Let the random variable \(X\) represent the number of damaged tapes in a sample of 15. The probability of a tape being damaged is \(p = 0.2\), and the probability of a tape not being damaged is \(1 - p = 0.8\).

We use the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\) where \(n = 15\).

Calculate \(P(X = 0)\):

\(P(0) = (0.8)^{15} = 0.03518\)

Calculate \(P(X = 1)\):

\(P(1) = \binom{15}{1} (0.2)^1 (0.8)^{14} = 15 \times 0.2 \times (0.8)^{14} = 0.1319\)

Calculate \(P(X = 2)\):

\(P(2) = \binom{15}{2} (0.2)^2 (0.8)^{13} = 105 \times (0.2)^2 \times (0.8)^{13} = 0.2309\)

The probability that at most 2 tapes are damaged is:

\(P(X \leq 2) = P(0) + P(1) + P(2) = 0.03518 + 0.1319 + 0.2309 = 0.398\)