# Thread: find P(X + Y < 1/2). My last problem.

1. ## find P(X + Y < 1/2). My last problem.

I thought I had done this question right, but I get a probability of 0.
If the joint probability density of X and Y is given by f(x,y) = { 24xy for 0<x<1, 0<y<1, x+y<1
0 elsewhere
find P(X + Y < 1/2).

I did the integral from 0 to 1/2 of the integral from x to (1/2 - x) of f(x) dy dx. Are my integrals correct? After the first integration, I get the integral from 0 to 1/2 of 12 xy^2 dx evaluated from x to (1/2 - x), which is the integral from 0 to 1/2 of (3x - 24x^3)dx. That equals (3/2x^2 - 6x^4) evaluated from 0 to 1/2 which = 0. What am I doing wrong? Sorry I don't know how to write it out using the actual math symbols.

2. Originally Posted by TheHolly
I thought I had done this question right, but I get a probability of 0.
If the joint probability density of X and Y is given by f(x,y) = { 24xy for 0<x<1, 0<y<1, x+y<1
0 elsewhere
find P(X + Y < 1/2).

I did the integral from 0 to 1/2 of the integral from x to (1/2 - x) of f(x) dy dx. Are my integrals correct? After the first integration, I get the integral from 0 to 1/2 of 12 xy^2 dx evaluated from x to (1/2 - x), which is the integral from 0 to 1/2 of (3x - 24x^3)dx. That equals (3/2x^2 - 6x^4) evaluated from 0 to 1/2 which = 0. What am I doing wrong? Sorry I don't know how to write it out using the actual math symbols.
I think your limits of integration are wrong

You want the triangle bounded by $\displaystyle y=-x+\frac{1}{2}$ and the x and y axis.

so you should get...

$\displaystyle \int_{0}^{1/2} \int_{0}^{1/2-x} 24(xy) dy dx=\frac{1}{16}$

Good luck.

3. ## Thanks

Thank you! I knew it wasn't that hard, but I couldn't figure out where my problem was.