The random variable X has a probability distribution P(X) of the following form, where k is some arbitary real number:
`P(X=x)={{:(k",","if "x=0), (2k",","if "x=1), (3k",","if "x=2), (0",", otherwise):}`
(i) Determine the value of k `" "` (ii) Find `P(X lt 2), P(X le 2), P(X ge 2)`

Answer

To solve the problem, we will break it down into two parts: determining the value of \( k \) and finding the required probabilities.

### Part (i): Determine the value of \( k \)

1. **Write the probability distribution**:
The probability distribution is given as:
\[ P(X=x) = \begin{cases} k & \text{if } x=0 \\ 2k & \text{if } x=1 \\ 3k & \text{if } x=2 \\ 0 & \text{otherwise} \end{cases} \]

2. **Use the property of probabilities**:
The sum of all probabilities must equal 1:
\[ P(X=0) + P(X=1) + P(X=2) = 1 \]
Substituting the values:
\[ k + 2k + 3k = 1 \]

3. **Combine like terms**:
\[ 6k = 1 \]

4. **Solve for \( k \)**:
\[ k = \frac{1}{6} \]

### Part (ii): Find \( P(X < 2) \), \( P(X \leq 2) \), and \( P(X \geq 2) \)

1. **Calculate \( P(X < 2) \)**:
\[ P(X < 2) = P(X=0) + P(X=1) = k + 2k = 3k \]
Substituting the value of \( k \):
\[ P(X < 2) = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

2. **Calculate \( P(X \leq 2) \)**:
\[ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = k + 2k + 3k = 6k \]
Substituting the value of \( k \):
\[ P(X \leq 2) = 6 \times \frac{1}{6} = 1 \]

3. **Calculate \( P(X \geq 2) \)**:
\[ P(X \geq 2) = P(X=2) + P(X>2) = 3k + 0 = 3k \]
Substituting the value of \( k \):
\[ P(X \geq 2) = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

### Final Answers:
- \( k = \frac{1}{6} \)
- \( P(X < 2) = \frac{1}{2} \)
- \( P(X \leq 2) = 1 \)
- \( P(X \geq 2) = \frac{1}{2} \)