Let $X$ be a discrete random variable taking values $1, 2, 3, 4$, and the relationships provided are:

$2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4)$

Let the probabilities of these events be expressed in terms of a common variable. Let:

$P(X = 4) = p$

Using the given relations, we can express the other probabilities in terms of $p$:

$P(X = 3) = 5p$ $P(X = 2) = \frac{5p}{3}$ $P(X = 1) = \frac{5p}{6}$

Now, since the total probability must sum to 1 (i.e., $P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$), we have:

$\frac{5p}{6} + \frac{5p}{3} + 5p + p = 1$

We can now solve for $p$. First, let's get a common denominator for the terms on the left-hand side:

$\frac{5p}{6} + \frac{10p}{6} + \frac{30p}{6} + \frac{6p}{6} = 1$

Simplifying the left-hand side:

$\frac{51p}{6} = 1$

Solving for $p$:

$p = \frac{6}{51} = \frac{2}{17}$

Now we can find the probabilities:

$P(X = 4) = \frac{2}{17}, \quad P(X = 3) = 5p = \frac{10}{17}, \quad P(X = 2) = \frac{5p}{3} = \frac{10}{51} = \frac{10}{17}, \quad P(X = 1) = \frac{5p}{6} = \frac{5}{17}$

Probability Distribution:

• $P(X = 1) = \frac{5}{17}$
• $P(X = 2) = \frac{10}{51}$
• $P(X = 3) = \frac{10}{17}$
• $P(X = 4) = \frac{2}{17}$

Cumulative Distribution Function (CDF):

To find the cumulative distribution function $F(X)$, we sum the probabilities up to each value of $X$:

• $F(1) = P(X = 1) = \frac{5}{17}$
• $F(2) = P(X = 1) + P(X = 2) = \frac{5}{17} + \frac{10}{17} = \frac{10}{71}$