A bag contains 24 green marbles, 22 blue marbles, 14 yellow marbles, and 12 red marbles. You select two marbles at random...find the probability of one marble being yellow and one marble being green with replacement of the first marble. Round your answer to the nearest tenth if necessary.

To solve this problem, let's break it down step-by-step:

1. **Understand the Problem Context**:  
   You have a bag containing four types of marbles:  
   - Green: 24 marbles  
   - Blue: 22 marbles  
   - Yellow: 14 marbles  
   - Red: 12 marbles  

   This gives a total of:

   \[
   \text{Total marbles} = 24 + 22 + 14 + 12 = 72
   \]

2. **Determine the Probability of Selecting a Yellow Marble**:  
   The probability of selecting a yellow marble on the first draw is:

   \[
   P(\text{Yellow}) = \frac{\text{Number of Yellow Marbles}}{\text{Total Marbles}} = \frac{14}{72}
   \]

   Simplify the fraction:

   \[
   P(\text{Yellow}) = \frac{7}{36}
   \]

3. **Determine the Probability of Selecting a Green Marble with Replacement**:  
   Since the first marble is replaced, the total number of marbles in the bag remains the same. The probability of selecting a green marble on the second draw is:

   \[
   P(\text{Green}) = \frac{\text{Number of Green Marbles}}{\text{Total Marbles}} = \frac{24}{72}
   \]

   Simplify the fraction:

   \[
   P(\text{Green}) = \frac{1}{3}
   \]

4. **Calculate the Combined Probability**:  
   The problem requires the probability of one marble being yellow and the other marble being green. Since the selection is with replacement, we consider both scenarios:
   
   - Yellow marble first, then green marble.  
   - Green marble first, then yellow marble.

   The probability of drawing a yellow marble first and then a green marble is:

   \[
   P(\text{Yellow then Green}) = P(\text{Yellow}) \times P(\text{Green}) = \frac{7}{36} \times \frac{1}{3} = \frac{7}{108}
   \]

   Similarly, the probability of drawing a green marble first and then a yellow marble is:

   \[
   P(\text{Green then Yellow}) = P(\text{Green}) \times P(\text{Yellow}) = \frac{1}{3} \times \frac{7}{36} = \frac{7}{108}
   \]

5. **Add the Two Probabilities Together**:  
   Since these two events are mutually exclusive (either one happens or the other), the total probability is the sum of these probabilities:

   \[
   P(\text{One Yellow and One Green}) = \frac{7}{108} + \frac{7}{108} = \frac{14}{108}
   \]

   Simplify the fraction:

   \[
   P(\text{One Yellow and One Green}) = \frac{7}{54}
   \]

6. **Convert to Decimal Form and Round to the Nearest Tenth**:  

   \[
   P(\text{One Yellow and One Green}) \approx 0.13
   \]

Thus, the probability of selecting one yellow marble and one green marble with replacement is **0.1 (or 10%)** when rounded to the nearest tenth.

### Would you like to see more details or have any questions? 

Here are some related questions to further explore this topic:
1. How would the probability change if the selection was without replacement?
2. What is the probability of selecting two marbles of the same color with replacement?
3. How do you calculate the probability of selecting one blue and one red marble?
4. What is the probability of drawing two yellow marbles in a row with replacement?
5. How can we calculate the expected number of green marbles in 10 draws with replacement?

**Tip:** Remember that when dealing with "with replacement" problems, the total number of items remains constant after each draw.

Find the probability of selecting one yellow marble and one green marble from a bag containing 24 green, 22 blue, 14 yellow, and 12 red marbles, with replacement. Detailed step-by-step solution for calculating probabilities with replacement, suitable for students in Grades 6-8.

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