<p>In a bulb manufacturing company probability that a bulb is defective is 0.01. The company sells the bulbs in a package of 10. After getting complains about defective bulb company decided to offer a money back guarantee that at most of the 10 from the package is defective. What percentage of the packages company has to replace?</p>
Step-by-Step Solution
Key Concept: Use the binomial distribution with n=10, p=0.01 to find P(X ≥ 2), where X is the number of defective bulbs. The company replaces packages when MORE than 1 bulb is defective (i.e., at least 2 defective bulbs).
<p><strong>Step 1:</strong> Identify the guarantee condition: At most 1 defective bulb means ≤1 defective. Company replaces when this is violated, i.e., when X ≥ 2 defective bulbs.</p><p><strong>Step 2:</strong> Set up binomial distribution: X ~ B(10, 0.01), where n=10, p=0.01, q=0.99</p><p><strong>Step 3:</strong> Calculate P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - [P(X=0) + P(X=1)]</p><p><strong>Step 4:</strong> P(X=0) = C(10,0)(0.01)⁰(0.99)¹⁰ = (0.99)¹⁰ ≈ 0.9044</p><p><strong>Step 5:</strong> P(X=1) = C(10,1)(0.01)¹(0.99)⁹ = 10(0.01)(0.99)⁹ ≈ 0.0914</p><p><strong>Step 6:</strong> P(X ≥ 2) = 1 - (0.9044 + 0.0914) = 1 - 0.9958 = 0.0042</p><p><strong>Step 7:</strong> Percentage of packages to replace = 0.0042 × 100% = 0.42% ≈ <strong>0.4% to 0.5%</strong></p><p>∴ Answer: A</p>
Correct Answer: A