Suppose that X and Y are continuous random variables with joint pdf f(x,y)=e−(x+y)0<x<∞ and 0<y<∞...

Step 1
Introduction:
The joint density function of two random variables X and Y is given below:
$f(x,y)=\left\{\begin{array}{c}{e}^{-(x+y)};0\le x\le \mathrm{\infty },0\le y<\mathrm{\infty }\\ 0;elsewhere\end{array}\right\}$
The marginal density function of X is,
${\displaystyle f\left(x\right)={\int }_0^{\mathrm{\infty }}{e}^{-(x+y)}dy}$
${\displaystyle =e^{-x}{\int }_0^{\mathrm{\infty }}{e}^{-y}dy}$
${\displaystyle =-e^{-x}{[-e^{-y}]}_0^{\mathrm{\infty }}}$
${\displaystyle =-e^{-x}[e^{-\mathrm{\infty }}-e^{-0}]}$
${\displaystyle =-e^{-x}[0,1]}$
${\displaystyle =e^{-x}}$
Step 2
The probability of P(X > 3) is obtained as 0.0498 from the calculation given below:
${\displaystyle P\left(X>3\right)={\int }_3^{\mathrm{\infty }}{e}^{-x}dx}$
${\displaystyle ={[-e^{-x}]}_3^{\mathrm{\infty }}}$
${\displaystyle =-[e^{-\mathrm{\infty }}-e^{-3}]}$
${\displaystyle =-[0-e^{-3}]}$
${\displaystyle =e^{-3}}$
=0.0498
Thus, the probability of P(X > 3) is 0.0498.