# Double Integral With probability

1. Dec 7, 2013

$$f(x,y)= \begin{cases} 4(x+y^2),&\text{if x > 0, y > 0, x + y < 1} \\ 0, &\text{elsewhere} \\ \end{cases}$$

Find two different integral expressions for P(y > x)

(1) $$\int_{0} ^{1/2}\int_{0} ^y 4(x+y^2),dx\,dy + \int_{1/2} ^{1}\int_{0} ^{1-y} 4(x+y^2),dx\,dy$$

(2) $$\int_{0} ^{1/2}\int_{x} ^{1-x} 4(x+y^2),dy\,dx$$
Are they right?

Second , find P(Y > 1/2 | X < 1/2). the set up I have is = P(Y > 1/2, X < 1/2)/P(X<1/2)= $$\int_{1/2} ^{1}\int_{0} ^{1-y} 4(x+y^2),dx\,dy / \int_{0} ^{1/2}\int_{0} ^{1-x} 4(x+y^2),dy\,dx$$ is this set up right?

Third, find P(Y > 1/2 | X > 1/2) I found that the prob should be 0 because there no intersection for Y > 1/2 and X < 1/2 ? Right?

Last edited: Dec 7, 2013
2. Dec 7, 2013

### vela

Staff Emeritus
Looks like you switched the roles of x and y in both cases.

If x + y < 1 and x > y, then y can't be greater than 1/2, right? Yet in (1), you have an integral for 1/2 < y < 1.

3. Dec 7, 2013

### vela

Staff Emeritus
Looks good.

I assuming that was a typo. Yes, your reasoning is correct.

4. Dec 7, 2013

Thank you for pointing it out

5. Dec 7, 2013

My typo it should P(y> x). With this prob., my (1) and (2) should be right?

6. Dec 7, 2013

### vela

Staff Emeritus
Looks good then.