Question

Let \(X \sim B(n, p)\) with mean \(\mu\) and variance \(\sigma^2\). If \(\mu = 2\sigma^2\) and \(\mu + \sigma^2 = 3\), then find \(P(X \leq 3)\).

• A. \(\frac{40}{49}\)
• B. \(\frac{40}{43}\)
• C. \(\frac{100}{101}\)
• D. \(\frac{15}{16}\)

Hint:

Use given relations to find \(p\) and \(n\), then calculate cumulative binomial probabilities using complement.

Answer 1

The Correct Option is D

Given, \[ \mu = np,
\sigma^2 = np(1-p). \] From \(\mu = 2\sigma^2\), \[ np = 2 np(1-p) \implies 1 = 2(1-p) \implies 1 = 2 - 2p \implies p = \frac{1}{2}. \] From \(\mu + \sigma^2 = 3\), \[ np + np(1-p) = 3 \implies np + np \times \frac{1}{2} = 3 \implies np \times \frac{3}{2} = 3 \implies np = 2. \] So, \(np = 2\), \(p = \frac{1}{2}\), so \(n = \frac{2}{p} = 4\). Calculate \[ P(X \leq 3) = \sum_{k=0}^3 \binom{4}{k} \left(\frac{1}{2}\right)^4 = 1 - P(X=4). \] \[ P(X=4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{16}. \] Hence, \[ P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}. \]