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Q: probability for 4 balls thrown into 4 boxes ( Answered 5 out of 5 stars,   0 Comments )
Subject: probability for 4 balls thrown into 4 boxes
Category: Science > Math
Asked by: emmag999-ga
List Price: $25.00
Posted: 29 Apr 2006 12:05 PDT
Expires: 29 May 2006 12:05 PDT
Question ID: 723962
Have been going through past exam papers in statistics and have got
completely stuck on this one.

Four balls are thrown at random into 4 boxes, with each ball having an
equal chance of going into every box. Work out the probability that
exactly m boxes are empty, where m = 0,1,2,3.

I managed m=0 and m=3 fine, and I have managed to work the other 2 out
the long way round by drawing it all out, but the answer given in the
exam report states probability that 1 box is empty is 12 x
[4!/(2!1!1!0!)]x 1/4squared x 1/4 x 1/4 = 36/64 and I have no idea how
they got this. Please help?
Subject: Re: probability for 4 balls thrown into 4 boxes
Answered By: leapinglizard-ga on 30 Apr 2006 02:50 PDT
Rated:5 out of 5 stars
Dear Emma,

Although in principle you can solve problems of this kind by enumerating
all possibilities on paper, in practice it is unlikely that an exam will
permit you enough time to do so. It is better to compute the probability
by an analysis that minimizes counting.

Among the various analytical methods one may employ to arrive at the
answer, the simplest begins by computing the total number of ways
there are to distribute four distinct balls among four distinct boxes.

Observe that we can place the first ball into one of four boxes, the
second ball again into one of four, the third into one of four, and the
fourth into one of four. This makes for a total of

	4*4*4*4  =  256


Next, we compute the number of cases in which exactly one box
is empty. Dividing this number by 256 will give us the desired
probability. There are four factors to consider.

1.  If exactly one box is empty, there are four possibilities as to
which one it is.

	factor: 4

2. Of the remaining three boxes, exactly one must contain two balls. This
gives a further three possibilities.

	factor: 3

3. How many possibilities are there for the two balls that go into one
box? This number is written C(4, 2), pronounced "4 choose 2", and is
conventionally computed as

	 4!      4*3*2*1     4*3
	----  =  -------  =  ---  =  6 .
	2!2!     2*1*2*1     2*1

This just says that we pick one of four balls and then one of three,
for a total of 4*3 = 12 permutations. But half of those permutations are
mirror images -- the same two balls in reverse order -- so we divide by
two, giving us 12/2 = 6 combinations.

	factor: 6

4. Finally, we consider the two remaining balls in their two individual
boxes. There are exactly two ways to order these two balls: left to right,
and right to left.

	factor: 2

We multiply the four factors to find that there are

	4*3*6*2  =  12*12  =  144

cases in which exactly one box is empty. Dividing by the total number
of cases, we obtain a probability of

        144      72     36     18     9
        ---  =  ---  =  --  =  --  = -- .
        256     128     64     32    16

So 36/64 is one way of expressing the probability, but the simplest
fraction is actually 9/16.


emmag999-ga rated this answer:5 out of 5 stars

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