Given the probability distribution of $X$ as follows:

$X: \quad 0 \quad 1 \quad 2 \quad 3 \quad 4$ $P(X = x): \quad K \quad 2K \quad 3K \quad 5K \quad 2K$

i) Finding the value of $K$:

The sum of all probabilities for a discrete random variable must equal 1:

$K + 2K + 3K + 5K + 2K = 1$

Simplifying this:

$K(1 + 2 + 3 + 5 + 2) = 1$

$K(13) = 1$

$K = \frac{1}{13}$

ii) Finding the probabilities:

1. $P(X \geq 2)$: This includes $X = 2$, $X = 3$, and $X = 4$.

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$

$P(X \geq 2) = 3K + 5K + 2K = 10K = 10 \times \frac{1}{13} = \frac{10}{13}$

2. $P(X < 2)$: This includes $X = 0$ and $X = 1$.

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X < 2) = K + 2K = 3K = 3 \times \frac{1}{13} = \frac{3}{13}$

3. $P(X \leq 3)$: This includes $X = 0$, $X = 1$, $X = 2$, and $X = 3$.

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X \leq 3) = K + 2K + 3K + 5K = 11K = 11 \times \frac{1}{13} = \frac{11}{13}$

Summary:

1. $K = \frac{1}{13}$
2. $P(X \geq 2) = \frac{10}{13}$
3. $P(X < 2) = \frac{3}{13}$
4. $P(X \leq 3) = \frac{11}{13}$

Would you like a more detailed explanation or have any questions?

Here are 5 related questions:

1. How would the probability distribution change if $P(X = 3)$ was doubled?
2. What is the expected value $E(X)$ for this distribution?
3. How do you calculate the variance of this probability distribution?
4. If another value $X = 5$ is added with probability $K$, how would $K$ change?
5. What is the cumulative distribution function (CDF) for this random variable $X$?

Tip: When dealing with probability distributions, always check that the total probability sums to 1 before making any calculations.