Probability Proportion Question

Hello I am having a bit of difficulty understanding how to go about this question.

A very large airline claims that 90% of their flight departures are on time.
If in fact the airline's claim is true, what is the exact probability that in a random sample of 8 flights that there would be as few as 6 departures on time.

What I did to try and solve this is I multiplied the percent of departures on time for this sample (.75) by the chance that the flights do NOT depart on time (.10). Am I on the right path? Any help is appreciated.
This is a Binomial Distribution

\( {n \choose k}{p^k}{(1-p)}^{n-k}\)

where n is the number of trials = 8
k is the number of successes = 6
p is the probability of success = .9
1 - p is the probability of failure = .1

\( {8 \choose 6}*{(.9)^6}*{(.1)}^{2}\)

= \( {\ 28}*{0.5314}*{.01}\)

\(\approx 0.1488\)

This is probability exactly six flights are delayed.

If you want the probability 6 or fewer, you could sum P(0), P(1) ... P(6)

or 1 - ( P(7) + P(8) )