Another probability problem

Observe one root is -2 . so other root is 4/a . now 4/a should lie between 1 and 2.
So 2<a<4 , so answer is 2/10

By the Quadratic Formula, the roots of the above inequality lie between:

$x = \frac{(4-2a) \pm \sqrt{(4-2a)^2 - 4(a)(-8)}}{2a} = \frac{(4-2a) \pm \sqrt{4a^2 + 16a + 16}}{2a} = \frac{(4-2a)\pm \sqrt{4(a+2)^2}}{2a} = \frac{(4-2a) \pm 2(a+2)}{2a} \Rightarrow -2 < x < \frac{4}{a}.$

For $a \in (0,2) \Rightarrow x$ has more than 3 integral solutions;

For $a \in [2, 4) \Rightarrow x$ has exactly 3 integral solutions;

For $a \in [4, 10) \Rightarrow x$ has exactly 2 integral solutions.

If $a$ is uniformly distributed over $(0, 10)$ , then its pdf is $f_{A}(a) = \frac{1}{10-0} = \frac{1}{10}$ . Calculating $P(2 \le a < 4)$ gives:

$P(2 \le a < 4) = \int_{2}^{4} \frac{1}{10} da = \frac{a}{10}|_{2}^{4} = \boxed{\frac{1}{5}}.$