document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1074771>Question 1074771</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a bag contains 6 white,4 red and 4 black balls. find the probability of getting at least two red balls if 4 balls are drawn.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #689486 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=689486\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Probability of 0 red balls: (10/14)(9/13)(8/12)(7/11), since 10 of 14 are not red. This is 0.2098.<BR>\n" );
document.write( "Probability of 1 red ball:  The denominator is the same.  The numerator is 4*10*9*8*4; the first 4 is the number of ways the 1 ball can appear.  The 10,9,8 are the non-red balls, and the 4 is the number of red balls present.  The order or the numbers in the numerator can change, but multiplication is commutative.<BR>\n" );
document.write( "That probability is 0.4795<BR>\n" );
document.write( "At least 2 is 1- the sum of these two probabilities or 0.3107.<BR>\n" );
document.write( "-------------------------<BR>\n" );
document.write( "Check by doing 2, which is 6 ways 2 red balls can be chosen out of 4, and the numerator is 6*4*3*10*9=6480.<BR>\n" );
document.write( "For 3, it is 4*10*4*3*2=960<BR>\n" );
document.write( "for 4, it is 4*3*2*1=24<BR>\n" );
document.write( "That sum is 7464/(14*13*12*11)=0.3107. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );