document.write( "Question 1197428: a quality control engineer inspects a random sample of three batteries from a lot of 30 cars batteries, which are ready to be shipped . if the such a lot contains ten batteries with slight defects, what then is the probability that the inspector's sample contains .
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\n" ); document.write( "2)at least one of the batteries with slight defects?
\n" ); document.write( "solve it by extended general addition rule depended event . \n" ); document.write( "

Algebra.Com's Answer #830700 by ikleyn(52780) $\"\"$ $\"About$

You can put this solution on YOUR website!
.
\n" ); document.write( "a quality control engineer inspects a random sample of three batteries from a lot of 30 cars batteries,
\n" ); document.write( "which are ready to be shipped . if the such a lot contains ten batteries with slight defects,
\n" ); document.write( "what then is the probability that the inspector's sample contains
\n" ); document.write( "1) none of the batteries with slight defects?
\n" ); document.write( "2) at least one of the batteries with slight defects?
\n" ); document.write( "solve it by extended general addition rule depended event.
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\n" ); document.write( "\n" ); document.write( "            The solution by @ewatrrr is WRONG not only numerically, but CONCEPTUALLY.\r
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\n" ); document.write( "\n" ); document.write( "            This problem IS NOT on binomial distribution.\r
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\n" ); document.write( "\n" ); document.write( "            I came to bring you a correct solution.\r
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document.write( "We have the set A of 30 items and its subset B of defective items consisting of 10 items.\r\n" );
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document.write( "We randomly select 3 items from the set A.\r\n" );
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document.write( "If no one of these 3 items is defective, it means that this triple of items is selected from the subset of 20 good items.\r\n" );
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document.write( "The probability to have good item at the first selection is  .\r\n" );
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document.write( "The probability to have good item at the second selection (having good item at the first selection) is  .\r\n" );
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document.write( "The probability to have good item at the third selection (having good items after the first and second selections) is  .\r\n" );
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document.write( "The probability to have good triple is thus  the product   =  =  =  = 0.2808  (rounded).\r\n" );
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document.write( "It is the answer to question (1).\r\n" );
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document.write( "The answer to question (2) is the COMPLEMENT  1 - 0.2808 = 0.7192  (rounded, too).\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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\n" ); document.write( "\n" ); document.write( "This problem is not for the first time at this forum: I solved it earlier by different method under this link\r
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\n" ); document.write( "\n" ); document.write( "https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1197318.html\r
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\n" ); document.write( "\n" ); document.write( "For your convenience, I copy-paste that solution here again.\r
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document.write( "(1) The probability in this problem is the ratio of two numbers.\r\n" );
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document.write( "    The denominator is the number of all possible triples of batteries from \r\n" );
document.write( "    the whole set of 30 batteries.  The number of such triples is the number of all combinations\r\n" );
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document.write( "         =  = 4060.\r\n" );
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document.write( "    The numerator of this fraction is the number of all triples comprising of good batteries.\r\n" );
document.write( "    It is the number of combinations   =  =  = 1140.\r\n" );
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document.write( "    Thus the probability  P = P(1) =  =  =  = 0.2808  (rounded).    ANSWER\r\n" );
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document.write( "(2)  For the second question, the probability is the COMPLEMENT to what we found  above\r\n" );
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document.write( "         P(2) = 1 - P(1) = 1 - 0.2808 = 0.7192   (rounded).    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved twice, by two different methods, for your better understanding.\r
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\n" ); document.write( "\n" ); document.write( "(Both answers are the same, naturally).\r
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