Smith has '4' bats namely B1, B2, B3, B4. He randomly selects
one out of the three bats which was not used in the previous match. If
for his first match the chosen bat was B1 and what is the probability
that he uses B1 again in the 6th Match? (3 attempts) We checked answer 4/27 and 1/3 bothare wrong .should we use bayes theorem.this is our last chance .kindly give the correct and accurate answer

To solve this problem, let's break it down step by step and use the correct approach.

### Problem Recap:
- Smith has 4 bats: B1, B2, B3, B4.
- In each match, Smith randomly selects one of the three bats **that were not used in the previous match**.
- The bat selected for the first match is B1.
- We need to find the probability that Smith uses B1 again in the 6th match.

### Step-by-Step Solution:
1. **Initial Match (1st Match):**
   - Bat chosen: B1.

2. **2nd Match:**
   - Since B1 was used in the 1st match, Smith can only choose from B2, B3, or B4.
   - Probability of choosing any of these bats is \( \frac{1}{3} \) for each.

3. **Subsequent Matches:**
   - For each subsequent match, the probability of choosing any bat except the one used in the previous match is \( \frac{1}{3} \).

4. **Key Observation:**
   - For Smith to use B1 in the 6th match, B1 should not have been chosen in the 5th match.
   - To get B1 in the 6th match, the following must happen:
     - **5th Match:** Any bat except B1 (Probability = \( \frac{2}{3} \))
     - **6th Match:** B1 (Probability = \( \frac{1}{3} \))

### Consider all possible sequences that lead to B1 being selected in the 6th match:
- Let’s denote B1 as \( x \) and any other bat as \( y \).
- There are several sequences, but the crucial ones are where B1 is selected in the 6th match.

Given that there are five matches preceding the 6th match, the sequence matters in terms of \( x \) and \( y \) selections, but due to the Markov property (only the immediate previous match affects the next), we don’t need to list out every possible sequence. Instead, we note that each cycle starting with a selection has the same probabilities. 

### 5th Match Selection Impact:
- **Probability that B1 is chosen in the 6th match**:
   - If the 5th match bat is \( y \) (which could be any of B2, B3, or B4), then the 6th match bat could be B1, with a probability of \( \frac{1}{3} \).

This situation simplifies to having \( y \) in the 5th match and selecting \( x \) (B1) in the 6th match. 

Given that the sequence can repeat in blocks for each match due to independent choices, the key outcome occurs when the 6th match comes after choosing \( y \) in the 5th match, ensuring that B1 is picked.

The probability that this occurs after 6 matches, given the rules, follows the recursive pattern and summing probabilities from each case.

Thus, the correct probability is **\( \frac{2}{9} \)**.

### Conclusion:
**The correct probability that Smith uses B1 in the 6th match is \( \frac{2}{9} \).**

Do you need more details or have any questions?

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Here are 5 related questions:

1. How does the Markov property simplify probability calculations in similar problems?
2. What are the effects of conditional probability in sequences with dependent events?
3. How does probability theory handle events where sequences are restricted by previous outcomes?
4. Why is it important to consider all possible sequences in such probability problems?
5. How does this problem compare to classical problems like the Gambler's Ruin in probability theory?

**Tip:** When solving multi-step probability problems, carefully track each stage's conditional probability.

Understand the probability problem involving Smith's bat selection across multiple matches. Learn the step-by-step solution and probability calculations related to conditional probability and Markov chains. Suitable for advanced high school students and those interested in probability theory.

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