Question

of the cartons produced by a company, 8% have a puncture, 5% have a smashed corner, and 0.8% 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, expressed as a percentage, is % (type an integer or a decimal. do not round.)

Explanation:

Step1: Recall the formula for \(P(A\cup B)\)

The formula for the probability of the union of two events \(A\) and \(B\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Let \(A\) be the event that a carton has a puncture and \(B\) be the event that a carton has a smashed corner.

Step2: Substitute the given values

We are given that \(P(A) = 8\%=0.08\), \(P(B)=5\% = 0.05\), and \(P(A\cap B)=0.8\%=0.008\).
Substitute these values into the formula: \(P(A\cup B)=0.08 + 0.05-0.008\).

Step3: Calculate the result

First, add \(0.08\) and \(0.05\): \(0.08+0.05 = 0.13\). Then subtract \(0.008\) from \(0.13\): \(0.13-0.008=0.122\).

Answer:

\(12.2\)