Question

Let \( X \sim {Bin}(2, \frac{1}{3}) \). Then \( 18 \cdot E(X^2) \) equals ___________ (answer in integer).

Hint:

For Binomial distributions, use the identity \( E(X^2) = {Var}(X) + (E(X))^2 \) to calculate second moments.

Answer 1

Correct Answer: 16

Step 1: Recall the properties of the Binomial distribution. For \( X \sim {Bin}(n, p) \), the expected value \( E(X) \) and the variance \( {Var}(X) \) are given by: \[ E(X) = np \quad {and} \quad {Var}(X) = np(1-p). \] For \( X \sim {Bin}(2, \frac{1}{3}) \), we have: \[ E(X) = 2 \times \frac{1}{3} = \frac{2}{3}, \quad {Var}(X) = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}. \] Step 2: Calculate \( E(X^2) \). Using the identity \( E(X^2) = {Var}(X) + (E(X))^2 \), we can compute \( E(X^2) \): \[ E(X^2) = \frac{4}{9} + \left( \frac{2}{3} \right)^2 = \frac{4}{9} + \frac{4}{9} = \frac{8}{9}. \] Step 3: Final Calculation. Now, we compute \( 18 \cdot E(X^2) \): \[ 18 \cdot E(X^2) = 18 \cdot \frac{8}{9} = 16. \] Thus, the answer is \( \boxed{16} \).