Ineed to know how to solve this binomial distribution word problem

We can either get the probabilities of getting 4-10 arrived on time or 7-10 not on time - I'll take the latter. Since the probability of getting not on time is
1-0.48=0.52, the probability of getting 7 not on time is
(0.52)^(7)(1-0.52)^(10-7) in the expression p^(k)*(1-p)^(n-k) where p is the probability of one flight or thing being one way, k is the amount of times you want it to happen, and n is the amount of things you're considering.

In addition, the total number of these outcomes is n!/(k!(n-k)!)=10!/(7!(10-7)!)
. Plugging these in for 7, 8, and 10 and then adding them up, we get around 0.21. Since this is the chance for it not to happen, we do 1-0.21=a 0.79 chance more than 3 arrive on time

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