document.write( "Question 885632: How many 5-digit numbers contain at least one 3? \n" ); document.write( "

Algebra.Com's Answer #535308 by Edwin McCravy(20056) $\"\"$ $\"About$

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document.write( "When you see \"at least one\" in a probability problem, you usually first\r\n" );
document.write( "first find the probability of the complement event, which is \"none at all\",\r\n" );
document.write( "and then subtract from 1.  That's what we do here.\r\n" );
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document.write( "First we find the denominator of the probability, which\r\n" );
document.write( "is the number of 5 digit numbers:\r\n" );
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document.write( "There are two ways to determine this.\r\n" );
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document.write( "First way:  The number of 5-digit numbers is 99999 minus all\r\n" );
document.write( "the numbers with fewer than 5 digits, which is 9999, So the\r\n" );
document.write( "number of 5-digit numbers is 99999-9999 = 90000\r\n" );
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document.write( "Second way:\r\n" );
document.write( "Choose the 1st digit 9 ways (can't choose 0).\r\n" );
document.write( "Choose the 2nd digit 10 ways (can choose 0).\r\n" );
document.write( "Choose the 3rd digit 10 ways.\r\n" );
document.write( "Choose the 4th digit 10 ways.\r\n" );
document.write( "Choose the 5th digit 10 ways.\r\n" );
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document.write( "9×10×10×10×10 = 90000\r\n" );
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document.write( "Now we find the numerator, which is all the 5-digit numbers\r\n" );
document.write( "which do not contain a 3.  That is, we choose only digits from\r\n" );
document.write( "these 9 {0,1,2,4,5,6,7,8,9}\r\n" );
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document.write( "Choose the 1st digit 8 ways (can't choose 0).\r\n" );
document.write( "Choose the 2nd digit 9 ways (can choose 0).\r\n" );
document.write( "Choose the 3rd digit 9 ways.\r\n" );
document.write( "Choose the 4th digit 9 ways.\r\n" );
document.write( "Choose the 5th digit 9 ways.\r\n" );
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document.write( "8×9×9×9×9 = 52488\r\n" );
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document.write( "Probability of complement event =  which\r\n" );
document.write( "reduces to \r\n" );
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document.write( "Desired probability =  \r\n" );
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document.write( "Edwin

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