document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1122708>Question 1122708</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose S={a,b,c,d,e} . <BR>\n" );
document.write( "If a three-element subset of S is selected at random, what is the probability it will contain one but not both of the elements b and c? <BR>\n" );
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document.write( "Enter your answer as a whole number or a fraction in lowest terms. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #738855 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=738855\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it can contain b or c, but not both.\r<BR>\n" );
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document.write( "i get 6/10 using brute force and then confirmed by formula.\r<BR>\n" );
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document.write( "brute force method.\r<BR>\n" );
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document.write( "a,b,c,d,e\r<BR>\n" );
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document.write( "sets of 3 are:\r<BR>\n" );
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document.write( "a,b,c<BR>\n" );
document.write( "a,b,d<BR>\n" );
document.write( "a,b,e<BR>\n" );
document.write( "a,c,d<BR>\n" );
document.write( "a,c,e<BR>\n" );
document.write( "a,d,e<BR>\n" );
document.write( "b,c,d<BR>\n" );
document.write( "b,c,e<BR>\n" );
document.write( "b,d,e<BR>\n" );
document.write( "c,d,e\r<BR>\n" );
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document.write( "that's 10 possible choices of getting 3 elements out of 5 where order is not important.\r<BR>\n" );
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document.write( "out of those, remove the ones that have both c and c in them.\r<BR>\n" );
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document.write( "those are:\r<BR>\n" );
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document.write( "a,b,c *****<BR>\n" );
document.write( "a,b,d<BR>\n" );
document.write( "a,b,e<BR>\n" );
document.write( "a,c,d<BR>\n" );
document.write( "a,c,e<BR>\n" );
document.write( "a,d,e<BR>\n" );
document.write( "b,c,d *****<BR>\n" );
document.write( "b,c,e *****<BR>\n" );
document.write( "b,d,e<BR>\n" );
document.write( "c,d,e\r<BR>\n" );
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document.write( "you will be left with:\r<BR>\n" );
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document.write( "a,b,d<BR>\n" );
document.write( "a,b,e<BR>\n" );
document.write( "a,c,d<BR>\n" );
document.write( "a,c,e<BR>\n" );
document.write( "a,d,e<BR>\n" );
document.write( "b,d,e<BR>\n" );
document.write( "c,d,e\r<BR>\n" );
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document.write( "now remove the ones that don't have b or don't have c in them.\r<BR>\n" );
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document.write( "the are:\r<BR>\n" );
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document.write( "a,b,d<BR>\n" );
document.write( "a,b,e<BR>\n" );
document.write( "a,c,d<BR>\n" );
document.write( "a,c,e<BR>\n" );
document.write( "a,d,e *****<BR>\n" );
document.write( "b,d,e<BR>\n" );
document.write( "c,d,e\r<BR>\n" );
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document.write( "you will be left with:\r<BR>\n" );
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document.write( "a,b,d<BR>\n" );
document.write( "a,b,e<BR>\n" );
document.write( "a,c,d<BR>\n" );
document.write( "a,c,e<BR>\n" );
document.write( "b,d,e<BR>\n" );
document.write( "c,d,e\r<BR>\n" );
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document.write( "that's 6 out of the original 10 possibilities.\r<BR>\n" );
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document.write( "therefore, the probability you will get b or c, but not both, is 6 out of 10.\r<BR>\n" );
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document.write( "by formula, you can do it a couple of ways.\r<BR>\n" );
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document.write( "the first way is:\r<BR>\n" );
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document.write( "probability of getting b only would be 1/5 * 3/4 * 2/3 * c(3,1) = 6/60 * 3 = 18/60\r<BR>\n" );
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document.write( "probability of getting c only would be 1/5 * 3/4 * 2/3 * c)3,1) = 6/60 * 3 = 18/60\r<BR>\n" );
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document.write( "the probability of getting b or c only would therefore be 18/60 + 18/60 = 36/60 = 6/10.\r<BR>\n" );
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document.write( "the formula works as follows;\r<BR>\n" );
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document.write( "the probability of getting b only on the first draw is 1/5.<BR>\n" );
document.write( "the probability of not getting c on the second draw is 3/4.<BR>\n" );
document.write( "the probability of not getting c on the third draw is 2/3.<BR>\n" );
document.write( "that gets you a probaiblity of 6/60 = 1/10.<BR>\n" );
document.write( "this can happen in 3 ways, i.e. b on the first draw, b on the second draw, b on the third draw, therefore 1/10 * 3 = 3/10.\r<BR>\n" );
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document.write( "same thing happens with c.\r<BR>\n" );
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document.write( "the other way is to use combination theory to get:\r<BR>\n" );
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document.write( "the number of ways you can get b only out of 3 is c(3,1).<BR>\n" );
document.write( "the number of ways you can get 3 out of 5 is c(5,3).\r<BR>\n" );
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document.write( "the probability is therefore c(3,1) / c(5,3) = 3/10\r<BR>\n" );
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document.write( "same probability to get c only.\r<BR>\n" );
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document.write( "probability of b or c is therefore 3/10 + 3/10 = 6/10\r<BR>\n" );
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document.write( "the c(3,1) means 3 ways to get b only.<BR>\n" );
document.write( "you would get b,x,x, or x,b,x or x,x,b<BR>\n" );
document.write( "same goes for c.\r<BR>\n" );
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document.write( "c(n,x) = n! / (x! * (n-x)!)\r<BR>\n" );
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document.write( "c(3,1) is therefore 3! / (1!*2!) = (3*2*1) / (1*2*1) = 3\r<BR>\n" );
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document.write( "c(5,3) is therefore 5! / (3! * 2!) = (5*4*3!) / (3!*2*1) = (5*4)/2 = 10\r<BR>\n" );
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document.write( "all 3 ways gets you the same result.\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );