Random variables X and Y have joint PDF fX,Y(x,y)={12e−(3x+4y), x≥0,y≥00, otherwise Find P[min(X,Y)≥2]

The value of P[min(X,Y)≥2] is obtained as given below: P[min(X,Y)≥2]=P[2≤min(X,Y)] =P(2≤X,2≤Y) =P(X≥2,Y≥2) =[1−P(X&lt;2)][1−P(Y&lt;2)] P(X&lt;2)=∫023e−3xdx =3[−e−3x3]02 =1−e−6 =0.9975 1-P(X&lt;2)=0.0025 f(y)=∫xf(x,y)dx =∫0∞12e−(3x+4y)dx =12e−4y[−e−3x3]0∞ f(y)=4e−4y P(Y&lt;2)=∫024e−4ydy =4[−e−4y4]02 =1−e−8 =0.9997 1-P(Y&lt;2)=0.0003 P[min(X,Y)≥2]=[1−P(X&lt;2)][1−P(y&lt;2)] =0.0025×0.0003 =0.0000075

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sibuzwaW

Answered question

2021-11-05

Random variables X and Y have joint PDF
$f_{X, Y} (x, y) = {\begin{cases} 12 e^{- (3 x + 4 y)}, x \geq 0, y \geq 0 \\ 0, o t h e r w i s e \end{cases}$
Find $P [min (X, Y) \geq 2]$

Answer & Explanation

sweererlirumeX

Skilled2021-11-06Added 91 answers

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