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: Understanding the Question:
This is a Binomial Distribution problem with \( n = 6 \) and \( p = \frac{1}{2} \). We need to calculate the probability for the range defined by \( |X - 2| \le 1 \).
Step 2: Key Formula or Approach:
1. \( |X - 2| \le 1 \implies -1 \le X - 2 \le 1 \implies 1 \le X \le 3 \).
2. \( P(X=k) = \binom{n}{k} p^k q^{n-k} \).
3. Since \( p = q = \frac{1}{2} \), \( P(X=k) = \binom{6}{k} \left(\frac{1}{2}\right)^6 = \frac{\binom{6}{k}}{64} \).
Step 3: Detailed Explanation:
We need \( P(1 \le X \le 3) = P(X=1) + P(X=2) + P(X=3) \).
\[ P(X=1) = \frac{\binom{6}{1}}{64} = \frac{6}{64} \] \[ P(X=2) = \frac{\binom{6}{2}}{64} = \frac{15}{64} \] \[ P(X=3) = \frac{\binom{6}{3}}{64} = \frac{20}{64} \] Summing them up:
\[ P(1 \le X \le 3) = \frac{6 + 15 + 20}{64} = \frac{41}{64} \] Step 4: Final Answer:
The probability is \( \frac{41}{64} \).