Question 189786
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If there is a 5% probability that a keyboard will be defective, then there is a 95% probability that it will have no defects.  So for a sample of six keyboards, whether or not any of them are defective is six independent events -- meaning that whether the first one is defective has no effect on whether any of the others are defective, and so on.


Hence, the probability that none of them will be defective is the product of the probabilities that each of them individually is not defective:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  .95 \times .95 \times .95 \times .95 \times .95 \times .95 = .95^6 \approx .735]


or 73.5%


Windows calculator: .95, "x^y" button, 6, "=" button.


Looking at it another way, the odds are better than 1 in 4 that one of the six <i><b>will</b></i> be defective.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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