Sure, let's solve this step-by-step.
Given:
- The bag contains 4 marbles: 1 red, 1 blue, and 2 green.
- Heather draws a marble, replaces it, and then draws another marble.
First, let's calculate the probability of drawing each marble during one draw:
1. Total number of marbles (N):
[tex]\[ N = 4 \][/tex]
2. Probability of drawing each marble:
- Red marble (P_red):
[tex]\[ \text{P(red)} = \frac{1 \text{ red marble}}{4 \text{ total marbles}} = \frac{1}{4} = 0.25 \][/tex]
- Blue marble (P_blue):
[tex]\[ \text{P(blue)} = \frac{1 \text{ blue marble}}{4 \text{ total marbles}} = \frac{1}{4} = 0.25 \][/tex]
- Green marble (P_green):
[tex]\[ \text{P(green)} = \frac{2 \text{ green marbles}}{4 \text{ total marbles}} = \frac{2}{4} = \frac{1}{2} = 0.5 \][/tex]
Next, we need to find the probability of drawing a red marble first, then a blue marble, and then a green marble in three successive draws with replacement.
Since each draw is independent and the marbles are replaced, we can multiply the probabilities of each event happening sequentially:
[tex]\[ P(\text{red then blue then green}) = P(\text{red}) \times P(\text{blue}) \times P(\text{green}) \][/tex]
3. Calculating the combined probability:
[tex]\[ P(\text{red then blue then green}) = (\frac{1}{4}) \times (\frac{1}{4}) \times (\frac{1}{2}) \][/tex]
[tex]\[ P(\text{red then blue then green}) = 0.25 \times 0.25 \times 0.5 \][/tex]
[tex]\[ P(\text{red then blue then green}) = 0.03125 \][/tex]
Finally, express the combined probability as a fraction:
[tex]\[ 0.03125 = \frac{1}{32} \][/tex]
So, the probability [tex]\( P \)[/tex] of drawing a red marble first, then a blue marble, and then a green marble is:
[tex]\[ \frac{1}{32} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\frac{1}{32}} \][/tex]
