document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=469414>Question 469414</A>: <I><SPAN STYLE=\"word-break: break-all;\"> six circles numbered one thru 6 <BR>\n" );
document.write( "1 is yellow<BR>\n" );
document.write( "2 is green<BR>\n" );
document.write( "3 is orange<BR>\n" );
document.write( "4 is orange<BR>\n" );
document.write( "5 is orange<BR>\n" );
document.write( "6 is yellow\r<BR>\n" );
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document.write( "assume that one cirlce is selected at random and each circle is equally likely to be selected. Determine the probability of selecting a 3, given that the circle is yellow. \r<BR>\n" );
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document.write( "how would I finish this? I just don't know how to work the problem out?\r<BR>\n" );
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document.write( "P=(3|yellow) my answer is 1/6 for selecting a 3 and 0 probability for selecting a yellow.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #322065 by </B><A HREF=/tutors/aboutme.mpl?userid=ccs2011><B>ccs2011(207)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ccs2011><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=ccs2011><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=322065\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> General formula:<BR>\n" );
document.write( "P(a|b) = P(ab)/P(b)<BR>\n" );
document.write( "where a is selecting number 3 and b is the circle is yellow.<BR>\n" );
document.write( "P(ab) = P(3 is selected AND circle is yellow) = 0<BR>\n" );
document.write( "P(b) = P(circle is yellow) = 2/6 = 1/3<BR>\n" );
document.write( "P(a|b) = 0/(1/3) = 0<BR>\n" );
document.write( "Therefore, probability selecting 3 given circle is yellow is 0. </SPAN>\n" );
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