When to use the multiplication rule in probability versus when to use a tree? So, as I have understood it, if you have two experiments and you want to know the probability of a set of two outcomes happening concurrently, then you multiply the chance of the first outcome by the chance of the second outcome and voila, you have your probability.
However, I am confused as to when this doesn't work. For example, I just did a problem:

James lives in San Francisco and works in Mountain View. In the
  morning, he has 3 transportation options (bus, cab, or train) to work,
  and in the evening he has the same 3 choices for his trip home. What is the probability that he uses the same of mode of transportation twice?

My first inclination was 1/9th but apparently I am wrong. I was told to use a tree to count the favorable outcomes. I did so, and see that the answer is 1/3, but for the life of me I can't see the difference between this question and the first type I mentioned. 
I am obviously missing some finer points or nuance in the question which should clue me in. What is it? 
 A: There are three ways to use the same transportation mode twice: Bus twice, cab twice, or train twice. So the probability of using, say, the bus twice is indeed $\frac{1}{9}$, but there is also the option of using the cab twice, or the train twice, so you have to consider those options as well.
$P(\text{same transportation twice}) = P(BB) + P(CC) + P(TT)$
$P(\text{same transportation twice}) = \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$
$P(\text{same transportation twice}) = 3 \cdot \frac{1}{9}$
$P(\text{same transportation twice}) = \frac{1}{3}$
Remember that "or" implies addition, whereas "and" implies multiplication.
A: 1/9 is the probability that a specific mode of transportation is used twice.  There are 3 different modes, so the answer is 1/9 + 1/9 +1/9 = 1/3.
A: You can just do it by counting.  You are correct that he has $3 \cdot 3=9$ choices of morning and evening transportation.  There are $3$ cases where he takes the same type both ways, giving a probability of $\frac 39=\frac 13$  
Alternately, you can reflect on the fact that whatever choice he makes in the morning, he can still match it.  In the evening he has three choices, one of which matches the morning.  Again $\frac 13$
