=2/9的交点为(1/3,2/3),(2/3,1/3).
它们的积不大于2/9的概率为
P2=2/9+(2/9)ln(9/2)+(1/2)(2/3+1/3)(2/3-1/3)
-∫(1/3到2/3)(2/9x)dx-1/2
=2/9+(2/9)ln(9/2)+1/6-(2/9)[ln(2/3)-ln(1/3)]-1/2
=(4/9)ln(3/2)-1/9