Questions: Of the cartons produced by a company, 4% have a puncture, 6% have a smashed corner, and 0.4% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner. The probability that a randomly selected carton has a puncture or a smashed corner %. (Type an integer or a decimal. Do not round.)

Solution Steps

• Probability of a carton having a puncture, $P(A) = 0.04$.
• Probability of a carton having a smashed corner, $P(B) = 0.06$.
• Probability of a carton having both a puncture and a smashed corner, $P(A \cap B) = 0.004$.

Use the formula for the probability of the union of two events: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Substitute the given probabilities into the formula: $$P(A \cup B) = 0.04 + 0.06 - 0.004$$

Final Answer