I would use binomial distribution to answer this, since a flight being late is an independent event.
So X ~ B(16, 0.2)
The above line states that the probability of X relates to the binomial distribution with 16 trials (the 16 flights) and a 20% (0.2) chance of "success" (in this case the "success" is beimg late, which is weird because being late is not a very good thing, but the question is about being late so allow it)
So the probability of 2 flights arriving late is P(X=2) = 16C2 × 0.2^2 × 0.8^14
(The formula is basically, trialsCsuccesses × (chance of success)^(number of successes) × (chance of failure)^(total trials - successes)
The calculation will give you a probability of 0.211. And that is it :)
