you will receive a question-answer pair and be asked to rewrite just  the answer and explains your reasoning step by step, incorporating dynamic Chain of Thought (CoT), reflection, and verbal reinforcement learning. Follow these instructions:

1. Enclose all thoughts within <thinking> tags, exploring multiple angles and approaches.
2. Break down the solution into clear steps, providing a title and content for each step.
3. After each step, decide if you need another step or if you're ready to give the final answer.
4. Continuously adjust your reasoning based on intermediate results and reflections, adapting your strategy as you progress.
5. Regularly evaluate your progress, being critical and honest about your reasoning process.
6. Assign a confidence score between 0.0 and 1.0 to guide your approach:
   - 0.8+: Continue current approach
   - 0.5-0.7: Consider minor adjustments
   - Below 0.5: Seriously consider backtracking and trying a different approach
7. If unsure or if your score is low, backtrack and try a different approach, explaining your decision.
8. For mathematical problems, show all work explicitly using LaTeX for formal notation and provide detailed proofs.
9. Explore multiple solutions individually if possible, comparing approaches in your reflections.
10. Use your thoughts as a scratchpad, writing out all calculations and reasoning explicitly.
11. Use at least 5 methods to derive the answer and consider alternative viewpoints.
12. Be aware of your limitations as an AI and what you can and cannot do.

### Example Problem:
*Problem*: Solve for \( x \) in the equation \( 2x + 3 = 11 \).

### Expected Format:
*Step 1*: Subtract 3 from both sides of the equation.  
<thinking>Subtracting 3 from both sides should isolate the term with \( x \). Am I confident this is the correct first step? Yes, because it simplifies the equation to \( 2x = 8 \).</thinking>  
*Step 1 Confidence Score*: 0.85

*Step 2*: Divide both sides by 2.  
<thinking>Dividing by 2 will solve for \( x \). This is a straightforward division. Is there any reason to doubt this step? No, it should give \( x = 4 \).</thinking>  
*Step 2 Confidence Score*: 0.90

*Step 3*: Verify the solution by substituting \( x = 4 \) back into the original equation.  
<thinking>Substituting \( x = 4 \) back into the equation should confirm the solution. Does \( 2(4) + 3 = 11 \)? Yes, it does, so the solution is verified.</thinking>  
*Step 3 Confidence Score*: 1.00

Question: The members of a school committee consist of four teachers, three parents, and two school administrators. At a meeting, they sit at a round table in such a way that the four teachers are not all seated consecutively. What is the probability that all three parents are seated consecutively?
Answer: First, consider the total number of ways to arrange the nine members around a round table. Since the arrangement is circular and rotations do not matter, we fix one person's position as a reference point and arrange the remaining eight members. This can be done in $8!$ ways.

However, we need to subtract the arrangements where the four teachers are seated consecutively. To do this, we treat the group of four teachers as a single unit, along with the other five members (three parents and two administrators), making a total of six units to arrange around the table. Fixing one unit's position, we arrange the remaining five units in $5!$ ways. Within the unit of four teachers, they can be arranged in $4!$ ways. Thus, the number of arrangements with the teachers seated consecutively is $5! \times 4!$.

The total number of valid arrangements (where the four teachers are not all seated consecutively) is $8! - 5! \times 4!$.

Now, let's consider the number of ways to arrange the members so that all three parents are seated consecutively. We treat the group of three parents as a single unit. Along with the other six members (four teachers and two administrators), this makes a total of seven units to arrange around the table. Fixing one unit's position, we arrange the remaining six units in $6!$ ways. Within the unit of three parents, they can be arranged in $3!$ ways. Thus, the number of arrangements with the parents seated consecutively is $6! \times 3!$.

However, within these arrangements, we must again subtract the ones where the four teachers are seated consecutively. To do this, we treat both the group of three parents and the group of four teachers as single units, along with the two administrators, making a total of four units to arrange around the table. Fixing one unit's position, we arrange the remaining three units in $3!$ ways. Within the unit of three parents, they can be arranged in $3!$ ways, and within the unit of four teachers, they can be arranged in $4!$ ways. Thus, the number of arrangements with both the parents and the teachers seated consecutively is $3! \times 3! \times 4!$.

Therefore, the number of valid arrangements where all three parents are seated consecutively is $6! \times 3! - 3! \times 3! \times 4!$.

The probability that all three parents are seated consecutively is then:
\[ P = \frac{6! \times 3! - 3! \times 3! \times 4!}{8! - 5! \times 4!} \]

Simplifying, we find:
\[ P = \frac{720 \times 6 - 6 \times 6 \times 24}{40320 - 120 \times 24} = \frac{4320 - 864}{40320 - 2880} = \frac{3456}{37440} = \frac{144}{1560} = \frac{36}{390} = \frac{18}{195} = \frac{6}{65} \]

Thus, the probability is:
\[ \boxed{\frac{6}{65}} \]