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.