A random variable $$X$$ has the following probability distribution:

$$X$$	0	1	2	3	4
$$P(X)$$	$$k$$	$$2k$$	$$4k$$	$$6k$$	$$8k$$

The value of $$P\left(\frac{1 < x < 4}{x \leq 2}\right)$$ is equal to

We are given a random variable $$X$$ with the probability distribution:

$$P(X=0) = k, \quad P(X=1) = 2k, \quad P(X=2) = 4k, \quad P(X=3) = 6k, \quad P(X=4) = 8k$$

Since the total probability must equal 1:

$$k + 2k + 4k + 6k + 8k = 1$$

$$21k = 1$$, so $$k = \frac{1}{21}$$

We need to find $$P\left(\frac{1 < X < 4}{X \leq 2}\right)$$, which is the conditional probability $$P(1 < X < 4 \mid X \leq 2)$$.

The event $$\{1 < X < 4\} = \{X = 2, X = 3\}$$

The event $$\{X \leq 2\} = \{X = 0, X = 1, X = 2\}$$

The intersection $$\{1 < X < 4\} \cap \{X \leq 2\} = \{X = 2\}$$

$$P(X = 2) = 4k = \frac{4}{21}$$

$$P(X \leq 2) = k + 2k + 4k = 7k = \frac{7}{21} = \frac{1}{3}$$

Therefore:

$$P(1 < X < 4 \mid X \leq 2) = \frac{P(\{1 < X < 4\} \cap \{X \leq 2\})}{P(X \leq 2)} = \frac{4/21}{7/21} = \frac{4}{7}$$

The correct answer is Option A.