A box contains three marbles: one red, one green, and one blue. Consider an experiment that consists of taking one marble from the box then replacing it in the box and drawing a second marble from the box. What is the sample space? If, at all times, each marble in the box is equally likely to be selected, what is the probability of each point in the sample space?

: To determine the sample space, we will list all possible combinations of drawing two marbles from the box, taking into account the replacement after the first draw: 1. Red, Red (RR) 2. Red, Green (RG) 3. Red, Blue (RB) 4. Green, Red (GR) 5. Green, Green (GG) 6. Green, Blue (GB) 7. Blue, Red (BR) 8. Blue, Green (BG) 9. Blue, Blue (BB) These 9 combinations represent the sample space for this experiment: Sample Space = {RR, RG, RB, GR, GG, GB, BR, BG, BB}

: Since there are 3 marbles in the box and at all times each marble has an equal chance to be selected, the probability of selecting any particular marble in each draw is \(\frac{1}{3}\). As the marbles are replaced after each draw, the draws are independent events. Remember that the probability of independent events occurring together is the product of their individual probabilities. Now, we will calculate the probability of each point in the sample space: 1. P(RR) = P(Red on 1st Draw) * P(Red on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 2. P(RG) = P(Red on 1st Draw) * P(Green on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 3. P(RB) = P(Red on 1st Draw) * P(Blue on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 4. P(GR) = P(Green on 1st Draw) * P(Red on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 5. P(GG) = P(Green on 1st Draw) * P(Green on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 6. P(GB) = P(Green on 1st Draw) * P(Blue on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 7. P(BR) = P(Blue on 1st Draw) * P(Red on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 8. P(BG) = P(Blue on 1st Draw) * P(Green on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 9. P(BB) = P(Blue on 1st Draw) * P(Blue on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) The probabilities of each point in the sample space are: P(RR) = P(RG) = P(RB) = P(GR) = P(GG) = P(GB) = P(BR) = P(BG) = P(BB) = \(\frac{1}{9}\)