Math Help - another one of Joint Probability density of X and Y

1. another one of Joint Probability density of X and Y

If the independent random variables X and Y have the mariginal densities

f(x)= 0.5 for 0<x<2 and zero elsewhere

g(y)=1/3 for 0<y<3 and zero elsewhere

find the value of P(X^2+Y^2 > 1)

I have the joint prob. density h(x,y)=1/6 for 0<x<2, 0<y<3 and zero elsewhere, it's because they are independent, so their joint p density is the product of those two. and don't know what to do next...

And I thank you all in advance, "Thank you"

2. Originally Posted by Judi
If the independent random variables X and Y have the mariginal densities

f(x)= 0.5 for 0<x<2 and zero elsewhere

g(y)=1/3 for 0<y<3 and zero elsewhere

find the value of P(X^2+Y^2 > 1)

I have the joint prob. density h(x,y)=1/6 for 0<x<2, 0<y<3 and zero elsewhere, it's because they are independent, so their joint p density is the product of those two. and don't know what to do next...

And I thank you all in advance, "Thank you"
If you draw the picture, you'll see it is easiest to evaluate P(X^2 + Y^2 < 1) = 1 - P(X^2 + Y^2 > 1). This is just the quarter circle with radius 1 with probability pi/24.

If you have to use an integral to evaluate this, write the inequality x^2 + y^2 < 1 as y < sqrt(1 - x^2) with 0 < x < 1. The iterated integral is

int (0,1) int (0,sqrt(1 - x^2)) 1/6 dy dx = int (0,1) sqrt(1 - x^2)/6 dx.

You can use The Integrator to find the antiderivative of sqrt(1 - x^2).

3. If you don't mind, could you replain how you got pi/24

thankyou

4. Originally Posted by Judi
If you don't mind, could you replain how you got pi/24

thankyou
1/6 of area of quarter circle. For more on converting area to probability, see my answer to your other post