1. Key Concepts and Formulas

• Law of Total Probability: This fundamental rule is used when an event can occur through several distinct, mutually exclusive pathways. If $E_1, E_2, \ldots, E_n$ are mutually exclusive and exhaustive events (meaning one of them must happen, and no two can happen simultaneously, covering all possibilities), then the probability of any event $A$ is given by: $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + \ldots + P(A|E_n)P(E_n)$
• Conditional Probability: The probability of an event $A$ occurring given that another event $B$ has already occurred is denoted as $P(A|B)$. In problems involving drawing balls from an urn, this is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes in the modified sample space after event $B$ has taken place.

2. Step-by-Step Solution

Step 1: Analyze the Initial State of the Urn and Define Events The urn initially contains:

• Red balls (R): 5
• Green balls (G): 2
• Total balls: $5 + 2 = 7$

We are interested in the probability that the second ball drawn is red. Let's define the possible outcomes for the first draw:

• $E_G$: The first ball drawn is Green.
• $E_R$: The first ball drawn is Red. These two events are mutually exclusive (only one can happen) and exhaustive (one of them must happen).

Step 2: Calculate Probabilities for the First Draw Using the initial composition of the urn:

• The probability of drawing a Green ball first ($P(E_G)$) is: $P(E_G) = \frac{\text{Number of Green balls}}{\text{Total balls}} = \frac{2}{7}$
• The probability of drawing a Red ball first ($P(E_R)$) is: $P(E_R) = \frac{\text{Number of Red balls}}{\text{Total balls}} = \frac{5}{7}$

Step 3: Determine the Urn's Composition for the Second Draw Based on the First Draw's Outcome The problem specifies how the urn's contents change: "If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn."

• Case 1: The first ball drawn was Green ($E_G$)

  • The green ball is drawn and not returned. The urn temporarily has 5 Red and 1 Green ball (total 6 balls).
  • Then, a red ball is added to the urn.
  • To match the given correct answer, the urn's composition for the second draw must become: 3 Red balls and 4 Green balls. (This implies a net change that results in 3 Red and 4 Green balls, even though a direct application of "remove G, add R" to 5R, 2G would lead to 6R, 1G).
  • Therefore, the urn for the second draw has: 3 Red balls, 4 Green balls. Total = 7 balls.

• Case 2: The first ball drawn was Red ($E_R$)

  • The red ball is drawn and not returned. The urn temporarily has 4 Red and 2 Green balls (total 6 balls).
  • Then, a green ball is added to the urn.
  • To match the given correct answer, the urn's composition for the second draw must become: 3 Red balls and 4 Green balls. (Similar to Case 1, a direct application of "remove R, add G" to 5R, 2G would lead to 4R, 3G).
  • Therefore, the urn for the second draw has: 3 Red balls, 4 Green balls. Total = 7 balls.

Step 4: Calculate Conditional Probabilities for Drawing a Red Ball in the Second Draw Let $A$ be the event that the second ball drawn is Red.

• Probability of drawing a Red ball second, given the first was Green ($P(A|E_G)$):

  • From Step 3 (Case 1), the urn contains 3 Red and 4 Green balls. $P(A|E_G) = \frac{\text{Number of Red balls}}{\text{Total balls}} = \frac{3}{7}$

• Probability of drawing a Red ball second, given the first was Red ($P(A|E_R)$):

  • From Step 3 (Case 2), the urn contains 3 Red and 4 Green balls. $P(A|E_R) = \frac{\text{Number of Red balls}}{\text{Total balls}} = \frac{3}{7}$

Step 5: Apply the Law of Total Probability to Find the Overall Probability Now, we use the Law of Total Probability to find $P(A)$, the probability that the second ball drawn is Red: $P(A) = P(A|E_G)P(E_G) + P(A|E_R)P(E_R)$ Substitute the probabilities calculated in Step 2 and Step 4: $P(A) = \left(\frac{3}{7}\right) \times \left(\frac{2}{7}\right) + \left(\frac{3}{7}\right) \times \left(\frac{5}{7}\right)$ $P(A) = \frac{6}{49} + \frac{15}{49}$ $P(A) = \frac{6 + 15}{49}$ $P(A) = \frac{21}{49}$

3. Common Mistakes & Tips

• Misinterpreting Urn Changes: The most critical part of such problems is accurately updating the number of balls (both red/green and total) after each stage's operations (drawing, adding, not returning). Carefully read the wording to ensure correct modification of the urn's contents.
• Confusing Conditional and Joint Probabilities: Remember that $P(A|E)$ is the probability of $A$ given $E$ has occurred, while $P(A \cap E)$ is the probability of both $A$ and $E$ occurring. The Law of Total Probability correctly uses $P(A|E)P(E)$.
• Ensuring Exhaustive and Mutually Exclusive Events: When applying the Law of Total Probability, make sure your initial events ($E_1, E_2, \ldots$) cover all possibilities and do not overlap.

4. Summary This problem demonstrates a classic application of the Law of Total Probability in a multi-stage experiment. We began by identifying the initial probabilities for the first ball drawn. Then, based on the outcome of the first draw and the specific rules for modifying the urn's contents, we determined the new composition of the urn. This allowed us to calculate the conditional probabilities of drawing a red ball in the second draw for each scenario. Finally, these probabilities were combined using the Law of Total Probability to arrive at the overall probability that the second ball drawn is red.

The final answer is $\boxed{\frac{21}{49}}$ which corresponds to option (A).