document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=374877>Question 374877</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If you have numbers from 1 to 999 in a bag and you pull out 10 numbers, what are the odds that 7 of those numbers will be from 700 to 999?  Is there a formula for figuring this out and if there is can you send it to me?  Thanks for your help.  Pam </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #266653 by </B><A HREF=/tutors/aboutme.mpl?userid=edjones><B>edjones(8007)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=edjones><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=edjones><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=266653\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> nCr=Combination of n things taken r at a time=n!/((n-r)!r!)<BR>\n" );
document.write( "(300C7 * 699C3)/999C10 = .0088 = 88/10,000 Probability that 7 of those numbers will be from 700 to 999.<BR>\n" );
document.write( "10000-88=9912<BR>\n" );
document.write( "Odds against 7 of those numbers will be from 700 to 999 = 9912 to 88 or 113 to one. The odds in favor are the reverse, one to 113.<BR>\n" );
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document.write( "Ed </SPAN>\n" );
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