Consider continuous bivariate random variables, X and Y, with the following joint PDF (where c is a constant): FXY(x,y)={x−cy1≤x≤2,0≤y≤10otherwise
Compute P(X≤1.25,Y≥0.75)
.

Answers

Answer 1

By integrating the joint PDF over the specified region.

How to calculate the given probability?

To compute the probability P(X ≤ 1.25, Y ≥ 0.75) using the given joint PDF, we need to integrate the joint PDF over the specified region.

The region of interest is defined as 1 ≤ x ≤ 1.25 and 0.75 ≤ y ≤ 10.

The joint PDF, FXY(x, y), is given by FXY(x, y) = (x - cy) for 1 ≤ x ≤ 2 and 0 ≤ y ≤ 10, and 0 otherwise.

To compute the probability, we integrate the joint PDF over the specified region:

P(X ≤ 1.25, Y ≥ 0.75) = ∫∫[FXY(x, y)]dydx

Breaking down the integral into two parts:

∫[∫[FXY(x, y)]dy]dx

First, integrate the inner integral with respect to y:

∫[FXY(x, y)]dy = ∫[(x - cy)]dy = xy - (cy[tex]^2[/tex])/2

Next, integrate the outer integral with respect to x over the given range:

∫[xy - (cy[tex]^2[/tex])/2]dx = ∫[(xy - (cy)[tex]^2[/tex]/2)]dx

Evaluate the integral over the range 1 ≤ x ≤ 1.25:

= [∫[(xy - (cy[tex]^2[/tex])/2)]dx] evaluated from 1 to 1.25

Substitute the limits of integration into the expression:

= [(1.25y - (cy[tex]^2[/tex])/2) - (y - (cy[tex]^2[/tex])/2)]

Simplifying the expression:

= 1.25y - (cy[tex]^2[/tex])/2 - y + (cy[tex]^2[/tex])/2

= 0.25y

Finally, evaluate the integral with respect to y over the range 0.75 ≤ y ≤ 10:

∫[0.25y]dy = (0.25/2)y[tex]^2[/tex] evaluated from 0.75 to 10

Substitute the limits of integration into the expression:

= (0.25/2)(10^2 - (0.75)^2)

= (0.25/2)(100 - 0.5625)

= (0.25/2)(99.4375)

= 12.4296875

Therefore, P(X ≤ 1.25, Y ≥ 0.75) is approximately equal to 12.4296875.

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Have a great day <3

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mag-aral ka mabuti para maging matalino

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