Use the counting techniques. A bag contains three red marbles, four green ones, one fluorescent pink one, four yellow ones, and four orange ones. Suzan grabs four at random. Find the probability of the indicated event.
She gets one of each color other than fluorescent pink, given that she gets the fluorescent pink one.

To solve this problem, we can apply counting techniques, specifically combinations, to determine the probability that Suzan grabs one marble of each color (red, green, yellow, and orange), given that she grabs the fluorescent pink marble.

### Step-by-step Solution:

1. **Total number of marbles:**
   - Red: 3
   - Green: 4
   - Fluorescent pink: 1
   - Yellow: 4
   - Orange: 4

   So, the total number of marbles in the bag is:
   \[
   3 + 4 + 1 + 4 + 4 = 16 \text{ marbles}.
   \]

2. **Given condition:**
   Suzan grabs the fluorescent pink marble. Therefore, she needs to choose 3 additional marbles from the remaining 15 marbles (since the fluorescent pink is already chosen).

3. **Desired outcome:**
   Suzan gets one red, one green, one yellow, and one orange marble. Since she already has the fluorescent pink marble, the remaining three marbles must include:
   - One red marble,
   - One green marble,
   - One yellow or orange marble.

4. **Computation of favorable outcomes:**
   To achieve this, we need to choose 1 red, 1 green, and 1 yellow or orange marble from the respective groups:
   
   - There are 3 red marbles, so the number of ways to choose 1 red marble is \( \binom{3}{1} = 3 \).
   - There are 4 green marbles, so the number of ways to choose 1 green marble is \( \binom{4}{1} = 4 \).
   - There are 4 yellow marbles and 4 orange marbles, so we need to choose one from this set. The number of ways to choose one yellow or one orange is \( \binom{4}{1} + \binom{4}{1} = 4 + 4 = 8 \).

   Therefore, the total number of favorable outcomes is:
   \[
   3 \times 4 \times 8 = 96.
   \]

5. **Total number of possible outcomes:**
   Suzan must pick 3 marbles from the remaining 15 marbles (since the fluorescent pink marble is already chosen). The number of ways to choose 3 marbles from 15 is given by the combination:
   \[
   \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455.
   \]

6. **Probability:**
   The probability that Suzan grabs one red, one green, and one yellow or orange marble (given that she already has the fluorescent pink marble) is the ratio of the favorable outcomes to the total possible outcomes:
   \[
   P(\text{one of each color other than fluorescent pink}) = \frac{96}{455}.
   \]

Thus, the probability is:
\[
P = \frac{96}{455}.
\]

Would you like further details or have any questions?

### Related questions:
1. How would the probability change if there were 5 orange marbles instead of 4?
2. What is the probability that Suzan grabs two green marbles, given that she gets the fluorescent pink one?
3. How many different sets of four marbles can Suzan grab from the bag without any other conditions?
4. If Suzan grabs five marbles instead of four, what is the probability that she gets the fluorescent pink marble and two green marbles?
5. What is the probability that Suzan grabs at least one red marble, given that she grabs the fluorescent pink one?

### Tip:
Always count the total possible outcomes first before identifying the specific favorable cases to make sure you're on track.

A bag contains three red marbles, four green ones, one fluorescent pink one, four yellow ones, and four orange ones. Suzan grabs four at random. Find the probability of the indicated event: She gets one of each color other than fluorescent pink, given that she gets the fluorescent pink one.

Explore how to solve a conditional probability problem involving combinations of colored marbles. Given that Suzan grabs a fluorescent pink marble, find the probability that she also grabs one red, one green, and one yellow or orange marble. This problem utilizes counting techniques, conditional probability, and the combination formula. Suitable for high school students in Grades 10-12.

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