Question 1048156
You should be commended for using your own head to solve a problem, especially if you understood the situation in the problem, and knew your answer was right without needing the teacher to tell you so.
I suspect that your teacher wants you to solve the problem the same way it was shown in class for a similar problem. Sometimes teachers even want to mention certain names and formulas taught in class, maybe using the same symbols in those formulas.
 
Teleporting at least once during the day, and not teleporting at all during the day are "mutually exclusive," or "complementary,"
so the probability that he teleports at least once during a day,
plus the probability that he does not teleport at all add to {{{1}}}.
 
Can you calculate the probability that he does not teleport at all?
 
The decision to use one or another mode of transportation for the trip to work is supposed to be a "random event", so each of the 3 outcomes (train, car, and teleport) is equally probable,
so the probability of choosing to teleport is {{{p(t)=1/3}}} .
The probability that he does not teleport for the trip to work {{{p(n)}}} is the complementary event, so the probabilities of teleporting or not teleporting add to {{{1}}} , and the probability of not teleporting to work is {{{p(n)=1-1/3=2/3}}} .
The same goes for thew second trip of the day (going home),
so you could say that the probability of not teleporting on the way home is also {{{p(n)=2/3}}} .
 
The trip to work and the trip home from work are "independent events,"
so the probabilities for the "compound event" made of the two trips for a day are the product of the probabilities for each trip.
So, the probability of not teleporting at all is
{{{(2/3)*(2/3)=4/9}}} .
 
Now it is clear that the probability that he teleports at least once during a day is {{{1-4/9=highlight(5/9)}}} .