Probability
Binomial Distribution
Grade 12
Question:
<p>If the mean and the variance of a binomial variate \(X\) are 2 and 1, respectively, then the probability that \(X\) takes a value greater than or equal to one is</p>
<p>\(\dfrac{1}{16}\)</p>
<p>\(\dfrac{9}{16}\)</p>
<p>\(\dfrac{3}{4}\)</p>
<p>\(\dfrac{15}{16}\)</p>
Step-by-Step Solution
Key Concept: For a binomial distribution, mean = np and variance = np(1-p). Use these two equations to find n and p, then calculate P(X ≥ 1) = 1 - P(X = 0).
<p><strong>Step 1:</strong> Set up equations from given information.</p><p>For binomial variate X ~ B(n,p):</p><p>Mean: np = 2 ... (1)</p><p>Variance: np(1-p) = 1 ... (2)</p><p><strong>Step 2:</strong> Find p by dividing equation (2) by equation (1).</p><p>np(1-p)/np = 1/2</p><p>1 - p = 1/2</p><p>p = 1/2</p><p><strong>Step 3:</strong> Find n from equation (1).</p><p>n × (1/2) = 2</p><p>n = 4</p><p><strong>Step 4:</strong> Calculate P(X ≥ 1).</p><p>P(X ≥ 1) = 1 - P(X = 0)</p><p>P(X = 0) = C(4,0)(1/2)⁴(1/2)⁰ = 1/16</p><p>P(X ≥ 1) = 1 - 1/16 = <strong>15/16</strong></p><p>∴ Answer: D</p>
Correct Answer: D