1. **State the problem:** We have a discrete probability distribution with values $x = \{1, 2, 3, 4, 5\}$ and probabilities $p(x) = \left\{\frac{1}{9}, \frac{2}{9}, \frac{1}{3}, \frac{2}{9}, \frac{1}{9}\right\}$. We need to find: 1) $p(x \geq 3)$ 2) $p(x < 5)$ 3) $p(\text{even})$ 4) $p(\text{even}) + p(\text{odd})$ 5) $p(\text{even}) - p(3)$ 2. **Recall the rules:** The sum of all probabilities must be 1. Probabilities of events are sums of probabilities of individual outcomes in those events. 3. **Calculate each:** 1) $p(x \geq 3) = p(3) + p(4) + p(5) = \frac{1}{3} + \frac{2}{9} + \frac{1}{9} = \frac{3}{9} + \frac{2}{9} + \frac{1}{9} = \frac{6}{9} = \frac{2}{3}$ 2) $p(x < 5) = p(1) + p(2) + p(3) + p(4) = \frac{1}{9} + \frac{2}{9} + \frac{1}{3} + \frac{2}{9} = \frac{1}{9} + \frac{2}{9} + \frac{3}{9} + \frac{2}{9} = \frac{8}{9}$ 3) Even values are $2$ and $4$, so $p(\text{even}) = p(2) + p(4) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$ 4) Odd values are $1, 3, 5$, so $p(\text{odd}) = p(1) + p(3) + p(5) = \frac{1}{9} + \frac{1}{3} + \frac{1}{9} = \frac{1}{9} + \frac{3}{9} + \frac{1}{9} = \frac{5}{9}$. Then, $$p(\text{even}) + p(\text{odd}) = \frac{4}{9} + \frac{5}{9} = 1$$ 5) $p(\text{even}) - p(3) = \frac{4}{9} - \frac{1}{3} = \frac{4}{9} - \frac{3}{9} = \frac{1}{9}$ 4. **Summary:** 1) $p(x \geq 3) = \frac{2}{3}$ 2) $p(x < 5) = \frac{8}{9}$ 3) $p(\text{even}) = \frac{4}{9}$ 4) $p(\text{even}) + p(\text{odd}) = 1$ 5) $p(\text{even}) - p(3) = \frac{1}{9}$ These results confirm the probabilities are consistent and sum correctly.