To solve this problem, we need to determine how many sets of four marbles from the bag include all three red marbles.

Total marbles in the bag:

• Red marbles: 3
• Green marbles: 5
• Lavender marbles: 1
• Yellow marbles: 3
• Orange marbles: 3

This gives a total of: $3 \,(\text{red}) + 5 \,(\text{green}) + 1 \,(\text{lavender}) + 3 \,(\text{yellow}) + 3 \,(\text{orange}) = 15 \, \text{marbles}.$

Step-by-step solution:

1. We need a set of 4 marbles that includes all 3 red marbles. That means 3 marbles are already chosen (the red ones).
2. We are left with $15 - 3 = 12$ marbles (non-red ones) to choose the fourth marble.

Conclusion:

We need to choose 1 marble from the remaining 12 non-red marbles. This can be done in: $\binom{12}{1} = 12 \, \text{ways}.$

Thus, there are 12 sets of 4 marbles that include all the red marbles.

Would you like more details on any step, or have any questions?

Here are 5 related questions:

1. How many total sets of 4 marbles can be chosen from the bag?
2. What if we want sets with at least 2 red marbles?
3. How many sets include exactly 1 red marble?
4. How many sets include no red marbles at all?
5. If we add another color to the bag, how does that change the combinations?

Tip: The binomial coefficient $\binom{n}{k}$ gives the number of ways to choose $k$ objects from $n$ without regard to the order.