Random variables X and Y have joint PDF fX,Y(x,y)={12e−(3x+4y), x≥0,y≥00, otherwise Find P[X+Y≤1]

The value of P[X+Y≤1] is obtained as given below: f(x,y)=12e−(3x+4y).x≥0 and y≥0 f(x)=∫yf(x,y)dy =∫0∞12e−(3x+4y)dy =12e−3x[−e−4y4]0∞ =12e−3x×14 f(x)=3e−3x P(X+Y≤1)=P(X≤1−Y) =∫01−yf(x)dx =∫01−y3e−3xdx =3[−e−3x3]01−y =−[e−3(1−y)−1] =[1−e−3(1−y)]

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avissidep

Answered question

2021-05-28

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 [X + Y \leq 1]

Answer & Explanation

Neelam Wainwright

Skilled2021-05-29Added 102 answers

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

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