Of the cartons produced by a company, 66% have a puncture, 44% have a smashed corner, and 0.50.5% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

0.095 or 9.5%

Step-by-step explanation:

Puncture → P(P) = 0.06

Smashed corner → P(S) = 0.04

Puncture and smashed corner → P(P and S) = 0.005

The probability that a randomly selected carton has a puncture or a smashed corner is given by the probability of a puncture, added to the probability of a smashed corner, subtracted by the probability of both:

The probability is 0.095 or 9.5%.