A random variable X has the following probability distribution: X 0...

To find $$P\left(\dfrac{1 < x < 4}{x \le 2}\right)$ or $P\left(\dfrac{A}{B}\right)$

We know that

$P\left(\dfrac{A}{B}\right) = \dfrac{P\left(A \cap B\right)}{P\left(B\right)}$

Given

$P\left(A\right) = \left\{2, 3\right\}$

$P\left(B\right) = \left\{0, 1, 2\right\}$

$P\left(A \cap B\right) = P\left(x = 2\right)$

$P\left(B\right) = P\left(x = 0\right) + P\left(x = 1\right) + P\left(x = 2\right)$

So $P\left(\dfrac{A}{B}\right) = \dfrac{P\left(x = 2\right)}{P\left(x = 0\right) + P\left(x = 1\right) + P\left(x = 2\right)}$

$= \dfrac{4k}{k + 2k + 4k} = \dfrac{4k}{7k} = \dfrac{4}{7}$

Hence option $A is correct.