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Out of 100 bicycles, 10 bicycles have punctures. What is the probability of not having any punctured bicycle in a sample of 5 bicycles?(a) \(\frac{1}{2^5}\)(b) \(\frac{1}{2^9}\)(c) \(\left(\frac{9}{10}\right)^5\)(d) \(\frac{1}{10^5}\) |
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Answer» Correct answer is (c) \(\left(\frac{9}{10}\right)^5\) |
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