The problem provided involves a probability distribution for a discrete random variable $X$. Let's analyze and solve the questions step by step.

The given table:

Information and Questions:

1. Condition 1: Probability Sum Property For any discrete random variable, the total probability must sum to 1: $\sum_{x} P(X = x) = 1$

2. Part (i): Solve for $k$.

3. Part (ii): Evaluate $P(X \geq 6)$ and $P(3 < X \leq 6)$.

Step 1: Solve for $k$

From the table: $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 1$ Substitute the probabilities: $0 + k + 2k + 2k + 3k + k + 2k + (7k - 2k) = 1$ Simplify: $k + 2k + 2k + 3k + k + 2k + 5k = 1$ $16k = 1$ $k = \frac{1}{16}$

Step 2: Evaluate $P(X \geq 6)$

This corresponds to: $P(X \geq 6) = P(X = 6) + P(X = 7)$ Substitute the probabilities: $P(X = 6) = 2k, \quad P(X = 7) = 5k$ $P(X \geq 6) = 2k + 5k = 7k$ Substitute $k = \frac{1}{16}$: $P(X \geq 6) = 7 \cdot \frac{1}{16} = \frac{7}{16}$

Step 3: Evaluate $P(3 < X \leq 6)$

This corresponds to: $P(3 < X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)$ Substitute the probabilities: $P(X = 4) = 3k, \quad P(X = 5) = k, \quad P(X = 6) = 2k$ $P(3 < X \leq 6) = 3k + k + 2k = 6k$ Substitute $k = \frac{1}{16}$: $P(3 < X \leq 6) = 6 \cdot \frac{1}{16} = \frac{6}{16} = \frac{3}{8}$

Final Answers:

1. $k = \frac{1}{16}$
2. $P(X \geq 6) = \frac{7}{16}$
3. $P(3 < X \leq 6) = \frac{3}{8}$

Would you like further details or an explanation of the steps?

Relative Questions:

1. What is the importance of the total probability summing to 1 for a probability distribution?
2. How do you calculate the cumulative probability $P(X \geq x)$?
3. What are the conditions for a valid probability distribution?
4. How does the value of $k$ affect the overall probabilities?
5. Can you generalize the method to find probabilities for other ranges of $X$?

Tip:

Always verify that the sum of all probabilities equals 1 when solving probability distribution problems!