【正确答案】P{max(X，Y)≠0}=1一P{max(X，Y)=0}=1一P(X=0，Y=0)
=1一P(X=0)P(Y=0)=1一e^一1e^一2=1一e^一3
P{min(X，Y)≠0}=1一P{min(X，Y)=0}，
令A={X=0}，B={Y=0}，则{min(X，Y)=0)=A+B，
于是P{min(X，Y)=0}=P(A+B)=P(A)+P(B)一P(AB)
=e^一1+e^一2一e^一1．e^一2=e^一1+e^一2一e^一3，
故P{min(X，Y}≠0)=1一e^一1一e^一2+e^一3．