9. A bag contains 6 marbles. It includes 2 blue, 3 red, and 1 white marble. You draw one marble. What is the probability of drawing a blue or a white marble? 10. A bag contains 6 marbles. It includes 2 blue, 3 red, and 1 white marble. You draw a marble and then replace it. Now you draw another marble. a. What is the probability of drawing a red and then a blue marble? b. What is the probability of drawing a white and then a red marble? c. What is the probability of drawing both red? d. What is the probability of drawing both blue?

Let's work through each problem step by step. 9. The bag has 6 marbles in total (2 blue, 3 red, 1 white). To find the probability of drawing a blue or a white marble, count the favorable outcomes:  • Blue marbles: 2  • White marbles: 1 Total favorable outcomes = 2 + 1 = 3 Probability = Favorable outcomes / Total outcomes = 3/6 = 1/2 10. In this problem, you draw a marble, replace it, and then draw again. Since the marble is replaced, the two draws are independent, and the probabilities remain the same each time. The probabilities for each marble are:  • Red = 3/6 = 1/2  • Blue = 2/6 = 1/3  • White = 1/6 a. Probability of drawing a red marble on the first draw and a blue marble on the second draw:  P(Red then Blue) = (Probability of red) × (Probability of blue)   = (1/2) × (1/3)   = 1/6 b. Probability of drawing a white marble on the first draw and a red marble on the second draw:  P(White then Red) = (Probability of white) × (Probability of red)   = (1/6) × (1/2)   = 1/12 c. Probability of drawing a red marble on both draws:  P(Red and Red) = (Probability of red) × (Probability of red)   = (1/2) × (1/2)   = 1/4 d. Probability of drawing a blue marble on both draws:  P(Blue and Blue) = (Probability of blue) × (Probability of blue)   = (1/3) × (1/3)   = 1/9 Summary: 9. Probability of blue or white = 1/2 10a. P(Red then Blue) = 1/6 10b. P(White then Red) = 1/12 10c. P(Both red) = 1/4 10d. P(Both blue) = 1/9