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We can choose 2 balls from a total of 13 balls (5 white and 8 red) in the urn. The number of ways we can do this is denoted as combination and can be represented as \({13 \choose 2}\).

There are 4 possible cases for \(X_{1}\) and \(X_{2}\): 1. Both \(X_{1}\) and \(X_{2}\) are white: There are \({5 \choose 2}\) ways to choose 2 white balls from 5 white balls and the probability is \(\frac{{5 \choose 2}}{{13 \choose 2}}\). 2. \(X_{1}\) is white and \(X_{2}\) is red: There are \({5 \choose 1}{8 \choose 1}\) ways to choose 1 white ball and 1 red ball. The probability is \(\frac{{5 \choose 1}{8 \choose 1}}{{13 \choose 2}}\). 3. \(X_{1}\) is red and \(X_{2}\) is white: Similarly, there are \({8 \choose 1}{5 \choose 1}\) ways and the probability is \(\frac{{8 \choose 1}{5 \choose 1}}{{13 \choose 2}}\). 4. Both \(X_{1}\) and \(X_{2}\) are red: There are \({8 \choose 2}\) ways to choose 2 red balls from 8 red balls and the probability is \(\frac{{8 \choose 2}}{{13 \choose 2}}\). We can represent the joint probability mass function of \(X_{1}, X_{2}\) as: \(P(X_{1}, X_{2}) = \begin{cases} \frac{{5 \choose 2}}{{13 \choose 2}} & \text{if both } X_{1} \text{ and } X_{2} \text{ are white,} \\ \frac{{5 \choose 1}{8 \choose 1}}{{13 \choose 2}} & \text{if } X_{1} \text{ is white and } X_{2} \text{ is red,} \\ \frac{{8 \choose 1}{5 \choose 1}}{{13 \choose 2}} & \text{if } X_{1} \text{ is red and } X_{2} \text{ is white,} \\ \frac{{8 \choose 2}}{{13 \choose 2}} & \text{if both } X_{1} \text{ and } X_{2} \text{ are red.} \end{cases}\) (b) Joint probability mass function of \(X_{1}\), \(X_{2}\), and \(X_{3}\)

We can choose 3 balls from a total of 13 balls in the urn. The number of ways we can do this is denoted as combination and can be represented as \({13 \choose 3}\).

There are 8 possible cases for \(X_{1}\), \(X_{2}\), and \(X_{3}\), and their respective probabilities are: 1. All three balls are white: \({5 \choose 3}\) ways and probability \(\frac{{5 \choose 3}}{{13 \choose 3}}\). 2. 2 white and 1 red: \({5 \choose 2}{8 \choose 1}\) ways and probability \(\frac{{5 \choose 2}{8 \choose 1}}{{13 \choose 3}}\). 3. 1 white and 2 red: \({5 \choose 1}{8 \choose 2}\) ways and probability \(\frac{{5 \choose 1}{8 \choose 2}}{{13 \choose 3}}\). 4. All three balls are red: \({8 \choose 3}\) ways and probability \(\frac{{8 \choose 3}}{{13 \choose 3}}\). 5. The other 4 cases are related to the order of balls being chosen. The complete joint probability mass function of \(X_{1}, X_{2}, X_{3}\) can be represented as a fraction for each of the 8 possible combinations with the events corresponding to white and red balls selected for \(X_{1}\), \(X_{2}\), and \(X_{3}\).