Re: 2004 question dse

Posted by Mike on Jun 09, 2016; 4:48pm
URL: http://discussion-forum.276.s1.nabble.com/2004-question-dse-tp7601345p7601355.html

p(x^2>1 & y^3>1) = p(x^2>1)*p(y^3>1) .........due to independence
                            = [p(x<-1) or p(x>1)]*p(y>1)
                            = [p+p]*p ............ since the standard normal distribution is symmetric
                            = 2p^2
so answer is (d)