document.write( "Question 1166471: A survey finds customers are charged incorrectly for 2 out of every 10 items. Suppose a customer purchases 14 items. Find the probability that the customer is charged incorrectly on at least 3 items.\r
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document.write( "In order to achieve this you have to add the binomial probability of 0, 1, 2, 3 together. Yet for some reason I keep turning up wrong... on my 3rd question with 3 attempts each.... I got:\r
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document.write( "3) .2501388953
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document.write( "2) .2501388953
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document.write( "3) .1539316279
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document.write( "4) .0439804651\r
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document.write( "=.6982\r
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document.write( "Please help.... \n" );
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Algebra.Com's Answer #790985 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! At least 3 means NOT 0,1,2 \n" ); document.write( "Try that. \n" ); document.write( "Probability is 0.4481 for those, so the answer is the complement or 0.5519 \n" ); document.write( " |