- #1
Askhwhelp
- 86
- 0
$$f(x,y)=
\begin{cases}
4(x+y^2),&\text{if x > 0, y > 0, x + y < 1} \\
0, &\text{elsewhere} \\
\end{cases}$$
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?
\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: