document.write( "Question 1189107: It is known that the computer disks produced by a company are defective with probability 0.02 independently of each other. Disks are sold in packs of 10. A money back guarantee is offered if a pack contains more than 1 defective disk. (a) What proportion of sales result in the customers getting their money back? (b) In a stock of 100 packs, how many packs are expected to result in the customers getting their money back? \n" ); document.write( "

Algebra.Com's Answer #820345 by ikleyn(52877) $\"\"$ $\"About$

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\n" ); document.write( "It is known that the computer disks produced by a company are defective with probability 0.02
\n" ); document.write( "independently of each other. Disks are sold in packs of 10.
\n" ); document.write( "A money back guarantee is offered if a pack contains more than 1 defective disk.
\n" ); document.write( "(a) What proportion of sales result in the customers getting their money back?
\n" ); document.write( "(b) In a stock of 100 packs, how many packs are expected to result in the customers getting their money back?
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document.write( "(a)  It is a Binomial Distribution probability problem.\r\n" );
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document.write( "     To answer question (a), we should calculate the probability having more than 1 defective disk \r\n" );
document.write( "     in a pack of 10 disks.\r\n" );
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document.write( "     It is the COMPLEMENT to the probability to have 0 (zero) or 1 (one) defective disk in the pack of 10 disks, which is\r\n" );
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document.write( "         P =  +  = 0.8171 + 0.16675 = 0.98385.\r\n" );
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document.write( "     So, in 1 - 0.98385 = 0.01615 = 1.615% cases the customers getting their money back.    ANSWER\r\n" );
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document.write( "(b)  In a stock of 100 packs,  1.615 packs are expected to result in the customers getting their money back.    ANSWER\r\n" );
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