document.write( "Question 1174700: A box contains 15 items, 4 of which are defective. Three items are selected. What is the probability that the first is good, the second is defective and third is also good. \n" ); document.write( "

Algebra.Com's Answer #800161 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "In the box, there are 11 good items and 4 defective.\r\n" );
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document.write( "The probability to select 1st item good is  .\r\n" );
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document.write( "If the 1st item was good, then the probability to get 2nd item defective is  .\r\n" );
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document.write( "If 1st item was good and 2nd item was defective, then the probability to get 3rd item good is  .\r\n" );
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document.write( "The overall probability under the problem's question is,  T H E R E F O R E,\r\n" );
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document.write( "    P =  =  =  = 0.1612 (rounded) = 16.12% (approximately).    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "If you ask me about the solution by respectful tutor Edwin, where and why it is wrong, I will answer you:\r
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document.write( "    It is wrong, BECAUSE Edwin mistakenly consider that 8 events in his list as EQUALLY LIKELIHOOD,\r\n" );
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document.write( "    while they ARE NOT.\r\n" );
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