document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1061544>Question 1061544</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Pease help me again:  A die is rolled 3 times.  What is the probability of?<BR>\n" );
document.write( "a.  no three turning up?<BR>\n" );
document.write( "b.  1 three?<BR>\n" );
document.write( "c.  3 three?\r<BR>\n" );
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document.write( "Thanks again.. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #676468 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=676468\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You use the binomial probability distribution<BR>\n" );
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document.write( "a) probability(Pr) of rolling a 3 on one roll is 1/6, therefore<BR>\n" );
document.write( "Pr ( k = 0 ) = 3C0 * (1/6)^0 * (5/6)^(3-0)<BR>\n" );
document.write( "Pr ( k = 0 ) = 1 * 1 * (5/6)^3 = 125/216 = 0.5787<BR>\n" );
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document.write( "b) Pr ( k = 1 ) = 3C1 * (1/6)^1 * (5/6)^(3-1)<BR>\n" );
document.write( "Pr ( k = 1 ) = 3 * (1/6) * (25/36) = 0.3472<BR>\n" );
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document.write( "c) Pr ( k = 3 ) = 3C3 * (1/6)^3 * (5/6)^(3-3)<BR>\n" );
document.write( "Pr ( k = 3 ) = 1 * (1/6)^3 * 1 = 1/216 = 0.0046<BR>\n" );
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document.write( "Note 3Ck = 3! / (k! * (3-k)!<BR>\n" );
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