Question

A fair coin is tossed three times. Let \( X \) be the number of tails appearing and its distribution is: \[ \begin{array}{|c|c|c|c|c|} \hline X = x & 0 & 1 & 2 & 3 \\ \hline P(X = x) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \end{array} \] Then the variance of \( X \) is:

• A. \( \frac{1}{4} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{5}{4} \)
• D. \( \frac{7}{4} \)

Hint:

Variance = \( E[X^2] - (E[X])^2 \). Carefully square the expected value and subtract from the second moment.

Answer 1

The Correct Option is B

Use the formula for variance: \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] First compute \( E[X] \): \[ E[X] = \sum x P(x) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} = \frac{0 + 3 + 6 + 3}{8} = \frac{12}{8} = 1.5 \] Then compute \( E[X^2] \): \[ E[X^2] = \sum x^2 P(x) = 0^2 \cdot \frac{1}{8} + 1^2 \cdot \frac{3}{8} + 4 \cdot \frac{3}{8} + 9 \cdot \frac{1}{8} = \frac{0 + 3 + 12 + 9}{8} = \frac{24}{8} = 3 \] \[ \text{Var}(X) = 3 - (1.5)^2 = 3 - 2.25 = 0.75 = \frac{3}{4} \]