document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1183890>Question 1183890</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Urn 1 contains 2 white, 3 blue, and 4 red balls. Urn 2 contains 3 white and 5 blue balls. A ball is randomly drawn from urn 1, without replacement,<BR>\n" );
document.write( "and then placed inside urn 2. A ball is then drawn randomly from urn 2 and placed back in urn 1. If a ball is to be randomly drawn from urn 1,<BR>\n" );
document.write( "what is the probability that the ball is red? Is there a change in the probability of red from before the experiment?  How would you explain your answer? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #814432 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=814432\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I have solved a similar problem earlier.  This other problem is different because urn 1 has extra 4 red balls.  The reasoning is similar, though, and \r<BR>\n" );
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document.write( "this problem should be EASY, although might take some time in sorting out the appropriate probabilities.\r<BR>\n" );
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document.write( "P(wwr) = (2/9)*(4/9)*(4/9) = 32/729<BR>\n" );
document.write( "P(wbr) = (2/9)*(5/9)*(4/9) = 40/729<BR>\n" );
document.write( "P(bwr) = (3/9)*(3/9)*(4/9) = 36/729<BR>\n" );
document.write( "P(bbr) = (3/9)*(6/9)*(4/9) = 72/729<BR>\n" );
document.write( "P(rwr) = (4/9)*(3/9)*(3/9) = 36/729<BR>\n" );
document.write( "P(rbr) = (4/9)*(5/9)*(3/9) = 60/729<BR>\n" );
document.write( "P(rrr) = (4/9)*(1/9)*(4/9) = 16/729\r<BR>\n" );
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document.write( "Hence. P(red 2nd time around) = sum of all preceding probabilities = 292/729 ~ 0.40, to 2 d.p.\r<BR>\n" );
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document.write( "The probability of 0.40 in drawing red the 2nd time around is LOWER than the probability of red (4/9 or 0.44 to 2 d.p.) in the first draw.\r<BR>\n" );
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document.write( "The lower probability comes from the fact that the first draw could be a red, but that same red may not be picked from the 2nd urn, <BR>\n" );
document.write( "thereby lowering the number of red balls in urn 1. </SPAN>\n" );
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