if X has exponential exp(lambda)and Y has normal distribution N(0,1) . X and Y are independent. how can one find p(X < Y)? My thoughts were to integrate :
$$\int_{-\infty}^{+\infty} F_X(y)f_Y(y) \,dy,$$
with function $F_X$ being the cdf of X and function $f_Y$ being the pdf of Y. I couldn't integrate it in a way that would give me a suitable answer without using the erf function. any help w ould be greatly appreciated.
edit: I mixed up X and Y's distributions
This addresses the original version of the question, before the OP edited it at roughly the same time this was posted. Maybe an opportunity to recall that one should not modify a question once, as in the present case, answers have been posted...
Note that, for every $x>0$, $$P(Y>x)=e^{-\lambda x}$$ hence $$P(X<Y)=P(X<0)+P(0<X<Y)=\frac12+E(e^{-\lambda X};X>0)$$ where, by definition, $$E(e^{-\lambda X};X>0)=\int_0^\infty\frac1{\sqrt{2\pi}}e^{-x^2/2}e^{-\lambda x}dx=\int_0^\infty\frac1{\sqrt{2\pi}}e^{-(x+\lambda)^2/2}e^{\lambda^2/2 }dx=e^{\lambda^2/2 }\int_\lambda^\infty\frac1{\sqrt{2\pi}}e^{-x^2/2}dx$$ Finally, $$P(X<Y)=\frac12+e^{\lambda^2/2 }(1-\Phi(\lambda))$$ where, for every $x$, $$\Phi(x)=P(X<x)=\int_{-\infty}^x\frac1{\sqrt{2\pi}}e^{-t^2/2}dt$$