# Probability after getting distribution from marginal distributions

This is a problem on Freund's Mathematical statistics book on page 101

If the independent random variables $$X$$ and $$Y$$ have the marginal densities

$$f(x) = \begin{cases} 1/2 \space \text{for} \space 0

$$\pi(y) = \begin{cases} 1/3 \space \text{for} \space 0

Find: $$\space$$ (a)the joint probability density of $$X$$ and $$Y$$ (b) the value of $$P(X^2 + Y^2 > 1)$$

I got the first answer since it's independent $$f(x,y) = \begin{cases} 1/6 \space \text{for} \space 0

How do I approach the second one, like I am not too sure of the limits of the integrals

• You need to integrate over the region of the joint density where this statement holds. Think of this as the continuous analog of "adding up all the probabilities." – dsaxton Dec 1 '18 at 17:39
• I get that, but I'm having trouble with the limits will it be $\int_{1}^{2} \int_{\sqrt(1-x)}^{3} 1/6 dy dx$ ? – Sumukh Sai Dec 1 '18 at 17:44
• Draw a graph, find the area such that $X^2 + Y^2 > 1$. Integral the joint pdf on the area, you get the probability. – user158565 Dec 1 '18 at 17:47
• Are you sure your limits in $\int_{1}^{2} \int_{\sqrt(1-x)}^{3}$ are correct? The answer is 1-$\pi/24$ – user158565 Dec 1 '18 at 17:51
• No, that's where I'm having trouble. Don't know if my limits are correct – Sumukh Sai Dec 1 '18 at 17:54 $$X^2 + Y^2 > 1$$ = area A
$$\Pr(A) = 1 - \Pr(-A)$$
$$\Pr (-A) = \int_{0}^{1} \int_{0}^{\sqrt{1-y^2}} \frac 16 dy dx$$
• Oh good grief! What you call the region "$-A$" (ugh!) has area $\pi/4$ and so the probability that $X^2+Y^2$ exceeds $1$ is just $1-\frac{\pi}{24}$, no mss, no fuss, no complicated integrals etc. – Dilip Sarwate Dec 1 '18 at 19:55