Let's solve the problem step by step:

We are given:

• 4 green marbles
• 6 red marbles
• 5 blue marbles

Step 1: Total marbles in the bag

The total number of marbles is: $4 + 6 + 5 = 15$

Step 2: First event (drawing a green marble)

The probability of drawing a green marble first is the ratio of green marbles to the total marbles: $P(\text{Green first}) = \frac{4}{15}$

Step 3: Second event (drawing a blue marble)

Since the marble is replaced after checking its color, the total number of marbles remains the same (15 marbles). Therefore, the probability of drawing a blue marble next is: $P(\text{Blue second}) = \frac{5}{15} = \frac{1}{3}$

Step 4: Combine the probabilities

To find the total probability of the first marble being green and the second being blue (with replacement), we multiply the probabilities of these independent events: $P(\text{Green first and Blue second}) = P(\text{Green first}) \times P(\text{Blue second}) = \frac{4}{15} \times \frac{1}{3} = \frac{4}{45}$

Thus, the probability that the first marble is green and the second is blue is: $\boxed{\frac{4}{45}} \quad \text{(Option A)}$

Would you like a more detailed explanation or have any questions?

Here are 5 related questions to expand your understanding:

1. What is the probability of drawing two green marbles in a row with replacement?
2. How would the probability change if the marbles were not replaced?
3. What is the probability of drawing a red marble followed by a blue marble?
4. If a third marble is drawn (with replacement), what is the probability that it will be red?
5. How does the concept of independent events apply to this problem?

Tip: For problems with replacement, remember that the total number of objects (like marbles) remains constant after each draw.