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The problem asks for the probability \( P(X = 2) \), where \( X \) is the number of rounds of two coin tosses needed to select one of the respondents \( A \), \( B \), or \( C \), based on specific outcomes from two unbiased coin tosses.

### Selection rules:
- \( A \) is selected if the toss results in two heads (HH).
- \( B \) is selected if the toss results in two tails (TT).
- \( C \) is selected if the toss results in a head followed by a tail (HT).
- If the outcome is tail followed by head (TH), the process is repeated.

We are tasked with finding \( P(X = 2) \), which means we want to find the probability that exactly 2 rounds of tossing are needed to select one of the respondents. This occurs if:
- The first round results in \( TH \), which triggers a repeat of the toss.
- The second round results in \( HH \), \( TT \), or \( HT \), which selects a respondent.

### Step-by-step solution:

1. **Possible outcomes for one round of two coin tosses**:
   - There are four possible outcomes from two tosses: \( HH \), \( HT \), \( TH \), and \( TT \). Each has a probability of \( \frac{1}{4} \), as the coin is unbiased.

2. **First round outcome: TH (repeat)**:
   - If the first round gives \( TH \), we repeat the process. The probability of this happening is \( P(\text{TH}) = \frac{1}{4} \).

3. **Second round outcome: non-TH (successful selection)**:
   - In the second round, we want any outcome other than \( TH \), so the selection is made with outcomes \( HH \), \( HT \), or \( TT \).
   - The probability of selecting a respondent in this round is \( P(\text{HH or HT or TT}) = \frac{3}{4} \).

4. **Total probability of selection in two rounds**:
   - To have two rounds, the first must be \( TH \) and the second must be one of \( HH \), \( HT \), or \( TT \).
   - Therefore, the probability is:
   \[
   P(X = 2) = P(\text{TH in 1st round}) \times P(\text{Selection in 2nd round})
   = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}.
   \]

5. **Numeric value up to three decimal places**:
   \[
   \frac{3}{16} = 0.1875 \quad \text{(rounded to three decimal places)} = 0.188.
   \]

Thus, the value of \( P(X = 2) \) is **0.188**.

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Would you like more details, or do you have any questions?

Here are 5 related questions to consider:
1. How would the probability change if a different outcome triggered a repeat?
2. What is the expected number of rounds to select a respondent?
3. How would the problem change if the coin was biased?
4. Can we generalize this problem for more than two coin tosses per round?
5. How does the probability distribution of \( X \) look in this case?

**Tip:** To solve such problems, break down the process into sequential events and calculate the probability for each step independently.

One of three respondents, A, B, and C, has to be selected randomly by a survey time, with the help of an unbiased coin. The coin is tossed twice. A is selected if there are two heads, B is selected if there are two tails, and C is selected if there is a head followed by a tail. If there is a tail followed by a head, the process is repeated. If X is the number of rounds of two tosses needed to make the selection, what is the value of P(X = 2), computed up to three places after decimal?

Learn how to calculate the probability P(X = 2) for selecting one respondent from A, B, or C using unbiased coin tosses. This problem involves key concepts from probability theory, including random experiments and combinatorics. Suitable for high school students in Grades 9-12.

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