Question

of the cartons produced by a company, 6% have a puncture, 10% have a smashed corner, and 1.1% 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 \(P(A) = 6\%=0.06\), \(P(B)=10\% = 0.1\), and \(P(A\cap B)=1.1\%=0.011\).
Substitute these values into the formula: \(P(A\cup B)=0.06 + 0.1-0.011\).

Step3: Calculate the result

\(P(A\cup B)=0.06+0.1 - 0.011=0.149\).

Answer:

\(14.9\)