# Simple question need answer #2

• Sep 7th 2009, 06:57 PM
ugling
2. Given the density f(x,y)=2(x+y), 0<x<1, 0<y<x, calculate
a) P(X<0.5, Y<0.5)
b) P(X<0.5)
c) P(Y<0.5)

Thanks~(Talking)
• Sep 8th 2009, 03:14 AM
TKHunny
2a)

$
\int_{0}^{a}\int_{0}^{x}2\cdot(x+y)\;dy\;dx\;=\;\f rac{1}{2}
$

Solve for 'a'.
• Sep 8th 2009, 05:47 AM
ugling
what's the difference between 2a and 2b?
what's the difference between 2a and 2b in this case?
• Sep 8th 2009, 06:14 AM
mr fantastic
Quote:

Originally Posted by ugling
2. Given the density f(x,y)=2(x+y), 0<x<1, 0<y<x, calculate
a) P(X<0.5, Y<0.5)
b) P(X<0.5)
c) P(Y<0.5)

Thanks~(Talking)

a. $\int_{x = 0}^{x = 1/2}\int_{y = 0}^{y = x}2 \cdot (x+y)\; dy \; dx$

b. Same as above.

c. $\int_{y = 0}^{y = 1/2}\int_{x = y}^{x = 1}2 \cdot (x+y)\; dx \; dy$

In each case it helps to draw the required region of integration to get the integral terminals.
• Sep 8th 2009, 02:53 PM
TKHunny
Quote:

Originally Posted by TKHunny
2a)

$
\int_{0}^{a}\int_{0}^{x}2\cdot(x+y)\;dy\;dx\;=\;\f rac{1}{2}
$

Solve for 'a'.

So, I'm just making up my own problems, today. Sorry about that.
• Sep 8th 2009, 08:11 PM
mr fantastic
Quote:

Originally Posted by TKHunny
So, I'm just making up my own problems, today. Sorry about that.

It's a good question though (Rofl)
• Sep 8th 2009, 09:36 PM
matheagle
Quote:

Originally Posted by ugling
what's the difference between 2a and 2b in this case?

shakespeare
2b or not 2b