document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=847210>Question 847210</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two dice are thrown once. Find the probability of getting a multiple of 3 on the first die or a total of 8 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #510270 by </B><A HREF=/tutors/aboutme.mpl?userid=swincher4391><B>swincher4391(1107)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=swincher4391><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=swincher4391><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=510270\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I apologize for last time. I made a slight error. \r<BR>\n" );
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document.write( "P[multiple of 3] = 1/3  =  12/36<BR>\n" );
document.write( "P[sum of 8] =  5/36\r<BR>\n" );
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document.write( "However as I said last time (3,5) is not the only roll that satisfies both, we also have (6,2). This means we need to take in account the 2 cases out of 36 where both events take place. \r<BR>\n" );
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document.write( "So we calculate  12/36 + 5/36 - 2/36 =   5/12  which is our answer. </SPAN>\n" );
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