If the random variable X has the following distribution:

The correct option is (A)- (IV), (B)- (III), (C)- (II), (D)- (I).

Step-by-step Explanation:

From the probability distribution table shown in the image, we have the random variable $X$ and its corresponding probability $P\left(X\right)$ as follows:
• For $X=0$, $P(X=0)=k$
• For $X=1$, $P(X=1)=2k$
• For $X=2$, $P(X=2)=3k$
• For any other value, $P\left(X\right)=0$

Step 1: Find the value of $k$
Since the sum of all probabilities in a probability distribution must equal 1, we can write:
$$\sum P\left(X\right)=1$$
$$P(X=0)+P(X=1)+P(X=2)=1$$
Substituting the given probability values:
$$k+2k+3k=1$$
$$6k=1$$
$$k=\frac{1}{6}$$
Therefore, (A) matches with (IV).

Step 2: Calculate $P(X<2)$
The probability that $X$ is less than 2 is given by:
$$P(X<2)=P(X=0)+P(X=1)$$
$$P(X<2)=k+2k=3k$$
Substituting $k=\frac{1}{6}$:
$$P(X<2)=3\left(\frac{1}{6}\right)=\frac{1}{2}$$
Therefore, (B) matches with (III).

Step 3: Calculate the Expectation $E\left(X\right)$
The expectation of the random variable $X$ is calculated as:
$$E\left(X\right)=\sum x_{i}P\left(x_i\right)$$
$$E\left(X\right)=0·P(X=0)+1·P(X=1)+2·P(X=2)$$
$$E\left(X\right)=0\left(k\right)+1\left(2k\right)+2\left(3k\right)$$
$$E\left(X\right)=2k+6k=8k$$
Substituting $k=\frac{1}{6}$:
$$E\left(X\right)=8\left(\frac{1}{6}\right)=\frac{8}{6}=\frac{4}{3}$$
Therefore, (C) matches with (II).

Step 4: Calculate $P(1\le X\le 2)$
The probability that $X$ lies between 1 and 2 inclusive is:
$$P(1\le X\le 2)=P(X=1)+P(X=2)$$
$$P(1\le X\le 2)=2k+3k=5k$$
Substituting $k=\frac{1}{6}$:
$$P(1\le X\le 2)=5\left(\frac{1}{6}\right)=\frac{5}{6}$$
Therefore, (D) matches with (I).

Conclusion:
Comparing the calculated values with List-II, we have:
• (A) matches (IV)
• (B) matches (III)
• (C) matches (II)
• (D) matches (I)