Probability tree diagrams illustrating various scenarios: rolling a die, picking shapes, and calculating sports win/loss probabilities.

Let's solve each problem step by step.

---

Problem 1: Probability of rolling a six, then not rolling a six

#### Given:
- A fair 6-sided die is rolled twice.
- The probability tree diagram is provided.

#### Task:
Find the probability of rolling a six on the first roll and not rolling a six on the second roll.

#### Solution:
1. From the probability tree diagram:
- The probability of rolling a six on the first roll is \( \frac{1}{6} \).
- If a six is rolled on the first roll, the probability of not rolling a six on the second roll is \( \frac{5}{6} \).

2. The combined probability of these two events happening in sequence is the product of their individual probabilities:
\[
P(\text{Six on first roll and not six on second roll}) = P(\text{Six on first roll}) \times P(\text{Not six on second roll})
\]
\[
P = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}
\]

#### Answer:
\[
\boxed{\frac{5}{36}}
\]

---

Problem 2: Tree diagram for picking shapes from a bag

#### Given:
- A bag contains shapes: 4 circles and 3 squares.
- A shape is picked randomly, replaced, and another shape is picked.
- The task is to complete the tree diagram showing all possible outcomes.

#### Task:
Complete the tree diagram and show the probabilities of all 4 possible outcomes.

#### Solution:
1. Step 1: Calculate the probabilities of picking each shape.
- Total number of shapes: \( 4 + 3 = 7 \).
- Probability of picking a circle: \( \frac{4}{7} \).
- Probability of picking a square: \( \frac{3}{7} \).

2. Step 2: Construct the tree diagram.
- First branch: Pick a circle or a square.
- Second branch: Pick a circle or a square again (since the shape is replaced).

The tree diagram will look like this:

```
Start
|
________|_________
| |
Circle Square
/ \ / \
Circle Square Circle Square
```

3. Step 3: Assign probabilities to each branch.
- Probability of picking a circle on the first draw: \( \frac{4}{7} \).
- Probability of picking a square on the first draw: \( \frac{3}{7} \).
- Since the shape is replaced, the probabilities remain the same for the second draw:
- Probability of picking a circle on the second draw: \( \frac{4}{7} \).
- Probability of picking a square on the second draw: \( \frac{3}{7} \).

4. Step 4: Calculate the probabilities for each outcome.
- Outcome 1: Circle, then Circle
\[
P(\text{Circle, then Circle}) = \frac{4}{7} \times \frac{4}{7} = \frac{16}{49}
\]
- Outcome 2: Circle, then Square
\[
P(\text{Circle, then Square}) = \frac{4}{7} \times \frac{3}{7} = \frac{12}{49}
\]
- Outcome 3: Square, then Circle
\[
P(\text{Square, then Circle}) = \frac{3}{7} \times \frac{4}{7} = \frac{12}{49}
\]
- Outcome 4: Square, then Square
\[
P(\text{Square, then Square}) = \frac{3}{7} \times \frac{3}{7} = \frac{9}{49}
\]

#### Answer:
The completed tree diagram and probabilities are:
\[
\boxed{
\begin{aligned}
&\text{Circle, then Circle: } \frac{16}{49} \\
&\text{Circle, then Square: } \frac{12}{49} \\
&\text{Square, then Circle: } \frac{12}{49} \\
&\text{Square, then Square: } \frac{9}{49}
\end{aligned}
}
\]

---

Problem 3: Probability of picking exactly one white shape

#### Given:
- A bag contains black and white shapes.
- The probability of picking a black shape is \( \frac{2}{5} \).
- The probability of picking a white shape is \( \frac{3}{5} \).
- A shape is picked, replaced, and another shape is picked.

#### Task:
Find the probability of picking exactly one white shape.

#### Solution:
1. Step 1: Identify the scenarios for picking exactly one white shape.
- Scenario 1: White on the first pick and Black on the second pick.
- Scenario 2: Black on the first pick and White on the second pick.

2. Step 2: Calculate the probability for each scenario.
- Scenario 1: White, then Black
\[
P(\text{White, then Black}) = P(\text{White}) \times P(\text{Black}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}
\]
- Scenario 2: Black, then White
\[
P(\text{Black, then White}) = P(\text{Black}) \times P(\text{White}) = \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}
\]

3. Step 3: Add the probabilities of the two scenarios.
\[
P(\text{Exactly one white shape}) = P(\text{White, then Black}) + P(\text{Black, then White})
\]
\[
P = \frac{6}{25} + \frac{6}{25} = \frac{12}{25}
\]

#### Answer:
\[
\boxed{\frac{12}{25}}
\]

---

Problem 4: Probability of losing in the final or semi-final

#### Given:
- Aufberg FC has 3 games left: Quarter Final (QF), Semi-Final (SF), and Final.
- Probabilities of winning:
- \( P(\text{win QF}) = 0.8 \)
- \( P(\text{win SF}) = \frac{3}{4} \)
- \( P(\text{win Final}) = 0.6 \)

#### Task:
Find the probability of losing in the final or semi-final.

#### Solution:
1. Step 1: Calculate the probabilities of losing each game.
- Probability of losing the Quarter Final:
\[
P(\text{lose QF}) = 1 - P(\text{win QF}) = 1 - 0.8 = 0.2
\]
- Probability of losing the Semi-Final:
\[
P(\text{lose SF}) = 1 - P(\text{win SF}) = 1 - \frac{3}{4} = \frac{1}{4}
\]
- Probability of losing the Final:
\[
P(\text{lose Final}) = 1 - P(\text{win Final}) = 1 - 0.6 = 0.4
\]

2. Step 2: Determine the scenarios for losing in the final or semi-final.
- To lose in the final or semi-final, Aufberg FC must win the Quarter Final but lose either the Semi-Final or the Final.

3. Step 3: Calculate the probability of winning the Quarter Final.
\[
P(\text{win QF}) = 0.8
\]

4. Step 4: Calculate the probability of losing in the Semi-Final given that they won the Quarter Final.
\[
P(\text{lose SF | win QF}) = P(\text{lose SF}) = \frac{1}{4}
\]

5. Step 5: Calculate the probability of reaching the Final (i.e., winning both QF and SF).
\[
P(\text{reach Final}) = P(\text{win QF}) \times P(\text{win SF}) = 0.8 \times \frac{3}{4} = 0.6
\]

6. Step 6: Calculate the probability of losing in the Final given that they reached the Final.
\[
P(\text{lose Final | reach Final}) = P(\text{lose Final}) = 0.4
\]

7. Step 7: Combine the probabilities.
- Probability of losing in the Semi-Final:
\[
P(\text{lose SF}) = P(\text{win QF}) \times P(\text{lose SF}) = 0.8 \times \frac{1}{4} = 0.2
\]
- Probability of reaching the Final and losing in the Final:
\[
P(\text{reach Final and lose Final}) = P(\text{reach Final}) \times P(\text{lose Final}) = 0.6 \times 0.4 = 0.24
\]

8. Step 8: Add the probabilities of the two scenarios.
\[
P(\text{lose in SF or Final}) = P(\text{lose SF}) + P(\text{reach Final and lose Final})
\]
\[
P = 0.2 + 0.24 = 0.44
\]

#### Answer:
\[
\boxed{0.44}
\]

---

Final Answers:

1. \(\boxed{\frac{5}{36}}\)
2. \(\boxed{\frac{16}{49}, \frac{12}{49}, \frac{12}{49}, \frac{9}{49}}\)
3. \(\boxed{\frac{12}{25}}\)
4. \(\boxed{0.44}\)