Probability
Binomial Distribution
Grade 12
Question:
<p>In a binomial distribution \(B(b,\, p = 1/4)\), if the probability of at least one success is greater than or equal to 9/10, then \(n\) is greater than</p>
<p>\(\dfrac{1}{\log_{10}4 - \log_{10}3}\)</p>
<p>\(\dfrac{1}{\log_{10}4 + \log_{10}3}\)</p>
<p>\(\dfrac{9}{\log_{10}4 - \log_{10}3}\)</p>
<p>\(\dfrac{4}{\log_{10}4 - \log_{10}3}\)</p>
Step-by-Step Solution
Key Concept: Use the complement rule: P(at least one success) = 1 - P(no success) = 1 - (1-p)^n. Set up the inequality 1 - (3/4)^n ≥ 9/10 and solve for n using logarithms.
<p><strong>Step 1:</strong> Set up the complement inequality. For binomial B(n, p = 1/4), P(at least one success) = 1 - P(X = 0) = 1 - (1 - 1/4)^n = 1 - (3/4)^n</p><p><strong>Step 2:</strong> Apply the given condition: 1 - (3/4)^n ≥ 9/10</p><p><strong>Step 3:</strong> Simplify: (3/4)^n ≤ 1/10</p><p><strong>Step 4:</strong> Take natural logarithm of both sides: n·ln(3/4) ≤ ln(1/10). Since ln(3/4) < 0, reverse the inequality when dividing: n ≥ ln(10)/ln(4/3) = ln(10)/(ln(4) - ln(3))</p><p><strong>Step 5:</strong> Calculate numerically: ln(10) ≈ 2.303, ln(4/3) ≈ 0.288, so n ≥ 2.303/0.288 ≈ 7.99</p><p><strong>Step 6:</strong> Since n must be a positive integer and n ≥ 7.99, we have n ≥ 8</p><p>∴ Answer: n is greater than 7 (or n ≥ 8)</p>
Correct Answer: A