document.write( "Question 565349: The probabilities that the serviceability of a new laser printer will be rated very difficult, difficult, average, easy or very easy are 0.11, 0.16, 0.35, 0.28, 0.10 respectively. Find the probability that the serviceability of a new laser printer will be rated (i) difficult or very difficult. (ii) difficult, average or easy. and (iii) average or better. \n" ); document.write( "

Algebra.Com's Answer #365767 by Theo(13342) $\"\"$ $\"About$

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probabilities for service ratings are:
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document.write( "very difficult        .11\r\n" );
document.write( "difficult             .16\r\n" );
document.write( "average               .35\r\n" );
document.write( "easy                  .28\r\n" );
document.write( "very easy             .10\r\n" );
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\n" ); document.write( "presumably these ratings are independent of each other, i.e. a laser printer can be difficult to service or it can be easy to service but it can't be both at the same time.
\n" ); document.write( "this makes p(a or b) equal to p(a) + p(b)
\n" ); document.write( "if they were not independent of each other, then p(a or b) would be equal to p(a) + p(b) - p(a and b)
\n" ); document.write( "as you may know, p(a and b) is equal to p(a) * p(b).
\n" ); document.write( "if you don't yet, you will soon enough.
\n" ); document.write( "anyway, assuming these are independent, then p(a or b) is equal to p(a) + p(b).
\n" ); document.write( "based on this, the answers to your questions are:
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document.write( "servicability rating                 calculation             probability\r\n" );
document.write( "difficult or very difficult          .11 + .16                   .27\r\n" );
document.write( "difficult or average or easy         .16 + .35 + .28             .79\r\n" );
document.write( "average or better                    .35 + .28 + .10             .73\r\n" );
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\n" ); document.write( "average or better presumes average or easy or very easy.\r
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