A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball?

Let $R_1$ and $G_1$ be the events that the first ball drawn is red and green, respectively. Initially, the pot contains 5 red and 2 green balls, total 7 balls.

Probability of drawing a red ball first: $P(R_1) = \dfrac{5}{7}$.
Probability of drawing a green ball first: $P(G_1) = \dfrac{2}{7}$.

Case 1: If the first ball is red ($R_1$), it is not replaced, and a green ball is added. The pot now contains $(5-1)$ red and $(2+1)$ green balls, i.e., 4 red and 3 green balls. Total balls = 7. The probability of drawing a red ball second given $R_1$ is $P(R_2 | R_1) = \dfrac{4}{7}$.

Case 2: If the first ball is green ($G_1$), it is replaced by a red ball. The pot now contains $(5+1)$ red and $(2-1)$ green balls, i.e., 6 red and 1 green ball. Total balls = 7. The probability of drawing a red ball second given $G_1$ is $P(R_2 | G_1) = \dfrac{6}{7}$.

Using the law of total probability:
$P(R_2) = P(R_1)P(R_2 | R_1) + P(G_1)P(R_2 | G_1)$
$P(R_2) = \left(\dfrac{5}{7} \times \dfrac{4}{7}\right) + \left(\dfrac{2}{7} \times \dfrac{6}{7}\right)$
$P(R_2) = \dfrac{20}{49} + \dfrac{12}{49} = \dfrac{32}{49}$

Answer: $\dfrac{32}{49}$