1. Initial composition of the urn

There are initially:

• $5$ red balls
• $2$ green balls

So total balls $= 7$.

We want the probability that the second ball drawn is red.

2. Case analysis based on the first draw

Since the process after the first draw depends on its color, we split into cases.

Case 1: First ball drawn is green

Probability of drawing green first: $P(G_1)=\frac{2}{7}$

If a green ball is drawn:

• that green ball is not returned
• one red ball is added

So new composition becomes:

• Red: $5+1=6$
• Green: $2-1=1$

Total remains $7$.

Hence, $P(R_2\mid G_1)=\frac{6}{7}$

So contribution from this case is: $P(G_1)P(R_2\mid G_1)=\frac{2}{7}\cdot\frac{6}{7}=\frac{12}{49}$

Case 2: First ball drawn is red

Probability of drawing red first: $P(R_1)=\frac{5}{7}$

If a red ball is drawn:

• that red ball is not returned
• one green ball is added

So new composition becomes:

• Red: $5-1=4$
• Green: $2+1=3$

Total again remains $7$.

Hence, $P(R_2\mid R_1)=\frac{4}{7}$

So contribution from this case is: $P(R_1)P(R_2\mid R_1)=\frac{5}{7}\cdot\frac{4}{7}=\frac{20}{49}$

3. Total probability

By the law of total probability, $P(R_2)=P(G_1)P(R_2\mid G_1)+P(R_1)P(R_2\mid R_1)$

Substitute values: $P(R_2)=\frac{12}{49}+\frac{20}{49}=\frac{32}{49}$

4. Match with options

$\frac{32}{49}$ corresponds to Option D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.