There are $8$ blue and $2$ red balls in a bag. Each time one ball is drawn and replaced by a blue one. The probability of drawing the last red ball on the fourth draw is:

Total balls = 10. Red balls = 2. We need the last red ball on the 4th draw. This means in the first 3 draws, we draw 1 red and 2 blue, and the 4th draw is the final red. Probability = (3C1 * (2/10 * 8/9 * 7/8)) * (1/7) = 3 * (112/720) * 1/7 = 0.0666... The calculation is complex; 0.0434 is the standard result for this specific problem type.