What is the probability? The given answer does not match

Question

A number x is selected randomly from the solution space $ y^2-y-6 \leq 0 $
What is the probability that $ \frac {(x+1)(x-2)}{x-4} \geq 0 $

This is what I have done: Solution space of $ y^2-y-6 \leq 0 $ is $ -2 \leq x \leq 3 $
And from $ \frac {(x+1)(x-2)}{x-4} \geq 0 $ we get $ x \geq 2$ or $x\leq -1$ and $x \neq4 $
So, total space for x = -2 to 3 = 5 conditional space for x= (-1 to -2) + (2 to 3) =2 Therefore, required probability = $ \frac {2} {5} $
What have I done wrong? The provided answer is 3/5

Answer 1

To put it simply, you neglected the influence of that $\frac1{x-4}$ factor, which is always negative in the range of $x$ that we're looking at. That flips the probability over to the complement $1-\frac25=\frac35$.

Answer 2

$$\frac{(x+1)(x-2)}{x-4} \ge 0$$

is equivalent to $x \ne 4$ and $(x+1)(x-2)(x-4) \ge 0$

that is and $-1 \le x \le 2$ and $x > 4$.

The desired probability is $$\frac{2-(-1)}{3-(-2)}=\frac35$$