# Probability distribution- standard normal dist.

suppose $$X$$ and $$Y$$ are independent random variables with standard normal distributions. The probability of $$X<-1$$ is some p which lies in the open interval (0,1). what is the probability of the event: $$X^2 > 1$$ and $$Y^3 > 1$$.

I have started with $$P(X<-1)= P(X>1) = p$$ (i think its right) But now stuck at how to proceed from here... I'm thinking of using cdf but not able to fit it in?

$$P((X^2 > 1) \cap Y^3 > 1)$$ $$=P(X^2 > 1)\times P(Y > 1)$$ $$=P(X > 1)\times P(Y>1) + P(X < -1)\times P(Y>1)$$
$$=(1-\Phi(1))(1-\Phi(1)) + \Phi(-1)\times (1-\Phi(1))$$
where $$\Phi(.)$$ denotes the CDF of $$N(0,1)$$
• In your Q, $p=\Phi(-1) = 1-\Phi(1)$. – Vizag Jun 4 at 17:19