On a web page here I see the following question:
The web page provides this answer:
But my own solution is different:
Who's right? What's the mistake in the wrong solution?
P.S. And by the way, what is the purpose in stating the formula as a fraction but still using a negative exponent? Why not write that factor in the denominator with a positive power?
You've computed the probability that exactly 3 taxis arrive, but the problem is asking for the probability that at least 3 taxis to arrive, so this is $$P(X \ge 3) = 1 - P(X=0) - P(X=1) - P(X=2) = 1 - e^{-2} - e^{-2} \cdot 2 - e^{-2} \cdot \frac{2^2}{2!} = 1-5e^{-2}.$$
Writing $e^{-\mu}$ instead of $\frac{1}{e^{\mu}}$ is a matter of personal preference, although I see $e^{-\mu}$ much more often for some reason (maybe because it has less visual clutter). I would not worry about this too much.