A random variable $X$ has the following probability distribution $\begin{array}{|r|c|c|c|c|c|} \hline X: & 0…

Probability Distribution:

$X: 0, 1, 2, 3, 4$
$P(X): k, 2k, 4k, 2k, k$

The total probability must sum to $1$: $k + 2k + 4k + 2k + k = 10k = 1$, yielding $k = \frac{1}{10}$.

The probabilities are thus $P(X=0) = \frac{1}{10}$, $P(X=1) = \frac{2}{10}$, $P(X=2) = \frac{4}{10}$, $P(X=3) = \frac{2}{10}$, $P(X=4) = \frac{1}{10}$.

For the conditional probability $P(1 \leqslant X < 4 \mid X \leqslant 2)$, define $A = \{1, 2, 3\}$ and $B = \{0, 1, 2\}$.

$P(B) = P(X \leqslant 2) = \frac{1}{10} + \frac{2}{10} + \frac{4}{10} = \frac{7}{10}$.

$A \cap B = \{1, 2\}$, so $P(A \cap B) = \frac{2}{10} + \frac{4}{10} = \frac{6}{10}$.

By the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{6/10}{7/10} = \frac{6}{7}$.

Final Answer: $\frac{6}{7}$