document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=790231>Question 790231</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have 5 identical balls and I have three buckets which are labelled A, B, C. (a) In how many different ways can I put the five balls into the buckets. (There are no constraints, e.g., a bucket can be empty.) (b) In how many different ways can I put the five balls into the buckets if each bucket must contain at least one ball? Now suppose I perform the following experiment. For each ball I pick one of the three buckets with equal probability and put the ball in that bucket. I claim that the probability that each bucket has at least one ball is 50/81. One might think this probability would be given by your answer to (b) divided by your answer to (a), which should work out to 2/7. (c) Explain why your answer to (b) divided by your answer to (a) is not the correct probability. (d) Derive my answer. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #479090 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=479090\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> =======arraingments possible<BR>\n" );
document.write( "5 0 0 ======3<BR>\n" );
document.write( "4 1 0 ======6<BR>\n" );
document.write( "3 1 1 ======3<BR>\n" );
document.write( "3 2 0 ======6<BR>\n" );
document.write( "2 2 1 ======3<BR>\n" );
document.write( "----------------sum<BR>\n" );
document.write( "===========21 (a)<BR>\n" );
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document.write( "b) 3+3=6<BR>\n" );
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document.write( "c)<BR>\n" );
document.write( "Because some arraignments are more likely than others. 5 in one bucket 3*(1/3)^5=1/81 while 3 in one and 2 in another equals 6*10*(1/3)^5 = 20/81 \r<BR>\n" );
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document.write( "d)<BR>\n" );
document.write( "Let a, b and c each = 1/3 They represent the chance of a ball going into each bucket.<BR>\n" );
document.write( "expand the trinomial (a+b+c)^5 The 5 represents the 5 balls.<BR>\n" );
document.write( "(a+b+c)^5 = a^5+5a^4b+5a^4c+10a^3b^2+20a^3bc+...This is a very long formula but you don't have to do it all because you know from part a) that there are three x^5 types, six 5x^4y types and six 10x^3y^2 types.<BR>\n" );
document.write( "3*a^5+6*5(a^4b)+6*10(a^3b^2)= 31/81 probability that one bucket has no ball.<BR>\n" );
document.write( "1 - 31/81 = 50/81<BR>\n" );
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document.write( "Ed </SPAN>\n" );
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