document.write( "Question 189484: Four salesmen play \"odd man out\" to see who pays for lunch. They each flip a coin, and if there is a salesman whose coin doesn't match the others he pays for lunch. (For instance, he might get heads while the other three get tails.) What is the probability that there is an \"odd man\" the first time they flip? \n" ); document.write( "
Algebra.Com's Answer #142207 by Edwin McCravy(20086)\"\" \"About 
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Four salesmen play \"odd man out\" to see who pays for lunch. They each flip a
\n" ); document.write( "coin, and if there is a salesman whose coin doesn't match the others he pays
\n" ); document.write( "for lunch. (For instance, he might get heads while the other three get tails.)
\n" ); document.write( "What is the probability that there is an \"odd man\" the first time they flip?
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document.write( "One way to do it is by listing all 16 possible cases:\r\n" );
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document.write( "        1st   2nd   3rd   4th     odd\r\n" );
document.write( "        man   man   man   man     man\r\n" );
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document.write( "case  1  H     H     H     H      NONE\r\n" );
document.write( "case  2  H     H     H     T    4th man\r\n" );
document.write( "case  3  H     H     T     H    3rd man\r\n" );
document.write( "case  4  H     H     T     T      NONE\r\n" );
document.write( "case  5  H     T     H     H    2nd man \r\n" );
document.write( "case  6  H     T     H     T      NONE\r\n" );
document.write( "case  7  H     T     T     H      NONE\r\n" );
document.write( "case  8  H     T     T     T    1st man \r\n" );
document.write( "case  9  T     H     H     H    1st man\r\n" );
document.write( "case 10  T     H     H     T      NONE\r\n" );
document.write( "case 11  T     H     T     H      NONE\r\n" );
document.write( "case 12  T     H     T     T    2nd man  \r\n" );
document.write( "case 13  T     T     H     H      NONE\r\n" );
document.write( "case 14  T     T     H     T    3rd man \r\n" );
document.write( "case 15  T     T     T     H    4th man \r\n" );
document.write( "case 16  T     T     T     T      NONE\r\n" );
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document.write( "Then we count and find there are 8 out of the 16\r\n" );
document.write( "cases in which there is an odd man.  So the\r\n" );
document.write( "probability is \"8%2F16\" or \"1%2F2\".\r\n" );
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document.write( "Method 2.  There are 4 coins. \r\n" );
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document.write( "We can choose 3 to assign\r\n" );
document.write( "heads to and choose 1 to assign tails to \r\n" );
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document.write( "OR\r\n" );
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document.write( "We can choose 3 to assign\r\n" );
document.write( "tails to and choose 1 to assign heads to\r\n" );
document.write( "OR\r\n" );
document.write( "We can choose 3 to assign\r\n" );
document.write( "heads to and choose 1 to assign tails to\r\n" );
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document.write( "OR indicates addition:\r\n" );
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document.write( "So there are (4C3)(4C1)+(4C3)(4C1) = (4)(1)+(4)(1) = 4+4 = 8\r\n" );
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document.write( "That's the numerator of the probability.\r\n" );
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document.write( "Then for the denominator, which is all the ways\r\n" );
document.write( "the coins can be tossed.\r\n" );
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document.write( "We can choose heads or tails for the first one 2 ways\r\n" );
document.write( "AND\r\n" );
document.write( "We can choose heads or tails for the second one 2 ways\r\n" );
document.write( "AND\r\n" );
document.write( "We can choose heads or tails for the third one 2 ways\r\n" );
document.write( "AND\r\n" );
document.write( "We can choose heads or tails for the fourth one 2 ways \r\n" );
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document.write( "AND indicated multiplication.\r\n" );
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document.write( "So that's (2)(2)(2)(2) = 16 ways\r\n" );
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document.write( "That's the denominator of the probability.\r\n" );
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document.write( "So that also gives us 8 out of 16 or \"8%2F16\" or \"1%2F2\"\r\n" );
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document.write( "You can also do this problem by a tree diagram.\r\n" );
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document.write( "Edwin
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