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put this solution on YOUR website!Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 12% of all visitors to the website are looking for other websites. Assuming that this estimate is correct, find the probability that, in a random sample of 5 visitors to the website, exactly 4 actually are looking for the website.
Round your response to at least three decimal places.
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It's a binomial problem with n = 5, p = 0.12 ; x = 4
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P(x=4) = 5C4(0.12)^4(0.88)^1 = 5*(0.00030736)(0.88) = 0.000912384..
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Cheers,
Stan H.
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