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To solve the problem:

We are given:
- 13 red marbles.
- 18 blue marbles.
- Total marbles = 13 + 18 = 31.
- We are drawing 2 marbles at random and want to find the probability that both are red.

### Step-by-Step Solution:

1. **Total number of ways to choose 2 marbles from 31:**

   This can be calculated using combinations:
   \[
   \binom{31}{2} = \frac{31 \times 30}{2} = 465
   \]

2. **Number of ways to choose 2 red marbles from 13:**

   Similarly, the number of ways to choose 2 red marbles from the 13 red marbles:
   \[
   \binom{13}{2} = \frac{13 \times 12}{2} = 78
   \]

3. **Probability that both marbles are red:**

   The probability is the ratio of favorable outcomes (choosing 2 red marbles) to total outcomes (choosing 2 marbles):
   \[
   P(\text{both red}) = \frac{\binom{13}{2}}{\binom{31}{2}} = \frac{78}{465} \approx 0.1677
   \]

Thus, the probability that both marbles are red is approximately **0.1677**, or **16.77%**.

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Would you like more details or clarification on any part of the solution? Here are some related questions to explore further:

1. How would the probability change if we added 5 more red marbles?
2. What is the probability of drawing two blue marbles?
3. How do combinations differ from permutations, and why do we use combinations here?
4. What if we drew 3 marbles instead of 2? How would the calculation change?
5. How does probability change if we are drawing the marbles one by one without replacement?

**Tip:** When calculating probabilities with combinations, always check if replacement is involved, as that would lead to a different calculation approach.

Suppose a jar contains 13 red marbles and 18 blue marbles. If you reach in the jar and pull out 2 marbles at random at the same time, find the probability that both are red.

Learn how to calculate the probability of drawing two red marbles from a jar containing 13 red and 18 blue marbles. This problem involves using combinations and basic probability concepts, suitable for students in Grades 9-11.

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