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