Joint density function

Hello friends!

Its great to be a part of this forum, and I am glad to make my first post here.
There is a question of the joint probability density function that I am unable to get even the idea how to solve.
Kindly help.

Question:
If x and y are two random variables having joint density function:
f(x,y)= 1/8(6-x-y) 0 < x < 2, 2 < y < 4
0 otherwise

Find:
1) P(x<1 ∩ y<3)
2) P(x+y<3)

Thanks in advance friends. :)

Quote:

---

Originally Posted by Sarmadi

Hello friends!

Its great to be a part of this forum, and I am glad to make my first post here.
There is a question of the joint probability density function that I am unable to get even the idea how to solve.
Kindly help.

Question:
If x and y are two random variables having joint density function:
f(x,y)= 1/8(6-x-y) 0 < x < 2, 2 < y < 4
0 otherwise

Find:
1) P(x<1 ∩ y<3)
2) P(x+y<3)

Thanks in advance friends. :)

---

Start by drawing the area defined by 0 < x < 2, 2 < y < 4 on a set of axes. Do it twice (once for each part). Now shade the part of the area corresponding to
1) x< 1 and y < 3.
2) x + y < 3.

Now use what you have drawn to set up the integral limits on the double integral of f(x, y).

How would I know what area corresponds to "x< 1 and y < 3" or "x + y < 3"?

Think of it like this: start by graphing y = 3. The area under this line would correspond to y < 3, while the area above the line would correspond to y > 3