Probability of $y-2x\geq-6$, if $x>y$

Question

We have an interval $[-2,4]$ and we select two random numbers $x$ and $y$.

a) What is the probability of $$y-2x\geq-6$$ if $x > y$.

I change $$y-2x\geq-6$$ into $$y\geq 2x-6$$ and get a graph and 'paint' everything above the line. The painted area's surface is $32$ and then get the probability
$$\frac{32}{36}=\frac{8}{9}$$Correct answer is $$\frac{7}{9}$$What am I doing wrong here?

Answer 1

This is a conditional probability, since we are given that $x < y$. The region defined by $x < y$ is precisely half of the $6 \times 6$ square. So we have: $$ \Pr[y \geq 2x - 6 \mid y < x] = \frac{\Pr[2x - 6 \leq y < x]}{\Pr[y < x]} = \frac{14/36}{18/36} = \frac{14}{18} = \frac{7}{9} $$

Answer 2

You are using the wrong textbook. The author of the text must have carelessly decided that the area of the triangle is base times height, or that $-6+2\cdot 4 = 4$. You will have an easier time understanding where you go wrong in problems if the answer presented is the right answer.