Check-out counter at a supermarket 
A check-out counter at a supermarket will service one customer per unit time
  if there is anyone in line. Customers arrive at the line: in each unit of time, the
  probability that a single new customer arrives is $\frac13$, the probability that two arrive
  is $\frac13$, and the probability that no new customer arrives is also $\frac13$. There are
  initially three customers in line. Find the probability that the line empties before
  it has ten persons in it.

 A: It's all about scenarios. All the theories you might want to use will always end up with :
Let's call $S_n$ the number of people in the queue after n units of time. You notice from the subject that :
$S_{n+1}=S_n+X_{n+1}- (1$ if $S_n>0)$
Where $X_n$ represents the number of new people that arrive, whose distribution is described.
for example, if after 2 units of time there are 3 people in the queue, after 3 units of time you know one will have been catered for, and between 0 and 2 will have arrived, thus there is a $\frac 1 3$ of chances that there are 2,3 and 4 people.
You can notice that the probabilities after $n+1$ steps only depend on what the situtation was after $n$ steps. So, this is the start of all the theories (Markov chain, martingales...) : You know the probabilities of transition from a state to another, and you look at diverse ways of predicting what will happen after a lot of these transitions have occured.
Here, markov chains would be very useful. They would mean for you to define the 11 states you want to track, to create a (mostly empty) 11x11 matrix of probability and then to solve a linear equation on this matrix to obtain the probabilities you want. Since you don't know markov chains, I won't get into the details now.
Martingales and stopping times would do even better : if you have between 1 and 9 customers in your queue, you know that the expected value of the number at the next step is the same as the number you have at this stage. A very strong martingale result is that this goes on to infinity, so if you start with n customers, the expected value of your number of customers will always be n. Since you know that eventually you will reach 0 or 10 with probabilities of $p_n$ and $1-p_n$, you know that : 
$n=0\times p_n+10\times(1-p_n) \iff p_n=1-n/10$
Also, you need to define more precisely what you mean by "empties" : if no customer ever comes, does it count? How many people are in the queue to begin with?
If we assume the counter starts when the first customer enters the shop, there is 50% probability that he arrives alone and 50% probability that he comes with another one, combining those probabilities with the above result would give you the result of 85%
A: As other have mentioned, markov chains would be able to solve this problem. Here would be the way:
We have our 11 states, 0-10 people in the line. Since at 10 people we have "lost" so to say, we may ignore all cases which originate from having 10 people in the line. Since at 0 we have satisfied the condition, we also shall not calculate further. 
Notice that in each state, we have a third chance to jump up two states, jump up one state, or remain at the same state. However, we will also deal with one customer in the same time unit, so in reality, we have a third chance for -1,0,+1 to occur each.
Our state matrix S will reflect both these things, with the column index being the state we are in, the row index the state we will jump to, and the entry being the probability of this happening:
$ S = $
$
\begin{bmatrix}
1 && 0 && 0 &&0&&0&&0&&0&&0&&0&&0&&0 \\
1/3&&1/3&&1/3&&0&&0&&0&&0&&0&&0&&0&&0\\
0&&1/3&&1/3&&1/3&&0&&0&&0&&0&&0&&0&&0\\
0&&0&&1/3&&1/3&&1/3&&0&&0&&0&&0&&0&&0\\
\vdots \\
0&&0&&0&&0&&0&&0&&0&&0&&1/3&&1/3&&1/3\\
0&&0&&0&&0&&0&&0&&0&&0&&0&&0&&1
\end{bmatrix}
 $
We now multiply this matrix by our initial state x to get to the next state. Notice that since initially we have 3 customers in line, x will look like this (with a length of 11, 1 per state):
$x_0 = \begin{bmatrix} 0&&0&&0&&1&&0&&0&&\cdots \end{bmatrix}$
So then:
$ x_{n+1} = x \cdot S $
Follows:
$ x_{n+j} = x \cdot S^j $
I leave it up to you to take the limit as j approaches infinity to see how the system performs over a long time. Your answer will be found in the state x as a first entry, as this will be the chance that we reach 0 customers.
