document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=746935>Question 746935</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two dice are rolled.  The probability that their sum is at least 4 equals _____. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #454599 by </B><A HREF=/tutors/aboutme.mpl?userid=timvanswearingen><B>timvanswearingen(106)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=timvanswearingen><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=timvanswearingen><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=454599\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There are 6 numbers on each die so the number of possibilities is 6*6=36\r<BR>\n" );
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document.write( "Now, the probability that the sum is at least 4, would be easier found by using<BR>\n" );
document.write( " 1-P(sum is 3 or less)\r<BR>\n" );
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document.write( "Sums of 3 or less would include rolls of    1,1   1,2    2,1<BR>\n" );
document.write( "therefore 3/36 or 1/12.  \r<BR>\n" );
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document.write( "So, P(sum is 4 or more) = 1-P(sum is 3 or less) = 1 - 1/12 = 11/12<BR>\n" );
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