Probability question : meaning of the sentence The following is the problem I am working on.

The probability of a passing car being an import is defined as $p(i)=1/4$ and the probability of it being domestic is $p(d)=3/4$. Find the probability that there will be at least 2 imports passing before the 3rd domestic car.

I am not quite understanding what they mean by the "3rd" domestic car.
Opinion 1), The probability I am looking for is $$p(\text{2i's 2ds and last d})+p(\text{3i's 2d's and last d})+\cdots p(\text{n i's 2d's and last d}))+\cdots$$
Opinion 2), The "3rd" car is domestic
$$p(\text{ddd})+p(\text{idd})+p(\text{did})+\cdots+p(()()d()()()\cdots)$$
But the book that I am working on explains as the only cases are
$$iidd,idid,iddi,ddid,didi,ddii, iiid,iidi,idii, ddii, iiii$$
I am very confused because first of all, i don't see a "3rd" domestic car in any of these cases and why is it assuming that there are exactly four cars passing by ?
Can someone explain me what is going on ?
 A: While you've already selected @drhab's answer and he gives excellent advice, I wanted to highlight the negative binomial distribution as a different way to approach the problem (and one that gives answers for numbers greater than 2nd import and 3rd domestic).
Consider the example of drawing red or black balls from a bucket (with replacement, so that the probabilities don't change). As you draw a ball out of the bucket, you record its color. Similar to your example, you continue this until the third black ball is encountered and then count the number of red balls.
When the number of balls drawn $n$ is chosen beforehand, you have the binomial distribution.
On the other hand, drawing until you see the first black ball is the geometric distribution.
Drawing until you see a given number $r$ of black balls is the negative binomial distribution. Notice that the geometric distribution is a special case with $r=1$.
For your question, let the random variable $X \sim NB(r=3, p=\frac{1}{4})$ be the number of imports seen (imports will be the red balls from above). Using formulas from Wikipedia and @drhab's advice to find the complement, we can find the answer to your question:
$$
\begin{align*}
P(X \ge 2)&=1 - \sum_{k=0}^1{P(X=k)} \\
&= 1 - \binom{2}{0}(\frac{3}{4})^3(\frac{1}{4})^0 -\binom{3}{1}(\frac{3}{4})^3(\frac{1}{4})^1 \\
&=1 - \frac{27}{64} - \frac{81}{256} \\
&=\frac{67}{256}
\end{align*}
$$
A: Just like you I am confused about the cases mentioned by the book. I tend to believe in opinion 1) and a solution for that problem is: 
Looking at the complement (i.e. the event that less than $2$ imports
pass before the $3$-rd domestic does) there are $4$ possibilities: $ddd$,
$iddd$, $didd$ and $ddid$
This with probability $\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}=\frac{27}{64}$
for $ddd$ and probability $\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4}=\frac{27}{256}$
for the others. Then there is a probability of $\frac{27}{64}+3\times\frac{27}{256}=\frac{189}{256}$
for the complement, hence a probability of $\frac{67}{256}$ for the
event.
It is a good custom to look at the complement first whenever you meet expressions like 'at least'.
