Problem in "Probability - An Introduction" by Grimmet and Wesh A fair die having two faces coloured blue, two red and two green, is thrown repeatedly. Find
the probability that not all colours occur in the first $k$ throws.
My attempt:
Denote by $G_k$ the number of results=green in $k-$tosses, and similarly $R_k$ and $B_k$. Then we have $\mathbb{P}(G_k=0)=(2/3)^k=\mathbb{P}(R_k=0)=\mathbb{P}(B_k=0)$. In this way the answer is $\mathbb{P}(G_kR_kB_k=0)=(2/3)^{3k}$.
 A: We use Inclusion/Exclusion.
The probability that blue does not occur in the $k$ throws is $(2/3)^k$. The same is true for red, and for green. So our first estimate is that the probability some colour does not occur is $3(2/3)^k$.
However, this double-counts the situations where blue and red does not occur, also the situations where blue and green does not occur, also red  and green. Each of these monochromatic situations has probability $(1/3)^k$.
Thus our required probability is $3(2/3)^k-3(1/3)^k$.
A: You can solve also in this way through the simple principles of combinatorial calculus. Consider three cases:
$1)$
How many are the possible combinations where the blue in these combinations isn't present and the yellow is present iff the green is it? 
There are $$\frac {2^k-1}{3^k}$$ combinations
$2)$
How many are the possible combinations where the yellow in these combinations isn't present and the green is present iff the blue is it? 
There are $$\frac {2^k-1}{3^k}$$ combinations 
$3)$
How many are the possible combinations where the green in these combinations isn't present and the blue is present iff the yellow is it? 
There are $$\frac {2^k-1}{3^k}$$ combinations 
We are done.
