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Total number of marbles = 4 (red) + 5 (blue) + 1 (green) = 10. The probability of drawing a red marble in one trial is $P(R) = \frac{4}{10} = \frac{2}{5}$.
The probability of not drawing a red marble in one trial is $P(R') = 1 - P(R) = 1 - \frac{2}{5} = \frac{3}{5}$.
Since the marble is replaced, the trials are independent. The probability of not drawing a red marble in 3 trials is: $$P(\text{No Red}) = \left(\frac{3}{5}\right)^3 = \frac{27}{125}$$
Using the complement rule, the probability of at least one red marble is: $$P(\text{At least one Red}) = 1 - P(\text{No Red}) = 1 - \frac{27}{125} = \frac{125 - 27}{125} = \frac{98}{125}$$
Final Answer: 98/125
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