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≥0andy≥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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Answered question

2021-11-07

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

ensojadasH

Skilled2021-11-08Added 100 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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