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Let's walk through the problem step-by-step.

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The given probability mass function is:

\[
P(X = x) = 
\begin{cases}
k(1 - x)^2 & \text{for } x = -1, 0, 1, 2, \\
0 & \text{otherwise}.
\end{cases}
\]

---

### (i) Show that \( k = \frac{1}{6} \).

Since this is a probability mass function, the total sum of all probabilities must be 1:

\[
\sum_{x = -1}^{2} P(X = x) = 1.
\]

We substitute the function \( P(X = x) = k(1 - x)^2 \) for the valid values of \( x \):

\[
k(1 - (-1))^2 + k(1 - 0)^2 + k(1 - 1)^2 + k(1 - 2)^2 = 1.
\]

Evaluating each term:

- When \( x = -1 \), \( (1 - (-1))^2 = (2)^2 = 4 \).
- When \( x = 0 \), \( (1 - 0)^2 = 1^2 = 1 \).
- When \( x = 1 \), \( (1 - 1)^2 = 0^2 = 0 \).
- When \( x = 2 \), \( (1 - 2)^2 = (-1)^2 = 1 \).

Thus, the equation becomes:

\[
k(4 + 1 + 0 + 1) = 1.
\]

\[
k(6) = 1 \quad \Rightarrow \quad k = \frac{1}{6}.
\]

So, \( k = \frac{1}{6} \) is correct.

---

### (ii) Find \( E(X) \).

The expected value \( E(X) \) is defined as:

\[
E(X) = \sum_{x = -1}^{2} x \cdot P(X = x).
\]

Using \( P(X = x) = \frac{1}{6}(1 - x)^2 \), we compute each term:

- For \( x = -1 \): \( -1 \cdot \frac{1}{6}(4) = -\frac{4}{6} = -\frac{2}{3}. \)
- For \( x = 0 \): \( 0 \cdot \frac{1}{6}(1) = 0. \)
- For \( x = 1 \): \( 1 \cdot \frac{1}{6}(0) = 0. \)
- For \( x = 2 \): \( 2 \cdot \frac{1}{6}(1) = \frac{2}{6} = \frac{1}{3}. \)

Thus:

\[
E(X) = -\frac{2}{3} + 0 + 0 + \frac{1}{3} = -\frac{1}{3}.
\]

---

### (iii) Find \( E(X^2) \).

The expected value \( E(X^2) \) is defined as:

\[
E(X^2) = \sum_{x = -1}^{2} x^2 \cdot P(X = x).
\]

We compute each term:

- For \( x = -1 \): \( (-1)^2 \cdot \frac{1}{6}(4) = 1 \cdot \frac{4}{6} = \frac{2}{3}. \)
- For \( x = 0 \): \( (0)^2 \cdot \frac{1}{6}(1) = 0. \)
- For \( x = 1 \): \( (1)^2 \cdot \frac{1}{6}(0) = 0. \)
- For \( x = 2 \): \( (2)^2 \cdot \frac{1}{6}(1) = 4 \cdot \frac{1}{6} = \frac{2}{3}. \)

Thus:

\[
E(X^2) = \frac{2}{3} + 0 + 0 + \frac{2}{3} = \frac{4}{3}.
\]

---

### (iv) Find the Variance \( \text{Var}(X) \).

The variance \( \text{Var}(X) \) is given by:

\[
\text{Var}(X) = E(X^2) - \left( E(X) \right)^2.
\]

From the previous parts, we know:

- \( E(X^2) = \frac{4}{3} \),
- \( E(X) = -\frac{1}{3} \).

Thus:

\[
\text{Var}(X) = \frac{4}{3} - \left( -\frac{1}{3} \right)^2 = \frac{4}{3} - \frac{1}{9}.
\]

Simplifying:

\[
\frac{4}{3} = \frac{12}{9}, \quad \frac{12}{9} - \frac{1}{9} = \frac{11}{9}.
\]

So:

\[
\text{Var}(X) = \frac{11}{9}.
\]

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### Final Answers:

1. \( k = \frac{1}{6} \).
2. \( E(X) = -\frac{1}{3} \).
3. \( E(X^2) = \frac{4}{3} \).
4. \( \text{Var}(X) = \frac{11}{9} \).

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Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How does the variance measure the spread of a probability distribution?
2. What is the significance of finding \( E(X) \) in a distribution?
3. How would the distribution change if we modified the function to \( k(1 - x)^3 \)?
4. Can the value of \( E(X) \) be used to predict outcomes? In what situations?
5. How is the variance affected if the random variable is shifted or scaled?

**Tip:** When solving probability problems, always ensure the probabilities sum to 1 to confirm you’ve used the correct constant \( k \).

A discrete random variable X has the distribution function: P(X = x) = k(1 - x)^2, where x = -1, 0, 1, 2. Find k, E(X), E(X^2), and Var(X).

Learn how to find k, E(X), E(X^2), and Var(X) for a discrete random variable with the distribution P(X = x) = k(1 - x)^2 for x = -1, 0, 1, 2. This problem covers essential concepts in probability theory, including expected value and variance.

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