To solve this, let's first express the probability distribution of the random variable $X$.

Step 1: Express the probabilities in terms of a common variable

Let the probability $P(X = 1) = P(X = 2) = p$.

From the given conditions: $P(X = 3) = \frac{p}{2}, \quad P(X = 4) = \frac{p}{3}.$

Now, the total probability must sum to 1: $P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1.$ Substitute the probabilities: $p + p + \frac{p}{2} + \frac{p}{3} = 1.$ Simplify the left-hand side: $2p + \frac{p}{2} + \frac{p}{3} = 1.$ To combine these, find the least common denominator, which is 6: $2p = \frac{12p}{6}, \quad \frac{p}{2} = \frac{3p}{6}, \quad \frac{p}{3} = \frac{2p}{6}.$ Now add the fractions: $\frac{12p + 3p + 2p}{6} = 1 \quad \Rightarrow \quad \frac{17p}{6} = 1.$ Solve for $p$: $p = \frac{6}{17}.$

Thus, the probabilities are: $P(X = 1) = P(X = 2) = \frac{6}{17}, \quad P(X = 3) = \frac{3}{17}, \quad P(X = 4) = \frac{2}{17}.$

Step 2: Find the required probabilities

(i) $P(X > 2)$

This is the probability that $X = 3$ or $X = 4$: $P(X > 2) = P(X = 3) + P(X = 4) = \frac{3}{17} + \frac{2}{17} = \frac{5}{17}.$

(ii) $P(1 < X < 4 \mid X > 2)$

We are asked to find the conditional probability $P(1 < X < 4 \mid X > 2)$.

First, find $P(1 < X < 4)$, which is the probability that $X = 3$: $P(1 < X < 4) = P(X = 3) = \frac{3}{17}.$ Now, the probability $P(X > 2)$ is already computed as $\frac{5}{17}$.

So the conditional probability is: $P(1 < X < 4 \mid X > 2) = \frac{P(1 < X < 4 \cap X > 2)}{P(X > 2)} = \frac{P(X = 3)}{P(X > 2)} = \frac{\frac{3}{17}}{\frac{5}{17}} = \frac{3}{5}.$

(iii) $P(X = 1 \text{ or } 2)$

This is the probability that $X = 1$ or $X = 2$: $P(X = 1 \text{ or } 2) = P(X = 1) + P(X = 2) = \frac{6}{17} + \frac{6}{17} = \frac{12}{17}.$

Final Answers:

1. $P(X > 2) = \frac{5}{17}$
2. $P(1 < X < 4 \mid X > 2) = \frac{3}{5}$
3. $P(X = 1 \text{ or } 2) = \frac{12}{17}$

---

Would you like further details on any part of the solution?

Here are some related questions you might find interesting:

1. How do you compute the expected value of $X$?
2. What is the variance of $X$?
3. How do conditional probabilities work in different types of probability distributions?
4. What is the cumulative distribution function (CDF) for $X$?
5. How would you solve if the probabilities were not equal at first?

Tip: Conditional probabilities are always calculated by dividing the joint probability by the probability of the given condition.