The correct answer is $\boxed{\frac{7}{16}}$.
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is drawing a red ball and leaving it out, and event B is drawing a green ball.
The probability of drawing a red ball is $\frac{8}{16}$, because there are 8 red balls and 16 total balls.
The probability of drawing a green ball after drawing a red ball is $\frac{8}{15}$, because there are 8 green balls left and 15 total balls left.
Therefore, the probability of drawing a red ball, then a green ball is $\frac{8}{16} \cdot \frac{8}{15} = \boxed{\frac{7}{16}}$.
Option A is incorrect because it is the probability of drawing a green ball, then a red ball. This is not the same as the probability of drawing a red ball, then a green ball.
Option B is incorrect because it is the probability of drawing a red ball on the first draw. This is not the same as the probability of drawing a red ball, then a green ball.
Option C is incorrect because it is the probability of drawing a green ball on the first draw, then a red ball on the second draw. This is not the same as the probability of drawing a red ball, then a green ball.
Option D is incorrect because it is the probability of drawing a red ball or a green ball on the first draw, then a red ball or a green ball on the second draw. This is not the same as the probability of drawing a red ball, then a green ball.