8. $\frac{4338}{91125}$

Analysis: According to the problem, the fourth draw is a red ball and exactly one of the first three draws is a red ball. First, calculate the probability of exactly one red ball in the first three draws. If the first draw is a red ball and the second and third draws are white balls, the probability is $P_{1}=\frac{2}{10} \times \frac{8}{9} \times \frac{8}{9}=\frac{64}{405}$; If the second draw is a red ball and the first and third draws are white balls, the probability is $P_{1}=\frac{8}{10} \times \frac{2}{10} \times \frac{8}{9}=\frac{32}{225}$; If the third draw is a red ball and the first and second draws are white balls, the probability is $P_{1}=\frac{8}{10} \times \frac{8}{10} \times \frac{2}{10}=\frac{16}{125}$; Therefore, the probability of exactly one red ball in the first three draws is $P_{1}+P_{2}+P_{3}=\frac{64}{405}+\frac{32}{225}+\frac{16}{125}=\frac{4338}{10125}$, and the probability of the fourth draw being a red ball is $\frac{1}{9}$, so the total probability is $\frac{4338}{10125} \times \frac{1}{9}=\frac{4338}{91125}$.
8. In a bag, there are 10 balls that are identical except for their color, with 8 white balls and 2 red balls. Each time, a ball is randomly drawn from the bag. If the ball is white, it is returned to the bag. If the ball is red, it is not returned to the bag. What is the probability that all the red balls are drawn exactly by the 4th draw? $\qquad$