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A bulb-factory has 5% defectives. In a random sample of 10 bulbs, the probability of **not more than one** defective is approximately:
A$(0.95)^{10}$
B$(0.95)^{10} + 10(0.05)(0.95)^9 \approx 0.914$
C$0.05 + 10(0.95)$
D$0.5$
Answer & Solution
Correct answer: B. $(0.95)^{10} + 10(0.05)(0.95)^9 \approx 0.914$
$P(X \leq 1) = P(0) + P(1) = (0.95)^{10} + 10(0.05)(0.95)^9 \approx 0.599 + 0.315 \approx 0.914$.
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