document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=7042>Question 7042</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Probability problem:\r<BR>\n" );
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document.write( "There are 5 red and 4 black balls in a box. If you pick out 3 balls without replacement, what is the probability of getiing at least one red ball?\r<BR>\n" );
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document.write( "I have tried so far: P(5/9) + P(4/9) - (3/9)= (6/9)&#8776; 2/3\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #3857 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><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=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=3857\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P(at least 1 red) is a lot of work. However, thankfully, we can think of it as 1 - P(3 black), which is a lot easier to compute.\r<BR>\n" );
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document.write( "P(BBB) = (4/9)*(3/8)*(2/7) = 1/21\r<BR>\n" );
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document.write( "so P(at least 1 red) = 1 - 1/21<BR>\n" );
document.write( "P(at least 1 red) = 20/21\r<BR>\n" );
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document.write( "jon.<BR>\n" );
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