document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=331751>Question 331751</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For a lottery drawing, balls numbered with the digits 1-9 are placed in three<BR>\n" );
document.write( "bins and one ball is selected from each bin in order. What is the probability<BR>\n" );
document.write( "that all three digits drawn will be odd?<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 #237888 by </B><A HREF=/tutors/aboutme.mpl?userid=jrfrunner><B>jrfrunner(365)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jrfrunner><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=jrfrunner><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=237888\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you have to consider all the possibilities of the three bins (say bin#1, bin#2 and bin#3).<BR>\n" );
document.write( "---<BR>\n" );
document.write( "You need to find P(all three digits are odd/draw one from each bin)<BR>\n" );
document.write( "lets look at that<BR>\n" );
document.write( "first, acknowledge that there are a total of 5 odd numbers out of the 9<BR>\n" );
document.write( "=======<BR>\n" );
document.write( "P(all three selected are odd, by selecting one from each of 3 bins) =\r<BR>\n" );
document.write( "\n" );
document.write( "P(all 3 selections are odd/Bin#1 has 0 odd balls) =0<BR>\n" );
document.write( "in fact any bin having no odd numbers makes the prob of getting all 3 odd numbers 0, since the 3 numbers are obtained one from each bin<BR>\n" );
document.write( "----<BR>\n" );
document.write( "P(all 3 numbers are odd / bin#1 has 1 odd,bin#2 has 0 odd 1,bin#3 has 3 odd)=  this is not possible since there exists 5 odd numbers and this makes the total of 4 odd.<BR>\n" );
document.write( "P(all 3 numbers are odd / bin#1 has 1 odd,bin#2 has 1 odd,bin#3 has 3 odd)=1/3*1/3*3/3=3/27<BR>\n" );
document.write( "P(all 3 numbers are odd / bin#1 has 1,bin#2 has 2,bin#3 has 2)=1/3*2/3*2/3=4/27<BR>\n" );
document.write( "P(all 3 numbers are odd / bin#1 has 1,bin#2 has 3,bin#3 has 1)=1/3*3/3*1/3= 3/27<BR>\n" );
document.write( "---<BR>\n" );
document.write( "P(all 3 numbers are odd / bin#1 has 2 odd,bin#2 has 0 odd,bin#3 has 2 odd) = 2/3*0/3*2/3=0<BR>\n" );
document.write( "P(all 3 numbers are odd/bin#1 has 2 odd,bin#2 has 1 odd,bin#3 has 2 odd)=2/3*1/3*2/3=4/27<BR>\n" );
document.write( "P(all 3 numbers are odd/bin#1 has 2,bin#2 has 2,bin#3 has 1)=2/3*2/3*1/3=4/27<BR>\n" );
document.write( "P(all 3 numbers are odd/bin#1 has 2,bin#2 has 3,bin#3 has 0) = 2/3*3/3*0= 0<BR>\n" );
document.write( "---<BR>\n" );
document.write( "P(all 3 numbers are odd/bin#1 has 3 odd,bin#2 has 0 odd,bin#3 has 2 odd) = 3/3*0/3*2/3=0<BR>\n" );
document.write( "P(all 3 numbers are odd/bin#1 has 3 odd,bin#2 has 1 odd,bin#3 has 1 odd)=3/3*1/3*1/3=3/27<BR>\n" );
document.write( "P(all 3 numbers are odd/bin#1 has 3,bin#2 has 2,bin#3 has 0)= 3/3*2/3*0/3= 0\r<BR>\n" );
document.write( "\n" );
document.write( "So, summing all these probabilities =3/27+4/27+3/27+4/27+4/27+3/27=21/27\r<BR>\n" );
document.write( "\n" );
document.write( "P(all three selected are odd, by selecting one from each bin) =21/27 or 7/9<BR>\n" );
document.write( "---<BR>\n" );
document.write( "which is much higher than if you just put all the balls in one bin and selected 3 balls and asked the prob( all 3 are odd ) which would be 5/9*4/8*3/7=60/540=0.119\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );