Of the cartons produced by a company, $9 \%$ have a puncture, $6 \%$ have a smashed corner, and $0.6 \%$ have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

Step 1 ：Given that the probability of a carton having a puncture is 9%, the probability of a carton having a smashed corner is 6%, and the probability of a carton having both a puncture and a smashed corner is 0.6%.

Step 2 ：We are asked to find the probability that a randomly selected carton has either a puncture or a smashed corner.

Step 3 ：This can be calculated using the formula for the probability of the union of two events: \(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\).

Step 4 ：In this case, A is the event that a carton has a puncture, B is the event that a carton has a smashed corner, and A and B is the event that a carton has both a puncture and a smashed corner.

Step 5 ：Substituting the given probabilities into the formula, we get \(P(A \text{ or } B) = 0.09 + 0.06 - 0.006 = 0.144\).

Step 6 ：Final Answer: The probability that a randomly selected carton has a puncture or a smashed corner is \(\boxed{0.144}\) or 14.4%.