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The problem asks for the probability that the first two candies drawn from a snack-size bag of M&Ms are green, and the third is red. The total number of candies in the bag and the distribution of colors are provided.

### Step 1: Calculate the total number of candies
There are:
- 12 red
- 12 blue
- 7 green
- 13 brown
- 3 orange
- 10 yellow

The total number of candies is:
\[
12 + 12 + 7 + 13 + 3 + 10 = 57
\]

### Step 2: Determine the desired sequence
The sequence we want is:
1. First draw: Green
2. Second draw: Green
3. Third draw: Red

### Step 3: Calculate the probabilities
Since the candies are drawn without replacement, the probabilities change with each draw.

1. **First draw (green)**:
   There are 7 green candies out of 57 total candies.
   \[
   P(\text{first green}) = \frac{7}{57}
   \]

2. **Second draw (green)**:
   After drawing one green candy, there are 6 green candies left, and now there are 56 candies remaining in total.
   \[
   P(\text{second green}) = \frac{6}{56}
   \]

3. **Third draw (red)**:
   After drawing two green candies, the number of red candies is still 12, but now there are 55 candies left in total.
   \[
   P(\text{third red}) = \frac{12}{55}
   \]

### Step 4: Multiply the probabilities
Since the events are dependent (no replacement), the overall probability is the product of the individual probabilities:
\[
P(\text{green, green, red}) = \frac{7}{57} \times \frac{6}{56} \times \frac{12}{55}
\]

Now, calculate the result:
\[
P = \frac{7 \times 6 \times 12}{57 \times 56 \times 55} = \frac{504}{175560} \approx 0.00287
\]

### Final Answer:
The probability that the first two candies drawn are green and the third is red is approximately **0.00287**, or about **0.287%**.

Would you like more details or have any further questions?

Here are some related questions for further exploration:
1. What is the probability of drawing two red candies followed by a green one?
2. How would the probability change if the drawing was done with replacement?
3. What is the probability that the first three candies drawn are all red?
4. How does the probability change if we draw five candies instead of three?
5. What is the expected number of green candies in a random draw of three?

**Tip:** Always reduce fractions before multiplying to simplify probability calculations!

A snack-size bag of M&Ms candies is opened. Inside there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement. What is the probability that the first two candies drawn are green and the third is red?

Calculate the probability of drawing two green M&Ms followed by one red M&M from a snack-size bag. This problem explores probability without replacement and involves concepts from combinatorics. Suitable for students in Grades 9-12.

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