Whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, two green ones, three white ones, and two purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms.
She has at most one green one.

To find the probability that Suzan grabs *at most one green marble* from a bag containing various marbles, we first need to understand the composition of the bag and the total number of ways she could select eight marbles.

### Step 1: Total Marbles in the Bag
The bag has the following marbles:
- 4 red marbles
- 2 green marbles
- 3 white marbles
- 2 purple marbles

The total number of marbles is:
\[
4 + 2 + 3 + 2 = 11
\]

### Step 2: Total Ways to Choose 8 Marbles
Suzan is grabbing 8 marbles out of 11. The total number of ways to choose 8 marbles from 11 is given by the binomial coefficient:
\[
\binom{11}{8} = \frac{11!}{8!(11 - 8)!} = \frac{11!}{8! \times 3!} = 165
\]

### Step 3: Favorable Outcomes (At Most One Green Marble)
To satisfy the condition of *at most one green marble*, Suzan can either:
1. Select **no green marbles**.
2. Select **one green marble**.

#### Case 1: No Green Marbles
If Suzan selects no green marbles, she will choose all 8 marbles from the remaining 9 non-green marbles (4 red, 3 white, and 2 purple). The number of ways to choose 8 marbles from these 9 non-green marbles is:
\[
\binom{9}{8} = 9
\]

#### Case 2: Exactly One Green Marble
If Suzan selects exactly one green marble, then she must choose the remaining 7 marbles from the 9 non-green marbles. The number of ways to do this is:
\[
\binom{2}{1} \times \binom{9}{7} = 2 \times 36 = 72
\]

### Step 4: Total Favorable Outcomes
The total number of favorable outcomes is the sum of the outcomes from both cases:
\[
9 + 72 = 81
\]

### Step 5: Probability Calculation
The probability that Suzan grabs at most one green marble is the ratio of the favorable outcomes to the total possible outcomes:
\[
\frac{81}{165} = \frac{27}{55}
\]

### Final Answer
The probability that Suzan grabs at most one green marble is:
\[
\frac{27}{55}
\]

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

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### Related Questions:
1. What if Suzan grabbed only 5 marbles instead of 8?
2. How would the probability change if there were three green marbles?
3. What is the probability that she grabs exactly two green marbles?
4. How would the calculation differ if we knew she picked 6 marbles with replacement?
5. Can we determine the probability of getting an equal number of red and white marbles?

### Tip:
When calculating probabilities involving "at most" or "at least" conditions, breaking down cases can simplify complex counting

Whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, two green ones, three white ones, and two purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has at most one green one.

Explore the probability of Suzan grabbing at most one green marble from a bag containing red, green, white, and purple marbles. This problem provides a detailed combinatorial approach, using probability and binomial coefficients. Suitable for students in Grades 10-12.

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