Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives

Arranging Marbles in Boxes

Date: 09/16/98 at 19:38:20
From: Chris
Subject: Probability

probability that each box contains two marbles? What is the formula 
used for this type of problem?

Date: 09/17/98 at 11:21:12
From: Doctor Anthony
Subject: Re: Probability

You can model the situation as follows:

    |****|  |**|      

where the |  | represent boxes and the * are marbles.

In the above diagram we have 4, 0, and 2 marbles in the three boxes, 

Now this diagram must begin and end with a  |  but otherwise the two 
remaining |'s and six *'s can be arranged in any order, and each of 
these orders is considered equally likely.

So we have 8 objects to arrange in every possible way, 2 of one kind 
being alike and 6 of a second kind being alike. This can be done in:

   ------  = C(8,2) = 28 ways.

[ If you haven't done permutations and combinations yet you should 
obtain a textbook on elementary probability and read up on the topic.  
It is essential knowledge for this type of question. There is also some 
information in this archive's FAQ to get you started.

  http://mathforum.org/dr.math/faq/faq.comb.perm.html    ]

So there are 28 possible arrangements of the marbles in the 3 boxes 
and only one of these will have two marbles in each box, so the 
required probability is:


- Doctor Anthony, The Math Forum

Date: 03/25/2014 at 14:15:53
From: Chris
Subject: Comments about "Arranging Marbles in Boxes," by Dr. Anthony

To my surprise, the answer given above is wrong.

That's because each possible distribution (number of marbles in each box)
is not equally likely.

Regard each marble as being distinct. Then there are 3^6 = 729 ways of
distributing the six marbles into the three boxes.

There are C(6,2) = 15 ways of selecting two of the marbles and placing
them in box 1.

For each of those, there are C(4,2) = 6 ways of selecting two marbles for
box 2.

And for each of those combinations, there is C(2,2) = 1 way of selecting
the remaining two marbles.

All together, that gives the probability of two marbles in each box not as
1/28, but rather as

   15*6*1 / 3^6 = 90/729 = 10/81

C(6,2) C(4,2) C(2,2) can be written more neatly as 6!/(2! 2! 2!).

Date: 03/25/2014 at 14:45:25
From: Doctor Anthony
Subject: Re: Comments about 

Yes, I realized this some ago. I had asked the administrators to correct
this mistake -- but your comment above provides at least as good a
starting point ... or a re-starting point!

We can only deal with equi-probable events if we're dividing number of
successes by total number of possibilities. As you said, these are not
equi-probable, because there are more ways of getting some distributions
rather than others.

In a general case of distributing, say, 6 identical balls into 3 boxes,
the distribution 3, 2, 1 could occur in a variety of orders, whereas 
6, 0, 0 could occur in only 1 order, and is much less likely.

So the diagram |***| ** | * | is not applicable except if you want the
number of DIFFERENT ways, but not if you want the probabilities where only
equi-probable events should be used.

To recap, the original problem was:

   probability that each box contains two marbles? What is the formula
   used for this type of problem?

For this type of problem, we MUST treat the 6 balls as distinct. This is
OK provided we treat both numerator and denominator using distinct balls.

The number of unrestricted ways of placing the 6 balls is 3^6.

The number of ways of dividing 6 DISTINCT objects into 3 sets of 2 is

   ---------- = 90
    2! 2! 2!
So required probability is  -----  =  10/81

Thank you for your feedback!

- Doctor Anthony, The Math Forum
Associated Topics:
High School Permutations and Combinations
High School Probability

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:


parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2015 The Math Forum