We are asked to find the probability of drawing 3 white balls in the first draw and 3 red balls in the second draw under two different scenarios: (i) when the balls are replaced after the first draw, and (ii) when the balls are not replaced.

Let's define the problem step by step for each case:

(i) With Replacement:

In this case, the balls are replaced after each draw, so the total number of balls remains the same for both draws.

• The total number of balls in the bag is $8 + 5 = 13$ balls.

First Draw (3 White Balls):

• The probability of drawing 1 white ball in the first draw is $\frac{5}{13}$.
• After drawing the first white ball, the balls are replaced, so the probability of drawing another white ball remains $\frac{5}{13}$, and similarly for the third ball.

Thus, the probability of drawing 3 white balls in the first draw is:

$P(\text{3 White in 1st Draw}) = \frac{5}{13} \times \frac{5}{13} \times \frac{5}{13} = \left( \frac{5}{13} \right)^3 = \frac{125}{2197}.$

Second Draw (3 Red Balls):

• After the first draw, the balls are replaced, so we again have 8 red and 5 white balls, a total of 13 balls.
• The probability of drawing 1 red ball is $\frac{8}{13}$, and since the balls are replaced, this probability remains the same for the next two draws.

Thus, the probability of drawing 3 red balls in the second draw is:

$P(\text{3 Red in 2nd Draw}) = \frac{8}{13} \times \frac{8}{13} \times \frac{8}{13} = \left( \frac{8}{13} \right)^3 = \frac{512}{2197}.$

Total Probability (with replacement):

The total probability of drawing 3 white balls in the first draw and 3 red balls in the second draw (with replacement) is:

$P(\text{Total with Replacement}) = P(\text{3 White in 1st Draw}) \times P(\text{3 Red in 2nd Draw}) = \frac{125}{2197} \times \frac{512}{2197} = \frac{64000}{4826809}.$

(ii) Without Replacement:

In this case, the balls are not replaced after the first draw, so the total number of balls changes after the first draw.

**First Draw (3 White Balls