The total number of pens in the box is $4 \text{ (green)} + 8 \text{ (blue)} + 3 \text{ (red)} = 15$.
The probability of picking a red pen in a single draw is $\frac{3}{15} = \frac{1}{5}$.
The probability of not picking a red pen in a single draw is $1 - \frac{1}{5} = \frac{4}{5}$.
Since the pen is replaced each time, the draws are independent. Therefore, the probability of not picking a red pen in 3 draws is $\left(\frac{4}{5}\right)^3 = \frac{64}{125}$.
The probability of picking at least one red pen is the complement of not picking any red pens in 3 draws. Therefore, the probability is $1 - \frac{64}{125} = \frac{125 - 64}{125} = \frac{61}{125}$.
Final Answer: \(\frac{61}{125}\)
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