A bag has 3 red, 3 blue and 3 green balls. 2 balls are picked at random without replacement. What is the probability that they are of the same colour?
I am thinking what is the easiest way to do this using sampling? The sample space has $72$ outcomes (ordered without replacement). Then for 1st ball, we have ${9}\choose{1}$ choices; since we want 2nd ball same colour as 1st, we need to use ${2}\choose{1}$, and hence we have the probability to be $\frac{9 \times 2}{72} = \frac{1}{4}$.
Is this the fastest way using ordered sampling?
Also the next option can be:
$$ \frac{\binom{3}{2}+\binom{3}{2}+\binom{3}{2}}{\binom{9}{2}}=\frac{1}{4}. $$