document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=975208>Question 975208</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the probability that 15 rolls of a fair die will show 5 threes?\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #597010 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=597010\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Probability of a 3 is (1/6)\r<BR>\n" );
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document.write( "This is a binomial with 15C5*(1/6)^5*(5/6)^10 = 0.0624\r<BR>\n" );
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document.write( "There are 15 rolls and 3003 different ways 5 threes can come up.<BR>\n" );
document.write( "The overall probability is that time (1/6)^5 times and (5/6)^10 times.\r<BR>\n" );
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document.write( "The expected value would be 2.5, so this is reasonable.\r<BR>\n" );
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document.write( "While the numbers are too small to use the normal approximation, the variance is np(1-p)=75/36<BR>\n" );
document.write( "The standard deviation would be about 1.35.\r<BR>\n" );
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document.write( "This is about 2 sd s above the mean, so again it is in the range of what one would expect.\r<BR>\n" );
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