1. **Problem Statement:** A discrete random variable $X$ has the probability distribution given by: $$X: 0, 1, 2, 3, 4, 5, 6, 7, 8$$ $$P(X): a, 3a, 5a, 7a, 9a, 11a, 13a, 15a, 17a$$ Find: (i) The value of $a$. (ii) $P(0 < X < 3)$. (iii) $P(X > 2)$. --- 2. **Formula and Rules:** - The sum of all probabilities must equal 1: $$\sum P(X) = 1$$ - Probability of an event is the sum of probabilities of the outcomes in that event. --- 3. **Step (i): Find $a$** Sum of probabilities: $$a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1$$ Calculate the sum of coefficients: $$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81$$ So, $$81a = 1 \implies a = \frac{1}{81}$$ --- 4. **Step (ii): Calculate $P(0 < X < 3)$** This means $X = 1$ or $X = 2$: $$P(1) + P(2) = 3a + 5a = 8a = 8 \times \frac{1}{81} = \frac{8}{81}$$ --- 5. **Step (iii): Calculate $P(X > 2)$** This means $X = 3,4,5,6,7,8$: $$7a + 9a + 11a + 13a + 15a + 17a = (7 + 9 + 11 + 13 + 15 + 17)a$$ Sum coefficients: $$7 + 9 + 11 + 13 + 15 + 17 = 72$$ So, $$P(X > 2) = 72a = 72 \times \frac{1}{81} = \frac{72}{81} = \frac{8}{9}$$ --- **Final answers:** (i) $a = \frac{1}{81}$ (ii) $P(0 < X < 3) = \frac{8}{81}$ (iii) $P(X > 2) = \frac{8}{9}$