91Ó°ÊÓ

The value of c

The given function is $f(x,y)=\left\{\begin{array}{l}\frac{x}{5}+cyâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 0<x<1,1<y<5\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{otherwise}\end{array}\right.$

The value of c can be calculated as

$\begin{array}{r}{∫}_0^1 {∫}_1^5 \left(\frac{x}{5}+Cy\right)dydx=1\\ {∫}_0^1 \left(\frac{4x}{5}+12C\right)=1\\ 12C+\frac{2}{5}=1\\ C=\frac{1}{20}\end{array}$

Hence the value of c is$\frac{1}{20}$

Are X and Y are independent

From the part(a) we know that c is $\frac{1}{20}$

Thus $\begin{array}{r}f(x,y)=\left\{\begin{array}{c}\frac{x}{5}+\frac{1}{20}y0<x<1,1<y<5\\ 0\text{Otherwise}\end{array}\right.\\ f(x,y)=\left\{\begin{array}{ll}\frac{4x+y}{20}& 0<x<1,1<y<5\\ 0\text{Otherwise}& \end{array}\right.\end{array}$

The marginal density X could be calculated as:

$\begin{array}{r}{f}_x\left(x\right)={∫}_1^5 \left(\frac{4x+y}{20}\right)dy\\ =\frac{1}{20}{\left[4xy+\frac{y^2}{2}\right]}_1^5\\ =\frac{1}{20}[16x+12]\\ =\frac{4x+3}{5}\end{array}$

The marginal density Y could be calculated as:

$\begin{array}{r}{f}_Y\left(y\right)={∫}_0^1 \left(\frac{4x+y}{20}\right)dy\\ =\frac{1}{20}{\left[\frac{4x^2}{2}+xy\right]}_0^1\\ =\frac{1}{20}[2+y]\\ =\frac{y+2}{20}\end{array}$

Now,

$\begin{array}{c}{f}_x\left(x\right)×f_y\left(y\right)=\left(\frac{4x+3}{5}\right)\left(\frac{y+2}{20}\right)\\ \frac{8x+4xy+6+3y}{100}≠\frac{4x+y}{20}\\ f_x\left(x\right)×f_y\left(y\right)≠f_{xy}(x,y)\end{array}$

Hence X and Y are not independent function.

$P[X+Y>3]$

The probability is
$\begin{array}{c}P[X+Y>3]={∫}_0^1 {∫}_{3−x}^5 \left(\frac{x}{5}+\frac{y}{20}\right)dydx\\ ={∫}_0^1 \left[(x+2)\frac{x}{5}+\frac{25}{40}−\frac{(3−x{)}^2}{40}\right]dx\\ =\frac{1}{5}+\frac{1}{15}+\frac{5}{8}−\frac{19}{20}\\ =\frac{11}{15}\end{array}$

Hence the probability is $\frac{11}{15}$