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
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 $$P[min(X,Y)\geq 2]$$

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