Out of a group of 30 honest people, 20 always speak the truth. Two person are selected at random form the group. Find the probability distribution of no. of selected persons who speak truth. Also find the mean.

Given a group of 30 people where 20 always speak the truth, two persons are selected at random from the group. We want to find the probability distribution of the number of selected persons who speak the truth and also find the mean of this distribution.

### Step 1: Define the Random Variable \( X \)

Let \( X \) be the random variable that represents the number of selected persons who speak the truth. \( X \) can take values \( 0 \), \( 1 \), or \( 2 \).

### Step 2: Calculate the Probability Distribution of \( X \)

1. **\( P(X = 0) \)**: The probability that neither of the selected persons speaks the truth.

   - The number of ways to choose 2 persons who do not speak the truth (i.e., from the remaining 10 people): \( \binom{10}{2} \)
   - The total number of ways to choose any 2 persons from the group: \( \binom{30}{2} \)

   \[
   P(X = 0) = \frac{\binom{10}{2}}{\binom{30}{2}} = \frac{\frac{10 \times 9}{2}}{\frac{30 \times 29}{2}} = \frac{45}{435} \approx 0.1034
   \]

2. **\( P(X = 1) \)**: The probability that exactly one of the selected persons speaks the truth.

   - The number of ways to choose 1 truth-teller from the 20 who speak the truth: \( \binom{20}{1} \)
   - The number of ways to choose 1 non-truth-teller from the remaining 10 people: \( \binom{10}{1} \)
   - The total number of ways to choose any 2 persons from the group: \( \binom{30}{2} \)

   \[
   P(X = 1) = \frac{\binom{20}{1} \binom{10}{1}}{\binom{30}{2}} = \frac{20 \times 10}{435} = \frac{200}{435} \approx 0.4598
   \]

3. **\( P(X = 2) \)**: The probability that both selected persons speak the truth.

   - The number of ways to choose 2 persons who speak the truth from the 20 truth-tellers: \( \binom{20}{2} \)
   - The total number of ways to choose any 2 persons from the group: \( \binom{30}{2} \)

   \[
   P(X = 2) = \frac{\binom{20}{2}}{\binom{30}{2}} = \frac{\frac{20 \times 19}{2}}{\frac{30 \times 29}{2}} = \frac{190}{435} \approx 0.4368
   \]

### Step 3: Probability Distribution of \( X \)

The probability distribution of \( X \) is:

\[
P(X = 0) \approx 0.1034, \quad P(X = 1) \approx 0.4598, \quad P(X = 2) \approx 0.4368
\]

### Step 4: Calculate the Mean (Expected Value) of \( X \)

The expected value \( E(X) \) is calculated using the formula:

\[
E(X) = \sum_{x=0}^{2} x \cdot P(X = x)
\]

Substituting the values we found:

\[
E(X) = 0 \cdot 0.1034 + 1 \cdot 0.4598 + 2 \cdot 0.4368 = 0 + 0.4598 + 0.8736 = 1.3334
\]

### Final Results:

1. **Probability Distribution:**
   - \( P(X = 0) \approx 0.1034 \)
   - \( P(X = 1) \approx 0.4598 \)
   - \( P(X = 2) \approx 0.4368 \)

2. **Mean (Expected Value):**
   - \( E(X) \approx 1.3334 \)

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Would you like to explore more details on this topic? Here are five related questions:

1. What is the variance of the random variable \( X \) in this case?
2. How would the probability distribution change if 3 persons were selected instead of 2?
3. Can you compute the probability that at least one of the selected persons speaks the truth?
4. How would the mean change if the group size increased to 40 people with the same ratio of truth-tellers?
5. How does this problem relate to the binomial distribution when sampling with replacement?

**Tip:** The expected value \( E(X) \) provides the average number of truth-tellers you would expect to find in repeated samples, making it a useful measure for understanding the distribution's central tendency.

Explore the probability distribution of truth-tellers among 30 people, where 20 always speak the truth. Calculate probabilities for selecting 0, 1, or 2 truth-tellers and find the mean of the distribution. Detailed solution with combinatorial calculations provided.

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