11. A bag contains 6 marbles. It includes 2 blue, 3 red, and 1 white marble. You draw a marble and you do not replace it. Now you draw another marble. What is the probability of: Drawing a red followed by a blue marble? b. Drawing a white followed by a red? c. Drawing both red? d. Drawing both blue?

To find the probabilities for the events described: a. **Drawing a red followed by a blue marble:** The probability of drawing a red marble first is 3/6 (since there are 3 red marbles out of 6 total). After drawing a red marble, there are now 5 marbles left (2 blue and 2 red). The probability of then drawing a blue marble is 2/5. Multiplying these probabilities gives: \( P(Red, then Blue) = \frac{3}{6} \times \frac{2}{5} = \frac{3 \cdot 2}{6 \cdot 5} = \frac{6}{30} = \frac{1}{5} \). b. **Drawing a white followed by a red:** The probability of drawing a white marble first is 1/6 (since there’s only 1 white marble). After that, there are still 3 red marbles remaining out of 5 total. Therefore, the probability of drawing a red marble after the white is 3/5. So, \( P(White, then Red) = \frac{1}{6} \times \frac{3}{5} = \frac{3}{30} = \frac{1}{10} \). c. **Drawing both red marbles:** The probability of drawing the first red marble is 3/6. If you successfully draw a red marble first, there are now 2 red marbles left out of a total of 5. Thus, the probability of drawing a second red marble is 2/5. So, \( P(Red, then Red) = \frac{3}{6} \times \frac{2}{5} = \frac{6}{30} = \frac{1}{5} \). d. **Drawing both blue marbles:** The probability of drawing a blue marble first is 2/6. If the first blue marble is drawn, only 1 blue marble remains out of a total of 5 marbles. So, the probability of drawing a second blue marble is now 1/5. Thus, \( P(Blue, then Blue) = \frac{2}{6} \times \frac{1}{5} = \frac{2}{30} = \frac{1}{15} \). So, in summary, the probabilities are: a. \( \frac{1}{5} \) b. \( \frac{1}{10} \) c. \( \frac{1}{5} \) d. \( \frac{1}{15} \)