To find the probability distribution and cumulative distribution function (CDF) of $X$, follow these steps:

Probability Distribution:

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

2. According to the problem:

   • $2P(X=1) = 3P(X=2)$ implies $P(X=1) = \frac{3}{2}p$.
   • $P(X=2) = p$.
   • $P(X=3) = p$.
   • $5P(X=4) = P(X=3)$ implies $P(X=4) = \frac{p}{5}$.

3. Using the total probability condition: $P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$ $\frac{3}{2}p + p + p + \frac{p}{5} = 1$ $\frac{15p}{10} + \frac{10p}{10} + \frac{10p}{10} + \frac{2p}{10} = 1$ $\frac{37p}{10} = 1$ $p = \frac{10}{37}$

4. Substitute $p$ back to find the probabilities:

   • $P(X=1) = \frac{3}{2}p = \frac{3}{2} \times \frac{10}{37} = \frac{15}{37}$
   • $P(X=2) = \frac{10}{37}$
   • $P(X=3) = \frac{10}{37}$
   • $P(X=4) = \frac{p}{5} = \frac{2}{37}$

Cumulative Distribution Function (CDF):

The CDF, $F(x) = P(X \leq x)$, is calculated as follows:

• $F(1) = P(X \leq 1) = P(X = 1) = \frac{15}{37}$
• $F(2) = P(X \leq 2) = P(X=1) + P(X=2) = \frac{15}{37} + \frac{10}{37} = \frac{25}{37}$
• $F(3) = P(X \leq 3) = P(X=1) + P(X=2) + P(X=3) = \frac{25}{37} + \frac{10}{37} = \frac{35}{37}$
• $F(4) = P(X \leq 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = \frac{35}{37} + \frac{2}{37} = 1$

Summary:

• Probability Distribution:

  • $P(X=1) = \frac{15}{37}$
  • $P(X=2) = \frac{10}{37}$
  • $P(X=3) = \frac{10}{37}$
  • $P(X=4) = \frac{2}{37}$

• Cumulative Distribution Function:

  • $F(1) = \frac{15}{37}$
  • $F(2) = \frac{25}{37}$
  • $F(3) = \frac{35}{37}$
  • $F(4) = 1$