Respuesta :
Answer:
a) [tex]C=\frac{1}{6}[/tex]
b) [tex] P(X\leq1, Y\leq 1) = \frac{1}{8}[/tex]
c) [tex]P(X+Y\leq 1) = \frac{5}{144}[/tex]
Step-by-step explanation:
Recall that given the joint density function f(x,y) it must happen that
[tex]\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) dx dy = 1[/tex].
a)Then,
[tex]\int_{0}^1 \int_{0}^4 Cx(1+y)dydx = C\int_{0}^1 x \cdot \left.\frac{(1+y)^2}{2}\right|_{0}^4dx = 12C\left\frac{x}{2}\right|_{0}^1 = 6C=1[/tex]
Which implies that [tex]C=\frac{1}{6}[/tex].
b)[tex]P(X\leq 1, Y\leq 1) = \int_0^1 \int_0^1 f(x,y) dy dx = \frac{1}{6} \int_0^1 \int_0^1 x(1+y) dy dx = \frac{1}{6}\int_0^1 x dx \cdot \int_0^1 (1+y)dy = \frac{1}{6}(\left.\frac{x^2}{2}\right|_{0}^{1}\cdot\left.\frac{(1+y)^2}{2}\right|_0^{1} = \frac{1}{6}(\frac{1}{2}\cdot\frac{3}{2}) = \frac{1}{8}[/tex]
c) If we have that[tex]X+Y\leq 1[/tex],so we have that [tex] Y \leq 1-X[/tex]. Then
[tex]P(X+Y\leq 1) = \frac{1}{6}\int_{0}^{1}\int_{0}^{1-x} x (1+y) dy dx = \frac{1}{6}\int_{0}^{1} x \left.\frac{(1+y)^2}{2}\right|_{0}^{1-x} dx = \frac{1}{6}\int_{0}^{1} x\frac{(2-x)^2-1}{2}dx = \frac{5}{144}[/tex]
