To find the probability of selecting a red marble first and then a blue marble second (denoted as P(red and blue)), we can approach the problem as follows:
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Calculate the total number of marbles.
- Red marbles: 3
- White marbles: 5
- Blue marbles: 4
- Total marbles = 3 (red) + 5 (white) + 4 (blue) = 12 marbles.
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Calculate the probability of selecting a red marble first. \[ P(\text{Red}) = \frac{\text{Number of Red Marbles}}{\text{Total Number of Marbles}} = \frac{3}{12} = \frac{1}{4} \]
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Since the first marble is returned to the bag, the probabilities remain the same for the second draw. Calculate the probability of selecting a blue marble second after replacing the first marble. \[ P(\text{Blue}) = \frac{\text{Number of Blue Marbles}}{\text{Total Number of Marbles}} = \frac{4}{12} = \frac{1}{3} \]
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Find the combined probability of both events occurring (selecting a red marble first and then a blue marble). Since these two events are independent (because of the replacement), we multiply the probabilities: \[ P(\text{Red and Blue}) = P(\text{Red}) \times P(\text{Blue}) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \]
Thus, the probability of drawing a red marble followed by a blue marble is \( \frac{1}{12} \).