What's the probability that $y\ge x+1$? "Two numbers, $x$ and $y$ are selected at random from the interval $(0,3)$. What is the probability that $y\ge x+1$?"
The answer key says $\frac 29$, but I can't figure out how to get to that answer.
 A: Consider coordinates
where $0 \le x, y \le 3$.
This has area $9$.
The points where
$y \ge x+1$
make up the part of this square
on or above the line
$y = x+1$.
This is a right triangle
with sides 2 and 2,
so its area is 2.
The probability wanted is
the ratio of these
or 2/9.
More generally,
suppose you want to find the probability
that $y \ge x+c$
where $0 \le c \le 3$.
The area where this is true is
$(3-c)^2/2$
(the sides of the
cut off triangle
have length $3-c$),
so the probability is
$(3-c)^2/18$.
A: Drawing a diagram is okay for two variables, but in general we can use integration.
For the problem:
$\displaystyle 
\mathbb{P}(y\ge x+1)=\dfrac{\displaystyle\int_0^2 \int_{x+1}^3 \, dy\, dx}{\displaystyle\int_0^3 \int_{0}^3 \, dy\, dx}=\dfrac{2}{9}$
A: Hint: Imagine a $3 \times 3$ square with coordinates $(0,0), (0,3), (3,0), (3,3)$. When yo pick $x$ and $y$, you are choosing any point on the interior of this square, each being equally likely. Find the intersection of this square with the inequality graph $y \geq x + 1$.
