Probability of Warcraft So I have this probability exercise from Khan Academy, which is about World of Warcraft ^^

Marvin lives in Stormwind city and works as an engineer in the city of
  Ironforge. In the morning, he has the choice to teleport, ride a
  dragon, or walk to work, and in the evening he has the same three
  choices for his trip home.
If Marvin randomly chooses his method of travel in the morning and in
  the evening, what is the probability that he teleports at least once?

It is easy to calculate the result if we write the possible outcomes and simply count which have at least one teleport, but thing is I can't be sitting and drawing stuff... So what I though of is that in 3 ways of transport then we have 3x3=9 possible outcomes, but I'm interested only in those that contain teleport, so I thought why don't remove teleport out of the equation so we have 2 ways of transport which gives us 4 possible outcomes, so now that we know this, from the 3 ways of transport (which is 9) we subtract the ones of 2 (which is 4) so we end up with 5 outcomes that we care about, namely the ones that contain teleport.. so we end up with 5/9... is that ideology correct?

As a side note,  for easier calculations, whenever you have a probability problem of the form "At least once", it's usually easier to calculate the probability of doing it "no times", and then do 1 minus that, as the two events are what are called "complements",  i.e. exactly one of the two occur, with no overlap, so their probability adds to 1.
So the odds of not teleporting are $\frac 2 3$, and each time is independent, so we multiply, the odds of not teleporting twice in a row are $(\frac 2 3)^2=\frac 4 9$,  then we subtract from 1 to get $\frac 5 9$.  It's effectively what you did, but the math is way easier when you get to larger numbers
For example,  what's the odds of getting at least one heads in 10 coin flips? With this method, you just do $1-(\frac 1 2)^{10}$, and you're done
