A bag contains 8 red ad 5 white balls. Two successive drawn of 3 balls each are made such
that (i) the balls are replaced before the second trial and (ii)the balls are not replaced before the
second trial. Find the probability that the first draw will give 3 white and the second, 3 red balls

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

A bag contains 8 red and 5 white balls. Two successive draws of 3 balls each are made such that (i) the balls are replaced before the second trial and (ii) the balls are not replaced before the second trial. Find the probability that the first draw will give 3 white and the second, 3 red balls.

Calculate the probability of drawing 3 white balls in the first draw and 3 red balls in the second draw, with and without replacement, from a bag containing 8 red and 5 white balls. This problem covers key probability concepts and formulas suitable for Grades 11-12 students.

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