1. **Problem statement:** A bag contains 5 red balls numbered 1, 2, 3, 4, 5 and 9 white balls numbered 6, 7, 8, 9, 10, 11, 12, 13, 14. We want to find: (a) The probability that the ball drawn is red and even-numbered. (b) The probability that the ball drawn is red or even-numbered. 2. **Formula and rules:** - Probability of an event $A$ is $\Pr(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$. - For two events $A$ and $B$: - $\Pr(A \cap B)$ is the probability both $A$ and $B$ occur. - $\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$. 3. **Total number of balls:** $$5 + 9 = 14$$ 4. **(a) Probability red and even-numbered:** - Red balls: $\{1,2,3,4,5\}$ - Even-numbered red balls: $\{2,4\}$ (2 balls) - So, $$\Pr(\text{red and even}) = \frac{2}{14} = \frac{1}{7}$$ 5. **(b) Probability red or even-numbered:** - Red balls: 5 balls - Even-numbered balls: even numbers from 1 to 14 are $2,4,6,8,10,12,14$ (7 balls) - Red and even balls (intersection): 2 balls (2 and 4) - Using the formula: $$\Pr(\text{red or even}) = \Pr(\text{red}) + \Pr(\text{even}) - \Pr(\text{red and even})$$ $$= \frac{5}{14} + \frac{7}{14} - \frac{2}{14} = \frac{10}{14} = \frac{5}{7}$$ **Final answers:** - (a) $\Pr(\text{red and even}) = \frac{1}{7}$ - (b) $\Pr(\text{red or even}) = \frac{5}{7}$