document.write( "Question 1129438: Tom has three coins. Two are fair and one is an unfair coin weighted so that
\n" ); document.write( "heads is five times as likely as tails. He selects one of the coins at random
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\n" ); document.write( "If it does come up tails, what is the probability it was a fair coin?
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Algebra.Com's Answer #746015 by Edwin McCravy(20060) $\"\"$ $\"About$

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Tom has three coins. Two are fair and one is an unfair coin weighted so that
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\n" ); document.write( "\n" ); document.write( "What is the probability it comes up tails?
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document.write( "We need to first calculate the probabilities for the unfair coin:  \r\n" );
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document.write( "P(H) = 5∙P(T)\r\n" );
document.write( "P(H) + P(T)=1\r\n" );
document.write( "P(T) = 1-P(H)\r\n" );
document.write( "P(T) = 1-5∙P(T)\r\n" );
document.write( "6P(T)= 1\r\n" );
document.write( "P(T) = 1/6\r\n" );
document.write( "P(H) = 5∙(1/6)\r\n" );
document.write( "P(H) = 5/6\r\n" );
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document.write( "There are two possibilities\r\n" );
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document.write( "1. He selects a fair coin and it comes up tails\r\n" );
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document.write( "P(fair coin) = 2/3\r\n" );
document.write( "P(tails on fair coin) = 1/2\r\n" );
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document.write( "P[(selects fair AND tails on fair coin) OR (selects unfair AND tails on unfair coin)]\r\n" );
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document.write( "AND means we multiply and OR means we add.\r\n" );
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document.write( "(2/3)(1/2) + (1/3)(1/6) \r\n" );
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document.write( "That calculates to be 7/18 \r\n" );
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\n" ); document.write( "If it does come up tails, what is the probability it was a fair coin?
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document.write( "P(fair|tail) = P(fair AND tail)/P(tail) = [(2/3)(1/2)]/(7/18) = 6/7\r\n" );
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document.write( "Edwin

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