A bag contains 3 red balls, 2 white balls, and 1 blue ball. Take out one ball, put it back into the bag, stir it well, and then take out another ball. Find the probability that the color of the ball taken the first time is the same as the color of the ball taken the second time.

Solution:
1. Total balls in the bag: 3 \, \text{(red)} + 2 \, \text{(white)} + 1 \, \text{(blue)} = 6

2. The probability of drawing each color on the first draw:
* Red: P(\text{Red}_1) = \frac{3}{6}
* White: P(\text{White}_1) = \frac{2}{6}
* Blue: P(\text{Blue}_1) = \frac{1}{6}

3. Since we put the ball back and draw again, the probability of drawing each color on the second draw remains the same:
* Red: P(\text{Red}_2) = \frac{3}{6}
* White: P(\text{White}_2) = \frac{2}{6}
* Blue: P(\text{Blue}_2) = \frac{1}{6}

4. Calculate the combined probability of drawing the same color twice (multiply probabilities for the first and second draws):
* Red-Red: P(\text{Red}_1 \cap \text{Red}_2) = \frac{3}{6} \times \frac{3}{6} = \frac{9}{36} = \frac{1}{4}
* White-White: P(\text{White}_1 \cap \text{White}_2) = \frac{2}{6} \times \frac{2}{6} = \frac{4}{36} = \frac{1}{9}
* Blue-Blue: P(\text{Blue}_1 \cap \text{Blue}_2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}

5. Add the probabilities of all favorable outcomes:
P(\text{Same Color}) = \frac{1}{4} + \frac{1}{9} + \frac{1}{36}

6. Find a common denominator (36) and sum:
P(\text{Same Color}) = \frac{9}{36} + \frac{4}{36} + \frac{1}{36} = \frac{9 + 4 + 1}{36} = \frac{14}{36} = \frac{7}{18}