document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=426302>Question 426302</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A committee of four people is to be selected from six Democrats and four Republicans. What is the probability that all are Democrats? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #296570 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=296570\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "If you consider chosing a Democrat for any one of the positions as a success (nothing to do with politics, it is simply a way of looking at the problem), then you want the probability of 4 successes in 4 trials where the probability of success on any given trial is 0.6.  I say 0.6 because there are 10 people to choose from and 6 are considered a success.\r<BR>\n" );
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document.write( "Alternatively, if your conservative politics simply won't allow you to consider selecting a Democrat as a success, you could calculate the probability of 4 failures (failure to pick a Republican) in 4 trials where the probability of success is 0.4.  You will find that it reduces to the exact same calculation.\r<BR>\n" );
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document.write( "The probability of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large k\"> successes in <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large n\"> trials where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large p\"> is the probability of success on any given trial is given by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}\">\r<BR>\n" );
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document.write( "Where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \left(n\cr k\right\)\"> is the number of combinations of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large n\"> things taken <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large k\"> at a time and is calculated by <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{n!}{k!(n\,-\,k)!}\">\r<BR>\n" );
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document.write( "So, for this problem you could either compute:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P_4(4,0.6)\ =\ \left(4\cr 4\right\)\left(0.6\right)^4\left(0.4\right)^{0}\">\r<BR>\n" );
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document.write( "or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P_4(0,0.4)\ =\ \left(4\cr 0\right\)\left(0.4\right)^0\left(0.6\right)^{4}\">\r<BR>\n" );
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document.write( "And remembering that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \left(n\cr n\right\)\ =\ \left(n\cr 0\right\)\ =\ 1\"> and  <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x^0\ =\ 1\">,\r<BR>\n" );
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document.write( "So either way, it reduces to\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P_4(4,0.6)\ =\ P_4(0,0.4)\ =\ \left(0.6\right)^{4}\">\r<BR>\n" );
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document.write( "You can do your own arithmetic.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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