document.write( "Question 487719: What is the probability of obtaining exacty seven heads in eight flips of a coin, given that at least one is a head. \n" ); document.write( "

Algebra.Com's Answer #333115 by Edwin McCravy(20055) $\"\"$ $\"About$

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What is the probability of obtaining exacty seven heads in eight flips of a coin, given that at least one is a head.
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document.write( "P(7 heads|1 or more is a head) = P(7 heads AND 1 or more is a head)/P(1 or more is a head)\r\n" );
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document.write( "First we find the numerator:\r\n" );
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document.write( "P(7 heads AND 1 or more is a head)\r\n" );
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document.write( "The \"AND 1 or more is a head\" is redundant because if you have 7 heads you automatically have 1 or more heads,\r\n" );
document.write( "so we just need the probability of 7 heads (out of 8 trials), which is\r\n" );
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document.write( "8C7(1/2)^7(1/2)1 = 8(1/2)^8 = 8(1/256) = 8/256 (we could reduce that but\r\n" );
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document.write( "we'll just leave it 8/256.\r\n" );
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document.write( "Next we find the denominator:\r\n" );
document.write( "P(1 or more is a head)\r\n" );
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document.write( "We get the probability of the complement event which is to get no heads, which is\r\n" );
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document.write( "8C0(1/2)^0(1/2)^8 = 1(1/2)^8 = (1/2)^8 = 1/256\r\n" );
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document.write( "So the probability of the complement event is (1/2)^8, so the probability of the\r\n" );
document.write( "event is 1-1/256 = 256/256-1/256 = 255/256\r\n" );
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document.write( "Now we go back to the original:\r\n" );
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document.write( "P(7 heads|1 or more is a head) = P(7 heads AND 1 or more is a head)/P(1 or more is a head)\r\n" );
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document.write( "(8/256)/(255/256) = (8/256)*(256/255) = 8/255\r\n" );
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document.write( "Edwin

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