Question

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]

Random variables
ANSWERED
asked 2021-05-28
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 \(P[min(X,Y)\geq 2]\)

Expert Answers (1)

2021-05-29

The value of \(P[min(X,Y)\geq 2]\) is obtained as given below:
\(P[min(X,Y)\geq 2]=P[2\leq min (X,Y)]\)
\(=P(2\leq X, 2\leq Y)\)
\(=P(X\geq 2, Y\geq 2)\)
\(=[1-P(X<2)][1-P(Y<2)]\)
\(P(X<2)=\int_{0}^{2}3e^{-3x}dx\)
\(=3[-\frac{e^{-3x}}{3}]_{0}^{2}\)
\(=1-e^{-6}\)
=0.9975
1-P(X<2)=0.0025
\(f(y)=\int_{x}f(x,y)dx\)
\(=\int_{0}^{\infty}12e^{-(3x+4y)}dx\)
\(=12e^{-4y}[\frac{-e^{-3x}}{3}]_{0}^{\infty}\)
\(f(y)=4e^{-4y}\)
\(P(Y<2)=\int_{0}^{2}4e^{-4y}dy\)
\(=4[-\frac{e^{-4y}}{4}]_{0}^{2}\)
\(=1-e^{-8}\)
=0.9997
1-P(Y<2)=0.0003
\(P[min(X,Y)\geq 2]=[1-P(X<2)][1-P(y<2)]\)
\(=0.0025\times 0.0003\)
=0.0000075

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