Let's break the problem into two parts: when the first ball is replaced and when it is not replaced before the second draw.

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Part (A): When the First Ball is Replaced

Here, after drawing the first ball, we put it back into the box. The number of balls and their proportions remain the same for the second draw.

• Total number of balls: $5 + 6 = 11$
• Probability that the second ball drawn is red, regardless of the first ball:
  • Probability of drawing a red ball on any draw = $\frac{5}{11}$

Since the replacement keeps the conditions for each draw identical, the probability that the second ball is red (given that the first ball was replaced) is simply: $P(\text{Second ball is red}) = \frac{5}{11}$

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Part (B): When the First Ball is Not Replaced

Here, the situation changes after drawing the first ball because we do not replace it, altering the total number of balls.

Two cases to consider:

1. The first ball drawn is red:

   • Now, there are 4 red balls and 6 white balls left.
   • Probability of drawing a red ball on the second draw: $P(\text{Second ball is red | First ball was red}) = \frac{4}{10} = \frac{2}{5}$

2. The first ball drawn is white:

   • Now, there are still 5 red balls and 5 white balls left.
   • Probability of drawing a red ball on the second draw: $P(\text{Second ball is red | First ball was white}) = \frac{5}{10} = \frac{1}{2}$

Now, we use the law of total probability to find the overall probability that the second ball is red. We need to account for both possible outcomes of the first draw (red or white).

• Probability of drawing a red ball first = $\frac{5}{11}$
• Probability of drawing a white ball first = $\frac{6}{11}$

Thus, the total probability that the second ball is red: $P(\text{Second ball is red}) = P(\text{First red}) \cdot P(\text{Second red | First red}) + P(\text{First white}) \cdot P(\text{Second red | First white})$ Substituting values: $P(\text{Second ball is red}) = \left( \frac{5}{11} \cdot \frac{2}{5} \right) + \left( \frac{6}{11} \cdot \frac{1}{2} \right)$ $P(\text{Second ball is red}) = \frac{10}{55} + \frac{12}{55} = \frac{22}{55}$

Thus, when the first ball is not replaced, the probability that the second ball is red is $\frac{22}{55}$.

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Final Answer:

• (A) With replacement: $\frac{5}{11}$
• (B) Without replacement: $\frac{22}{55}$

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Would you like further details on any part of the solution?

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Here are 5 related questions you might find useful:

1. How would the probabilities change if there were an equal number of red and white balls in the box?
2. What is the probability that both balls drawn are red (with and without replacement)?
3. How does the probability change if three balls are drawn instead of two (with and without replacement)?
4. What is the probability that at least one red ball is drawn (with and without replacement)?
5. How would you calculate the probability if there were more than two colors of balls?

Tip: When dealing with "without replacement" scenarios, always account for the change in total number of balls after each draw.