To find the probability of choosing one red marble and one blue marble when you select from the bag, we can break the problem down into two scenarios:
- Draw a red marble first and then a blue marble.
- Draw a blue marble first and then a red marble.
Let's calculate the probability for each scenario and then sum them up.
Total marbles in the bag:
- Red marbles: 3
- White marbles: 3
- Blue marbles: 5
- Total marbles = 3 + 3 + 5 = 11
Scenario 1: Red first, then blue
- Probability of drawing a red marble first: \[ P(\text{Red first}) = \frac{3}{11} \]
- After returning the red marble, draw a blue marble: \[ P(\text{Blue second}) = \frac{5}{11} \]
- Combined probability for this scenario: \[ P(\text{Red first, Blue second}) = P(\text{Red first}) \times P(\text{Blue second}) = \frac{3}{11} \times \frac{5}{11} = \frac{15}{121} \]
Scenario 2: Blue first, then red
- Probability of drawing a blue marble first: \[ P(\text{Blue first}) = \frac{5}{11} \]
- After returning the blue marble, draw a red marble: \[ P(\text{Red second}) = \frac{3}{11} \]
- Combined probability for this scenario: \[ P(\text{Blue first, Red second}) = P(\text{Blue first}) \times P(\text{Red second}) = \frac{5}{11} \times \frac{3}{11} = \frac{15}{121} \]
Total probability
Now we add the probabilities of both scenarios: \[ P(\text{Red and Blue}) = P(\text{Red first, Blue second}) + P(\text{Blue first, Red second}) = \frac{15}{121} + \frac{15}{121} = \frac{30}{121} \]
Therefore, the probability of selecting one red marble and one blue marble in two draws (with replacement) is: \[ \boxed{\frac{30}{121}} \]