The mathematical model in this problem is used in all lottery games. 

There are 5 red and 4 white, numbered, marbles in a bowl, well mixed. A sample of 3 marbles is taken.

How many different samples are possible? 
How many samples consist of 3 red and 0 white marbles? 
How many samples consist of 2 red and 1 white marbles? 
How many samples consist of 1 red and 2 white marbles? 
How many samples consist of 0 red and 3 white marbles? 
The sum of the four numbers from B to E = 
 . How does it compare to A?

To solve this problem, we will use combinations since the order of drawing the marbles does not matter.

### A. Total number of different samples of 3 marbles:
The total number of marbles is \(5\) red + \(4\) white = \(9\) marbles. We need to choose \(3\) out of these \(9\).

\[
\text{Total samples} = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
\]

### B. Samples consisting of 3 red and 0 white marbles:
We need to choose \(3\) red marbles from \(5\):

\[
\text{Samples (3 red, 0 white)} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

### C. Samples consisting of 2 red and 1 white marbles:
We need to choose \(2\) red marbles from \(5\) and \(1\) white marble from \(4\):

\[
\text{Samples (2 red, 1 white)} = \binom{5}{2} \times \binom{4}{1} = \frac{5 \times 4}{2 \times 1} \times 4 = 10 \times 4 = 40
\]

### D. Samples consisting of 1 red and 2 white marbles:
We need to choose \(1\) red marble from \(5\) and \(2\) white marbles from \(4\):

\[
\text{Samples (1 red, 2 white)} = \binom{5}{1} \times \binom{4}{2} = 5 \times \frac{4 \times 3}{2 \times 1} = 5 \times 6 = 30
\]

### E. Samples consisting of 0 red and 3 white marbles:
We need to choose \(3\) white marbles from \(4\):

\[
\text{Samples (0 red, 3 white)} = \binom{4}{3} = \frac{4!}{3!(4-3)!} = 4
\]

### Summary of Results:
- **A**: Total samples = \(84\)
- **B**: Samples (3 red, 0 white) = \(10\)
- **C**: Samples (2 red, 1 white) = \(40\)
- **D**: Samples (1 red, 2 white) = \(30\)
- **E**: Samples (0 red, 3 white) = \(4\)

### Sum of B to E:
\[
\text{Sum} = 10 + 40 + 30 + 4 = 84
\]

### Comparison of the Sum to A:
The sum of the numbers from B to E (\(84\)) is equal to the total number of different samples (A = \(84\)).

### Questions for Further Exploration:
1. How would the total number of samples change if there were more white marbles?
2. What is the probability of drawing 2 red and 1 white marble?
3. If you were to draw 4 marbles instead, how would the calculations differ?
4. How can this model be applied to other games of chance?
5. What happens if the number of red and white marbles is equal?

**Tip:** Understanding combinations is crucial in probability problems, especially in scenarios involving draws or selections.

The mathematical model in this problem is used in all lottery games. There are 5 red and 4 white, numbered, marbles in a bowl, well mixed. A sample of 3 marbles is taken. How many different samples are possible? How many samples consist of 3 red and 0 white marbles? How many samples consist of 2 red and 1 white marbles? How many samples consist of 1 red and 2 white marbles? How many samples consist of 0 red and 3 white marbles? The sum of the four numbers from B to E = . How does it compare to A?

Explore the combinatorial analysis of drawing marbles from a bowl containing 5 red and 4 white marbles. This detailed solution covers the total number of samples, and the breakdown of samples based on different combinations of red and white marbles, highlighting the concept of combinations. Suitable for Grades 7-10.

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