 # Random variables X and Y have joint PDF f_{X,Y}(x,y)=\begin{cases}12e^{-(3x+4y)},\ x \geq 0, y \geq 0\\0,\ otherwise\end{cases} Find P[min(X,Y)\geq 2] chillywilly12a 2021-05-28 Answered
Random variables X and Y have joint PDF

Find $P\left[min\left(X,Y\right)\ge 2\right]$
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The value of $P\left[min\left(X,Y\right)\ge 2\right]$ is obtained as given below:
$P\left[min\left(X,Y\right)\ge 2\right]=P\left[2\le min\left(X,Y\right)\right]$
$=P\left(2\le X,2\le Y\right)$
$=P\left(X\ge 2,Y\ge 2\right)$
$=\left[1-P\left(X<2\right)\right]\left[1-P\left(Y<2\right)\right]$
$P\left(X<2\right)={\int }_{0}^{2}3{e}^{-3x}dx$
$=3\left[-\frac{{e}^{-3x}}{3}{\right]}_{0}^{2}$
$=1-{e}^{-6}$
=0.9975
1-P(X<2)=0.0025
$f\left(y\right)={\int }_{x}f\left(x,y\right)dx$
$={\int }_{0}^{\mathrm{\infty }}12{e}^{-\left(3x+4y\right)}dx$
$=12{e}^{-4y}\left[\frac{-{e}^{-3x}}{3}{\right]}_{0}^{\mathrm{\infty }}$
$f\left(y\right)=4{e}^{-4y}$
$P\left(Y<2\right)={\int }_{0}^{2}4{e}^{-4y}dy$
$=4\left[-\frac{{e}^{-4y}}{4}{\right]}_{0}^{2}$
$=1-{e}^{-8}$
=0.9997
1-P(Y<2)=0.0003
$P\left[min\left(X,Y\right)\ge 2\right]=\left[1-P\left(X<2\right)\right]\left[1-P\left(y<2\right)\right]$
$=0.0025×0.0003$
=0.0000075