Question

If \( X \sim B(6, \frac{1}{2}) \), then \( P(|X - 2| \le 1) = \)}

• A. \( \frac{31}{32} \)
• B. \( \frac{41}{64} \)
• C. \( \frac{51}{64} \)
• D. \( \frac{63}{64} \)

Hint:

$|X-a| \le b$ means $X$ is within the range $[a-b, a+b]$.

Answer 1

The Correct Option is B

Step 1: Expand Absolute Inequality
$|X - 2| \le 1 \implies -1 \le X - 2 \le 1$.
$1 \le X \le 3$. So we need $P(X=1) + P(X=2) + P(X=3)$.
Step 2: Binomial Formula
$P(X=k) = \binom{6}{k} (1/2)^6 = \frac{\binom{6}{k}}{64}$.
Step 3: Calculation
$P = \frac{\binom{6}{1} + \binom{6}{2} + \binom{6}{3}}{64} = \frac{6 + 15 + 20}{64}$.
Step 4: Conclusion
$P = \frac{41}{64}$.
Final Answer:(B)