The random variable $x$ has a probability distribution $p(x)$ of the form $P(x=r)=\begin{cases} rk, & \text{if } r \le 2, \\ (r-1)k, & \text{if } 2<r\le 4, \\ 0, & \text{otherwise,} \end{cases}$ where $r \in \mathbb{N}\cup\{0\}$ and $k\in\mathbb{R}$, where $\mathbb{N}$ is the set of natural numbers. Then:

(A) $k=\frac{1}{9}$

(B) $P(2\le x\le 3)=\frac{1}{2}$

(C) $P(x=4)=\frac{1}{3}$

(D) $P(x>1)=\frac{7}{8}$

Choose the correct answer from the options given below.

The random variable takes values $r \in \{0, 1, 2, 3, 4\}$. Let's compute $P(x = r)$ for each:

$P(x = 0) = 0 \cdot k = 0$

$P(x = 1) = 1 \cdot k = k$

$P(x = 2) = 2 \cdot k = 2k$

$P(x = 3) = (3-1) \cdot k = 2k$

$P(x = 4) = (4-1) \cdot k = 3k$

Since all probabilities must sum to 1:

$0 + k + 2k + 2k + 3k = 1$

$8k = 1$

$k = \frac{1}{8}$

So the complete distribution is:

$P(x = 0) = 0, \\ P(x = 1) = \frac{1}{8}, \\ P(x = 2) = \frac{2}{8}, \\ P(x = 3) = \frac{2}{8}, \\ P(x = 4) = \frac{3}{8}$

Checking option (A): $k = \frac{1}{9}$

We found $k = \frac{1}{8}$, so this is incorrect. ❌

Checking option (B): $P(2 \leq x \leq 3) = \frac{1}{2}$

$P(2 \leq x \leq 3) = P(x=2) + P(x=3)$

$= \frac{2}{8} + \frac{2}{8} = \frac{4}{8} = \frac{1}{2}$

This is correct. ✅

Checking option (C): $P(x = 4) = \frac{1}{3}$

$P(x = 4) = \frac{3}{8} \neq \frac{1}{3}$

This is incorrect. ❌

Checking option (D): $P(x > 1) = \frac{7}{8}$

$P(x > 1) = P(x=2) + P(x=3) + P(x=4)$

$= \frac{2}{8} + \frac{2}{8} + \frac{3}{8} = \frac{7}{8}$

This is correct. ✅

Options (B) and (D) are correct.