Probability
Binomial Distribution
Grade 12
Question:
<p>The mean and the variance of a binomial distribution are 4 and 2, respectively. Then the probability of 2 successes is</p>
<p>\(\dfrac{37}{256}\)</p>
<p>\(\dfrac{219}{256}\)</p>
<p>\(\dfrac{128}{256}\)</p>
<p>\(\dfrac{28}{256}\)</p>
Step-by-Step Solution
Key Concept: From mean np = 4 and variance npq = 2, we can find n, p, q. Then use P(X=2) = C(n,2)p²q^(n-2) with these values.
<p><strong>Step 1:</strong> Use given conditions for binomial distribution:</p><p>Mean: np = 4</p><p>Variance: npq = 2</p><p><strong>Step 2:</strong> Find q by dividing variance by mean:</p><p>npq/np = 2/4 → q = 1/2</p><p><strong>Step 3:</strong> Therefore p = 1 - q = 1/2</p><p><strong>Step 4:</strong> Find n from np = 4:</p><p>n(1/2) = 4 → n = 8</p><p><strong>Step 5:</strong> Calculate P(X = 2):</p><p>P(X = 2) = C(8,2) × (1/2)² × (1/2)⁶</p><p>= 28 × (1/2)⁸</p><p>= 28/256 = 7/64</p><p>∴ Answer: D</p>
Correct Answer: D