Probability tree diagram illustrating the chances of drawing two sweets of specific colors from a bag, showing outcomes and probabilities for green-green, green-blue, blue-green, and blue-blue combinations.

Problem Analysis:

The problem involves calculating the probabilities of different outcomes when two sweets are taken from a bag, given that the composition of the sweets changes after the first sweet is taken. The scenario is described using a probability tree diagram.

#### Key Information:
1. Initial Composition:
- After 1 green sweet is taken, there are 20 sweets left: 11 green and 9 blue.
- After 1 blue sweet is taken, there are 20 sweets left: 12 green and 8 blue.

2. Probability Tree:
- The tree diagram shows the probabilities of drawing a green (G) or blue (B) sweet in the first and second draws.
- The probabilities are calculated based on the changing composition of the sweets after each draw.

3. Outcomes:
- (G, G): Both sweets are green.
- (G, B): First sweet is green, second sweet is blue.
- (B, G): First sweet is blue, second sweet is green.
- (B, B): Both sweets are blue.

#### Objective:
We need to verify the probabilities calculated in the tree diagram and ensure they are correct.

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Step-by-Step Solution:

#### 1. First Draw:
- There are two possible outcomes for the first draw: drawing a green sweet (G) or a blue sweet (B).
- The probabilities for the first draw are:
- Probability of drawing a green sweet (G): \( \frac{12}{21} \)
- Probability of drawing a blue sweet (B): \( \frac{9}{21} \)

#### 2. Second Draw (Given the First Draw):
- The composition of the sweets changes after the first draw, so the probabilities for the second draw depend on the outcome of the first draw.

##### Case 1: First Sweet is Green (G)
- If the first sweet drawn is green, there are now 20 sweets left: 11 green and 9 blue.
- Probabilities for the second draw:
- Probability of drawing a green sweet (G): \( \frac{11}{20} \)
- Probability of drawing a blue sweet (B): \( \frac{9}{20} \)

- Outcomes:
- (G, G): Probability = \( \frac{12}{21} \times \frac{11}{20} = \frac{132}{420} = \frac{11}{35} \)
- (G, B): Probability = \( \frac{12}{21} \times \frac{9}{20} = \frac{108}{420} = \frac{9}{35} \)

##### Case 2: First Sweet is Blue (B)
- If the first sweet drawn is blue, there are now 20 sweets left: 12 green and 8 blue.
- Probabilities for the second draw:
- Probability of drawing a green sweet (G): \( \frac{12}{20} \)
- Probability of drawing a blue sweet (B): \( \frac{8}{20} \)

- Outcomes:
- (B, G): Probability = \( \frac{9}{21} \times \frac{12}{20} = \frac{108}{420} = \frac{9}{35} \)
- (B, B): Probability = \( \frac{9}{21} \times \frac{8}{20} = \frac{72}{420} = \frac{6}{35} \)

#### 3. Verification of Probabilities:
- Sum of all probabilities should equal 1:
\[
P(G, G) + P(G, B) + P(B, G) + P(B, B) = \frac{11}{35} + \frac{9}{35} + \frac{9}{35} + \frac{6}{35} = \frac{35}{35} = 1
\]
- The probabilities are consistent and correctly calculated.

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Final Answer:

The probabilities calculated in the tree diagram are correct. The final probabilities for the outcomes are:
\[
\boxed{\frac{11}{35}, \frac{9}{35}, \frac{9}{35}, \frac{6}{35}}
\]