A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and the second draw 3 red balls.

Answer

To solve the problem, we need to find the probability that the first draw yields 3 white balls and the second draw yields 3 red balls from a bag containing 8 red and 5 white balls.

### Step-by-Step Solution:

1. **Identify the Total Number of Balls:**
The bag contains a total of \(8 + 5 = 13\) balls (8 red and 5 white).

2. **Calculate the Probability of Drawing 3 White Balls in the First Draw:**
We need to find the number of ways to choose 3 white balls from the 5 available. This can be calculated using combinations:
\[ P(A) = \frac{\text{Number of ways to choose 3 white balls}}{\text{Total ways to choose 3 balls from 13}} \]
This can be expressed as:
\[ P(A) = \frac{5C3}{13C3} \]

3. **Calculate \(5C3\) and \(13C3\):**
Using the combination formula \(nCr = \frac{n!}{r!(n-r)!}\):
\[ 5C3 = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10 \]
\[ 13C3 = \frac{13!}{3! \cdot 10!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286 \]

4. **Calculate \(P(A)\):**
\[ P(A) = \frac{10}{286} = \frac{5}{143} \]

5. **Adjust the Total Balls After the First Draw:**
After drawing 3 white balls, the remaining balls are:
- Red balls: 8
- White balls: 5 - 3 = 2
- Total remaining balls: \(8 + 2 = 10\)

6. **Calculate the Probability of Drawing 3 Red Balls in the Second Draw:**
Now we need to find the number of ways to choose 3 red balls from the 8 available:
\[ P(B|A) = \frac{8C3}{10C3} \]

7. **Calculate \(8C3\) and \(10C3\):**
\[ 8C3 = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
\[ 10C3 = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

8. **Calculate \(P(B|A)\):**
\[ P(B|A) = \frac{56}{120} = \frac{7}{15} \]

9. **Calculate the Combined Probability \(P(A \cap B)\):**
The probability of both events occurring is given by:
\[ P(A \cap B) = P(A) \times P(B|A) = \frac{5}{143} \times \frac{7}{15} \]

10. **Final Calculation:**
\[ P(A \cap B) = \frac{5 \times 7}{143 \times 15} = \frac{35}{2145} \]
Simplifying \(2145\) gives \(429\) (since \(2145 = 5 \times 429\)):
\[ P(A \cap B) = \frac{7}{429} \]

### Final Answer:
The probability that the first draw yields 3 white balls and the second draw yields 3 red balls is \(\frac{7}{429}\).