Question 790231: 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.
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! =======arraingments possible
5 0 0 ======3
4 1 0 ======6
3 1 1 ======3
3 2 0 ======6
2 2 1 ======3
----------------sum
===========21 (a)
.
b) 3+3=6
.
c)
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
d)
Let a, b and c each = 1/3 They represent the chance of a ball going into each bucket.
expand the trinomial (a+b+c)^5 The 5 represents the 5 balls.
(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.
3*a^5+6*5(a^4b)+6*10(a^3b^2)= 31/81 probability that one bucket has no ball.
1 - 31/81 = 50/81
.
Ed
|
|
|
