document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1094068>Question 1094068</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that you have 9 cards. 5 are green and 4 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 4 yellow cards are numbered 1, 2, 3, and 4. The cards are well shuffled. Suppose that you randomly draw two cards, one at a time, and without replacement.<BR>\n" );
document.write( "• G1 = first card is green<BR>\n" );
document.write( "• G2 = second card is green<BR>\n" );
document.write( "Enter the probability as a fraction.<BR>\n" );
document.write( "P(at least one green) = </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #708688 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=708688\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The easiest way to solve is to first compute P(no green).  <BR>\n" );
document.write( "Then the probability of at least one green is<BR>\n" );
document.write( "P(at least one green) = 1 - P(no green)<BR>\n" );
document.write( "P(no green) = 4/9*3/8 = 12/72 = 1/6<BR>\n" );
document.write( "Thus P(at least one green) = 1 - 1/6 = 5/6 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );