document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=254695>Question 254695</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A bag contains 7 red balls numbered 1,2,4,5,6,7 and 3 white balls numbered 3,8,9. If a ball is drawn at random, what is the probability the ball is:\r<BR>\n" );
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document.write( "A. red or odd numbered?<BR>\n" );
document.write( "B. white or even numbered> </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #187027 by </B><A HREF=/tutors/aboutme.mpl?userid=drk><B>drk(1908)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drk><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=187027\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If we look at the given information, there are 6 red and 3 white, not 7 and 3. I will use the 6 and 3 numbering:<BR>\n" );
document.write( "(i) P(red or odd) means <BR>\n" );
document.write( "(ii) P(red) + p(odd)<BR>\n" );
document.write( "P(red) = 6/9<BR>\n" );
document.write( "P(odd) = 5/9<BR>\n" );
document.write( "however, some red are also odd and need to be subtracted, so we adjust (ii) to get<BR>\n" );
document.write( "(iii) P(red) + p(odd) - p(red and odd).<BR>\n" );
document.write( "So, we get<BR>\n" );
document.write( "P(red) + p(odd) - p(red AND odd) = 6/9 + 5/9 - 3/9 = 8/9<BR>\n" );
document.write( "--<BR>\n" );
document.write( "(B) P(white or even) = P(white) + p(even) - P(white AND even)<BR>\n" );
document.write( "P(white or even) = 3/9 + 4/9 - 1/9 = 6/9 = 2/3 </SPAN>\n" );
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