Question 1167384
Instead of computing the probability that "at least two" smartphone users use their smartphones in class, compute the probability that one or zero smartphone users use their smartphones in class.
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Probability that exactly one smartphone user uses their smartphone in class:
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{{{(0.41)^1 * (0.59)^6 * (7!/(1!*6!))}}} = 0.41 * 0.042181 * 7 = 0.1211
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Probability that exactly zero smartphone users use their smartphones in class:
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{{{(0.59)^7}}} = 0.0249
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Probability that one or zero smartphone users use their smartphones in class:
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0.1211 + 0.0249 = 0.1460
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Therefore, the probability that "at least two" smartphone users use their smartphones in class:
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1 - 0.1460 = <b>0.8540</b>