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### Solving the Problem Using Poisson Distribution

Given the nature of the problem, it's not directly aligned with the Poisson distribution, as the Poisson distribution is typically used for modeling the number of times an event occurs in a fixed interval of time or space. However, if you are referring to a different context within Poisson's application or another question, please provide more details. Otherwise, let's proceed with solving the given questions:

### Question 4

#### (a) Disco lights arrangement:

##### (i) If all nine lights are used:
We use the multinomial coefficient to determine the number of different arrangements of nine lights with 2 green, 3 blue, and 4 red lights:
\[ \text{Number of arrangements} = \frac{9!}{2! \cdot 3! \cdot 4!} \]

Calculating:
\[ 9! = 362880 \]
\[ 2! = 2 \]
\[ 3! = 6 \]
\[ 4! = 24 \]

So:
\[ \frac{9!}{2! \cdot 3! \cdot 4!} = \frac{362880}{2 \cdot 6 \cdot 24} = \frac{362880}{288} = 1260 \]

Thus, there are 1260 different arrangements if all nine lights are used.

##### (ii) If at least eight lights are used:
This requires considering the cases of using 8 lights and using all 9 lights:

1. **Using all 9 lights**: We have already calculated this to be 1260.
2. **Using exactly 8 lights**: We calculate the arrangements for each combination of 8 lights taken from 2 green, 3 blue, and 4 red lights:
   - Removing one green light:
     \[ \frac{8!}{1! \cdot 3! \cdot 4!} = \frac{40320}{6 \cdot 24} = \frac{40320}{144} = 280 \]
   - Removing one blue light:
     \[ \frac{8!}{2! \cdot 2! \cdot 4!} = \frac{40320}{2 \cdot 2 \cdot 24} = \frac{40320}{96} = 420 \]
   - Removing one red light:
     \[ \frac{8!}{2! \cdot 3! \cdot 3!} = \frac{40320}{2 \cdot 6 \cdot 6} = \frac{40320}{72} = 560 \]

So, the total number of arrangements using at least 8 lights:
\[ 1260 + 280 + 420 + 560 = 2520 \]

#### (b) Discrete random variable \( X \):

##### (i) Find the value of \( k \):
The sum of all probabilities must equal 1:
\[ k + 3k + 4k + 5k = 1 \]
\[ 13k = 1 \]
\[ k = \frac{1}{13} \]

So, \( k = \frac{1}{13} \).

##### (ii) Find the mean and variance of \( X \):

**Mean (Expected Value) \( E(X) \)**:
\[ E(X) = \sum_{x} x \cdot P(X = x) \]
\[ E(X) = 0 \cdot \frac{1}{13} + 1 \cdot \frac{3}{13} + 2 \cdot \frac{4}{13} + 3 \cdot \frac{5}{13} \]
\[ E(X) = 0 + \frac{3}{13} + \frac{8}{13} + \frac{15}{13} \]
\[ E(X) = \frac{3 + 8 + 15}{13} = \frac{26}{13} = 2 \]

**Variance \( Var(X) \)**:
\[ Var(X) = E(X^2) - [E(X)]^2 \]

First, calculate \( E(X^2) \):
\[ E(X^2) = \sum_{x} x^2 \cdot P(X = x) \]
\[ E(X^2) = 0^2 \cdot \frac{1}{13} + 1^2 \cdot \frac{3}{13} + 2^2 \cdot \frac{4}{13} + 3^2 \cdot \frac{5}{13} \]
\[ E(X^2) = 0 + \frac{3}{13} + \frac{16}{13} + \frac{45}{13} \]
\[ E(X^2) = \frac{3 + 16 + 45}{13} = \frac{64}{13} \]

Now, find the variance:
\[ Var(X) = \frac{64}{13} - (2)^2 \]
\[ Var(X) = \frac{64}{13} - 4 \]
\[ Var(X) = \frac{64}{13} - \frac{52}{13} = \frac{12}{13} \]

So, the mean of \( X \) is 2 and the variance of \( X \) is \( \frac{12}{13} \).

Would you like more details or have any questions on this topic?

Here are some questions you might want to ask next:
1. How do we interpret the mean and variance in a probability distribution?
2. Can you explain multinomial coefficients in more detail?
3. What are the properties of discrete random variables?
4. How do you find the standard deviation from variance?
5. What are some applications of probability distributions?

**Tip:** When working with probability distributions, always ensure that the sum of probabilities equals 1 to verify the correctness of the distribution.

Learn how to solve a problem using Poisson distribution and multinomial coefficients. This detailed solution covers calculating probabilities and arrangements involving lights, as well as finding the mean and variance of a discrete random variable. Suitable for advanced high school and college-level students.

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