2004 question dse

[Ansh malhotra]

How to do these type of questions

[Ansh malhotra]

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[Mike]

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)

[Ansh malhotra]