document.write( "Question 1203658: Assume the probability of a boy being born is the same as the probability of a girl being born.
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Algebra.Com's Answer #839408 by math_helper(2461)\"\" \"About 
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document.write( "There are two possible outcomes on each birth, boy (B) or girl (G).  So for 4 children, the number of total possible outcomes is \"2%5E4\" = \"16\"\r
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document.write( "This is small enough to enumerate all of them.  After that, you
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document.write( "should be able to easily complete the problems (I'll do the first
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document.write( "part as a template).\r
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document.write( "1  BBBB
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document.write( "2  BBBG
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document.write( "3  BBGB
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document.write( "4  BBGG
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document.write( "5  BGBB
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document.write( "6  BGBG
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document.write( "7  BGGB
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document.write( "8  BGGG
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document.write( "9  GBBB
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document.write( "10 GBBG
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document.write( "11 GBGB
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document.write( "12 GBGG
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document.write( "13 GGBB
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document.write( "14 GGBG
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document.write( "15 GGGB
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document.write( "16 GGGG\r
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document.write( "(1) P(two boys and two girls) = 6/16 = 3/8 
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document.write( "To see this, notice the number of outcomes with 2B and 2G is 6 (lines 4,6,7,10,11, and 13).  The probability is the count of the outcomes of interest, divided by the total number of possible outcomes.total possible outcomes.\r
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document.write( "As an Algebraic solution, you would use combinations.  The probability of 2 boys and 2 girls is equivalent to the probability of 2 boys out of 4 children total: 
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document.write( "    Using nCr = (n!)/((n-r)!r!)  (\"n choose r\")  
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document.write( "  P(2 boys and 2 girls) = P(2 boys out of 4 children) = 4C2 / 16 = 6/16 = 3/8
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document.write( "The other parts are solved similarly, just adjust how you select the outcomes of interest (or compute the combinations).   \n" );
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