1. **Problem statement:** We have a box with 5 red and 3 white balls (total 8 balls). Two balls are drawn in succession. 2. **Goal:** Find the probability that the second ball drawn is red given two scenarios: (A) The first ball is replaced before the second draw. (B) The first ball is not replaced before the second draw. 3. **Important rule:** When a ball is replaced, the total number of balls remains the same for the second draw. When not replaced, the total decreases by one. --- ### Part (A): First ball replaced before second draw 4. Since the first ball is replaced, the composition of the box remains the same for the second draw. 5. The probability that the second ball is red is simply the probability of drawing a red ball from the original box: $$P(\text{second red}) = \frac{5}{8}$$ 6. Therefore, the answer for part (A) is $\frac{5}{8}$. --- ### Part (B): First ball not replaced before second draw 7. The first ball drawn was red or white, but we are given the first ball was drawn (no color specified here, but the problem asks for the probability the second ball is red given the first ball was drawn and not replaced). 8. Since the first ball is not replaced, the total number of balls for the second draw is 7. 9. We want $P(\text{second red} \mid \text{first ball drawn and not replaced})$. 10. The problem states "given that the first ball was drawn" but does not specify the color of the first ball. However, the problem is standard and usually means the first ball was drawn (any color), and we want the probability the second ball is red. 11. To find this, we consider the two cases for the first ball: - First ball red: probability $\frac{5}{8}$, then second draw has 4 red and 3 white balls left, so $P(\text{second red} \mid \text{first red}) = \frac{4}{7}$. - First ball white: probability $\frac{3}{8}$, then second draw has 5 red and 2 white balls left, so $P(\text{second red} \mid \text{first white}) = \frac{5}{7}$. 12. Using total probability: $$P(\text{second red}) = P(\text{first red}) \times P(\text{second red} \mid \text{first red}) + P(\text{first white}) \times P(\text{second red} \mid \text{first white})$$ $$= \frac{5}{8} \times \frac{4}{7} + \frac{3}{8} \times \frac{5}{7} = \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}$$ 13. Interestingly, the probability is the same as in part (A), $\frac{5}{8}$. --- **Final answers:** - (A) $\frac{5}{8}$ - (B) $\frac{5}{8}$