Problem

Of the cartons produced by a company, $9 \%$ have a puncture, $6 \%$ have a smashed corner, and $0.6 \%$ have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

Solution

Step 1 :Given that the probability of a carton having a puncture is 9%, the probability of a carton having a smashed corner is 6%, and the probability of a carton having both a puncture and a smashed corner is 0.6%.

Step 2 :We are asked to find the probability that a randomly selected carton has either a puncture or a smashed corner.

Step 3 :This can be calculated using the formula for the probability of the union of two events: \(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\).

Step 4 :In this case, A is the event that a carton has a puncture, B is the event that a carton has a smashed corner, and A and B is the event that a carton has both a puncture and a smashed corner.

Step 5 :Substituting the given probabilities into the formula, we get \(P(A \text{ or } B) = 0.09 + 0.06 - 0.006 = 0.144\).

Step 6 :Final Answer: The probability that a randomly selected carton has a puncture or a smashed corner is \(\boxed{0.144}\) or 14.4%.

From Solvely APP
Source: https://solvelyapp.com/problems/8508/

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