(a) The probability that the flight does not arrive late is the sum of the probabilities that it arrives early or on time:

\(P( ext{not late}) = 0.15 + 0.55 = 0.7\)

The probability that on each of 3 days the flight does not arrive late is:

\((0.7)^3 = 0.343\)

(b) We need to find the probability that the flight arrives early at least 3 times out of 9 days. This is a binomial probability problem where \(n = 9\), \(p = 0.15\), and we want \(P(X \geq 3)\).

First, calculate \(P(X < 3)\):

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:

\(P(X = 0) = \binom{9}{0} (0.15)^0 (0.85)^9 = 0.231617\)

\(P(X = 1) = \binom{9}{1} (0.15)^1 (0.85)^8 = 0.367862\)

\(P(X = 2) = \binom{9}{2} (0.15)^2 (0.85)^7 = 0.259667\)

Thus,

\(P(X < 3) = 0.231617 + 0.367862 + 0.259667 = 0.859146\)

Therefore,

\(P(X \geq 3) = 1 - P(X < 3) = 1 - 0.859146 = 0.140854\)

Rounding to three decimal places, the probability is 0.141.