Let x and y have joint density function $p ( x , y ) = \left\{ \begin{array} { l l }
x + y & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \
0 & \text { otherwise }
\end{array} \right.$ Find the probability that x y +0.4.

Question

Let x and y have joint density function $p ( x , y ) = \left\{ \begin{array} { l l } 
x + y & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \
0 & \text { otherwise }
\end{array} \right.$ Find the probability that x > y +0.4.

Quizplus · Accepted Answer

The Answer of Let x and y have joint density function $p ( x , y ) = \left\{ \begin{array} { l l } 
x + y & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \
0 & \text { otherwise }
\end{array} \right.$ Find the probability that x > y +0.4.

Let X and Y Have Joint Density Function p(x,y)={x+y if 0≤x≤1,0≤y≤10 otherwise p ( x , y ) = \left\{ \begin{array} { l l } x + y & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \\0 & \text { otherwise }\end{array} \right.p(x,y)={x+y0​ if 0≤x≤1,0≤y≤1 otherwise ​

Let X and Y Have Joint Density Function $p ( x , y ) = \left\{ \begin{array} { l l } x + y & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \\0 & \text { otherwise }\end{array} \right.$