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.)

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.)

Transcript text: 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 $\square$ $\%$. (Type an integer or a decimal. Do not round.)

Solution

Solution Steps

Step 1: Identify Given Probabilities

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 $.

Step 2: Apply the Formula for the Union of Two Events

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

Step 3: Substitute the Given Values into the Formula

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