Here is the correct computation of the question.
The future lifetimes (in months) of two components of a machine have the following joint density function:
[tex]f(x,y) =\left \{ { {\dfrac{6}{12,500}(50-x-y)} \atop {0}} \right.[/tex] for 0 < x < 50 - y < 50, otherwise.
Write down a single integral representing the probability that both components are still functioning in 20 months from now.
Answer:
[tex]\mathbf{ P{(x>20) \cap(Y>20)} } =0.0008}[/tex]
Step-by-step explanation:
From the given information;
[tex]f(x,y) =\left \{ { {\dfrac{6}{12,500}(50-x-y)} \atop {0}} \right.[/tex] for 0 < x < 50 - y < 50, otherwise.
We can assert that the probability is the integral of the given density over the part of the range 0 ≤ x ≤ 50 - y ≤ 50 in which both x and y are greater than 20.
From the attached file below; their shows a probability density graph illustrating the above statement being said.
Now; to determine the probability that illustrates the integral of the density ; we have : P[(X > 20)∩(Y > 20)]
In addition to that:
From the image attached below;
We look into the region where the joint density under study is said to be positive and the triangle limits by the line axis x+y = 50
∴
[tex]P{(x>20) \cap(Y>20)} } = \dfrac{6}{125000}\int\limits^{30}_{20}\int\limits^{50-x}_{20}(50-x-y)dydx[/tex]
[tex]P{(x>20) \cap(Y>20)} } = \dfrac{6}{125000}\int\limits^{30}_{20} \dfrac{1}{2}(x-30^2)dx[/tex]
[tex]P{(x>20) \cap(Y>20)} } = \dfrac{6}{125000} ( \, \dfrac {500}{3})[/tex]
[tex]P{(x>20) \cap(Y>20)} } = \dfrac{6*500}{125000*3}[/tex]
[tex]P{(x>20) \cap(Y>20)} } = \dfrac{3000}{375000}[/tex]
[tex]\mathbf{ P{(x>20) \cap(Y>20)} } =0.0008}[/tex]
Thus; the single integral representing the probability that both components are still functioning in 20 months from now is [tex]\mathbf{ P{(x>20) \cap(Y>20)} } =0.0008}[/tex]
