document.write( "Question 1157709: A family has 6 children. Assume that each child is as likely to be a boy as it is to be a girl. Find the
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Algebra.Com's Answer #780581 by Edwin McCravy(20056)\"\" \"About 
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A family has 6 children. Assume that each child is as likely to be a boy as it is to be a girl. Find the
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document.write( "                               P(all girls AND at least 1 girl)     P(all girls)\r\n" );
document.write( "P(all girls|at least 1 girl) = -------------------------------- = ------------------\r\n" );
document.write( "                                     P(at least 1 girl)           P(at least 1 girl)\r\n" );
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document.write( "The oldest could be a boy or girl, that's 2 ways.\r\n" );
document.write( "The next to oldest could be a boy or girl, that's 2 ways.\r\n" );
document.write( "The 3rd to oldest could be a boy or girl, that's 2 ways. \r\n" );
document.write( "The 3rd to youngest could be a boy or girl, that's 2 ways.\r\n" );
document.write( "The next to youngest could be a boy or girl, that's 2 ways.\r\n" );
document.write( "The youngest could be a boy or girl, that's 2 ways.\r\n" );
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document.write( "That's 2∙2∙2∙2∙2∙2 = 26 = 64 ways the 6 children could be.\r\n" );
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document.write( "P(all girls) = 1/64\r\n" );
document.write( "P(at least 1 girl) = 1 - P(all boys) = 1 - 1/64 = 63/64\r\n" );
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document.write( "                                  P(all girls)       1/64\r\n" );
document.write( "P(all girls|at least 1 girl) = ------------------ = ------- = (1/64)(63/64) = 1/63\r\n" );
document.write( "                               P(at least 1 girl)    63/64\r\n" );
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document.write( "Edwin
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