Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = xe−x(1 + y) x ≥ 0 and y ≥ 0 0 otherwise (a) What is the probability that the lifetime X of the first component exceeds 5? (Round your answer to three decimal places.)

Step-by-step explanation:

Given that,

f(x, y) = xe^−x(1 + y) x ≥ 0 and y ≥ 0

It is f(x, y)= 0 otherwise

To find probability of life time, we will take the double integral of the function with respect to X and Y

Y ranges from 0 to ∞

X exceed 5, so it ranges from 5 to ∞

∫ ∫ f(x, y) dxdy

∫ ∫ xe^−x(1 + y) dxdy

∫ ∫ x•e^(−x -xy) dxdy

Separating the exponential

∫ ∫ x•e^(−x) • e^(-xy) dxdy

Integrating with respect to y and keeping x as a constant.

∫ x•e^(−x) • e^(-xy) / -x dx

∫ - e^(−x) • e^(-xy) dx y = 0 to y=∞

∫ - e^(−x) • [e^(-∞) - e^(0) ]dx

∫ - e^(−x) • [ 0 - 1] dx

∫ - e^(−x) • -1 dx

∫ e^(−x) dx

Now integrating this

e^(-x) /-1 from x=5 to x=∞

-e^(-x) from x=5 to x=∞

- [ e^(-∞) - e^(-5)]

- (0-e^(-5))

- - e^(-5)

e^(-5) = 0.006738

To 3d.p =0.007

Then probability that the life time X exceed 5 is 0.007

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.

is there any other information given for this problem? this is not enough information.