Let's re-evaluate the calculations for Divisions 5 and 6, particularly focusing on the number of remaining numbers to consider after selecting some of the winning numbers and the supplementary numbers.

### Division 5: 3 Winning Numbers + 1 or 2 Supplementary Numbers

- **Condition:** You must match exactly 3 out of the 6 winning numbers. The remaining numbers must include 1 or 2 supplementary numbers.

**Case 1: 3 Winning Numbers + 1 Supplementary Number**
- You choose 3 out of 6 winning numbers: \(\binom{6}{3}\)
- You choose 1 out of 2 supplementary numbers: \(\binom{2}{1}\)
- The remaining 2 numbers must come from the 37 non-winning, non-supplementary numbers: \(\binom{37}{2}\)

**Case 2: 3 Winning Numbers + 2 Supplementary Numbers**
- You choose 3 out of 6 winning numbers: \(\binom{6}{3}\)
- You choose 2 out of 2 supplementary numbers: \(\binom{2}{2}\)
- The remaining 1 number must come from the 37 non-winning, non-supplementary numbers: \(\binom{37}{1}\)

- **Total combinations for Division 5:**
\[
\binom{6}{3} \times \left[\binom{37}{2} \times \binom{2}{1} + \binom{37}{1} \times \binom{2}{2}\right]
\]

- **Probability for Division 5:**
\[
P(\text{Division 5}) = \frac{\binom{6}{3} \times \left[\binom{37}{2} \times \binom{2}{1} + \binom{37}{1} \times \binom{2}{2}\right]}{\binom{45}{6}}
\]

### Division 6: 1 or 2 Winning Numbers + 2 Supplementary Numbers

- **Condition:** You must match exactly 1 or 2 out of the 6 winning numbers. The remaining numbers must include both supplementary numbers.

**Case 1: 1 Winning Number + 2 Supplementary Numbers**
- You choose 1 out of 6 winning numbers: \(\binom{6}{1}\)
- You choose 2 out of 2 supplementary numbers: \(\binom{2}{2}\)
- The remaining 3 numbers must come from the 37 non-winning, non-supplementary numbers: \(\binom{37}{3}\)

**Case 2: 2 Winning Numbers + 2 Supplementary Numbers**
- You choose 2 out of 6 winning numbers: \(\binom{6}{2}\)
- You choose 2 out of 2 supplementary numbers: \(\binom{2}{2}\)
- The remaining 2 numbers must come from the 37 non-winning, non-supplementary numbers: \(\binom{37}{2}\)

- **Total combinations for Division 6:**
\[
\binom{6}{1} \times \binom{37}{3} \times \binom{2}{2} + \binom{6}{2} \times \binom{37}{2} \times \binom{2}{2}
\]

- **Probability for Division 6:**
\[
P(\text{Division 6}) = \frac{\binom{6}{1} \times \binom{37}{3} \times \binom{2}{2} + \binom{6}{2} \times \binom{37}{2} \times \binom{2}{2}}{\binom{45}{6}}
\]

### Summary:
- For **Division 5**, you consider 37 numbers for the remaining positions after selecting the winning and supplementary numbers.
- For **Division 6**, the same logic applies; you select from the 37 remaining numbers for the non-winning and non-supplementary positions.

Would you like to see the exact numerical probabilities calculated for each division, or do you have any further questions?

Explore the detailed calculations for Division 5 and Division 6 probabilities in lottery games. Understand the combinations and probabilities involved in matching specific numbers and supplementary numbers. Suitable for advanced mathematics enthusiasts and lottery strategy analysts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation