document.write( "Question 1080530: Suppose that a die is weighted such that prime numbers are twice as likely to
\n" ); document.write( "appear as non-primes. What is the probability of rolling 5 at most 2 times in 15
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Algebra.Com's Answer #694640 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The primes are 2,3,5, the non-primes are 1,4,6\r\n" );
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document.write( "Let x = probability of a non-prime, then\r\n" );
document.write( "2x = probability of a prime\r\n" );
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document.write( "P(1)=P(4)=P(6)=x, P(2)=P(3)=P(5)=2x\r\n" );
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document.write( "P(1 or 4 or 6) = x+x+x = 3x\r\n" );
document.write( "P(2 or 3 or 5) = 2x+2x+2x = 6x \r\n" );
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document.write( "   3x + 6x = 1\r\n" );
document.write( "        9x = 1\r\n" );
document.write( "         x = 1/9\r\n" );
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document.write( "P(1 or 4 or 6) = 3x = 3(1/9) = 1/3\r\n" );
document.write( "P(2 or 3 or 5) = 6x = 6(1/9) = 6/9 = 2/3 \r\n" );
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document.write( "At most 2 times means 0 times, 1 time, or 2 times. \r\n" );
document.write( "P(5) = 2x = 2(1/9) = 2/9\r\n" );
document.write( "P(not 5) = 1-2/9 = 9/9-2/9 = 7/9 \r\n" );
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document.write( "P(no times) = (15C0)(2/9)^0(7/9)^15 = (1)(1)(0.0230586012) = 0.0230586012\r\n" );
document.write( "P(1 time) = (15C1)(2/9)^1(7/9)^14 = (15)(2/9)(0.029646773) = 0.0988225767\r\n" );
document.write( "P(2 times) = (15C2)(2/9)^2(7/9)^13 = (105)(4/81)(0.0381172796) = 0.1976461533\r\n" );
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document.write( "P(rolling five 0,1, or 2 times) = sum of those = 0.3195263312\r\n" );
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document.write( "Or you can do it on a TI-83 or 84 this way:\r\n" );
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document.write( "Press 2ND VARS then scroll down to binomcdf( \r\n" );
document.write( "press ENTER        \r\n" );
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document.write( "         if you have wizard  [trials:15, p:1/9, x value: 2, \r\n" );
document.write( "         scroll to Paste]\r\n" );
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document.write( "either way you should have this:  \r\n" );
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document.write( "binomcdf(15,1/9,2)\r\n" );
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document.write( "press ENTER\r\n" );
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document.write( "see answer .3195263312\r\n" );
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document.write( "Expected value = 1(1/9)+2(2/9)+3(2/9)+4(1/9)+5(2/9)+6(1/9) =\r\n" );
document.write( "               = 1/9+4/9+6/9+4/9+10/9+6/9\r\n" );
document.write( "               = 31/4 = 7 3/4 = 7.75\r\n" );
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document.write( "Now even though you'll never expect a value of 7 3/4 ever :), \r\n" );
document.write( "if you rolled the die 15 times every day for a year, and averaged \r\n" );
document.write( "up the value you got each day, you would get an average\r\n" );
document.write( "very close to 7.75.  So \"expected value\" really means \"expected\r\n" );
document.write( "AVERAGE value if the experiment were repeated many times\".\r\n" );
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document.write( "Edwin
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