Problem 627

Solution 1:

The probability that at least one colour is not used is equal to 1
minus the probability that all three colours are used.

Each of the three doors can be painted one of three different colours,
and so there are $3\times3\times3=27$
different ways to paint the doors with no restrictions.

If all $3$ colours are used, then the
first door can be painted $3$
different colours, the second door can be painted $2$ different colours, and the third door
must be painted the $1$ remaining
colour.

The probability that all three colours are used is thus $\dfrac{3\times2\times1}{27}=\dfrac{6}{27}$,
and so the probability that at least one colour is not used is $1-\dfrac{6}{27}=\dfrac{27}{27}-\dfrac{6}{27}=\dfrac{21}{27}$,
which is equal to $\dfrac{7}{9}$.

Solution 2:

The probability that at least one colour is not used is equivalent to
the probability that at most two different colours are used.

Thus, the probability that at least one colour is not used is equal to
the sum of the probability that exactly one colour is used and the
probability that exactly two different colours are used.

We begin by determining the probability that exactly one colour is
used.

Each of the three doors can be painted one of three different colours,
and so there are $3\times3\times3=27$
different ways to paint the doors with no restrictions.

If exactly one colour is used, there are 3 choices for the colour and
each door must be painted with that colour, and so the probability that
exactly one colour is used is $\dfrac{3}{27}$.

If exactly two colours are used, then two doors are painted the same
colour and the remaining door is painted a different colour.

There are $3$ choices for the first
colour and then $2$ choices for the
second colour, and thus $3\times2=6$
ways to choose the two colours. Once the two colours are chosen, the
doors can be painted with these colours in $3$ different ways. For example, if the
colour chosen for two of the doors is black (B) and the colour for the
remaining door is white (W), the doors can be painted BBW, BWB or
WBB.

Therefore, the number of ways to paint the doors with exactly two
different colours is $6\times3=18$,
and so the probability that exactly two different colours are used is
$\dfrac{18}{27}$.

Finally, the probability that at least one colour is not used is
$\dfrac{3}{27}+\dfrac{18}{27}=\dfrac{21}{27}=\dfrac{7}{9}$.