The random variable X can take only the values 0, 1, 2. Given that P(X=0) = P(X= 1) = p and `E(X^2) = E(X)` , then find the value of p

Answer

To solve the problem, we need to find the value of \( p \) given that the random variable \( X \) can take values 0, 1, and 2, with the probabilities \( P(X=0) = P(X=1) = p \) and \( P(X=2) = x \). We also know that \( E(X^2) = E(X) \).

### Step-by-Step Solution:

1. **Define the probabilities:**
\[ P(X=0) = p, \quad P(X=1) = p, \quad P(X=2) = x \]
Since the total probability must equal 1, we have:
\[ p + p + x = 1 \implies 2p + x = 1 \]

2. **Express \( x \) in terms of \( p \):**
\[ x = 1 - 2p \]

3. **Calculate \( E(X) \):**
The expected value \( E(X) \) is calculated as follows:
\[ E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) = 0 \cdot p + 1 \cdot p + 2 \cdot (1 - 2p) \]
Simplifying this gives:
\[ E(X) = p + 2(1 - 2p) = p + 2 - 4p = 2 - 3p \]

4. **Calculate \( E(X^2) \):**
The expected value \( E(X^2) \) is calculated as follows:
\[ E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) + 2^2 \cdot P(X=2) = 0 \cdot p + 1 \cdot p + 4 \cdot (1 - 2p) \]
Simplifying this gives:
\[ E(X^2) = p + 4(1 - 2p) = p + 4 - 8p = 4 - 7p \]

5. **Set up the equation from the condition \( E(X^2) = E(X) \):**
\[ 4 - 7p = 2 - 3p \]

6. **Solve for \( p \):**
Rearranging the equation:
\[ 4 - 2 = 7p - 3p \implies 2 = 4p \implies p = \frac{1}{2} \]

### Final Answer:
The value of \( p \) is \( \frac{1}{2} \).