Of the cartons produced by a company, 9% have a puncture, 3% have a smashed corner, and 1.2% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corne The probability that a randomly selected carton has a puncture or a smashed corner (Type an integer or decimal. Do not round.)

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**Understanding Probability in Quality Control**

In the context of quality control for a company, it's essential to evaluate the likelihood of defects in products. Consider the following scenario regarding cartons produced by a company:

- **9%** of the cartons have a **puncture**.
- **3%** have a **smashed corner**.
- **1.2%** have **both a puncture and a smashed corner**.

To determine the **probability** that a randomly selected carton has either a puncture or a smashed corner, apply the principle of inclusion-exclusion:

\[ P(\text{puncture or smashed corner}) = P(\text{puncture}) + P(\text{smashed corner}) - P(\text{both}) \]

By substituting the given percentages:

\[ P(\text{puncture or smashed corner}) = 9\% + 3\% - 1.2\% \]

Therefore, the probability that a carton has a puncture or a smashed corner is calculated as:

\[ 9 + 3 - 1.2 = 10.8\% \]

This analysis helps in understanding and improving product quality by identifying and reducing defects.
Transcribed Image Text:**Understanding Probability in Quality Control** In the context of quality control for a company, it's essential to evaluate the likelihood of defects in products. Consider the following scenario regarding cartons produced by a company: - **9%** of the cartons have a **puncture**. - **3%** have a **smashed corner**. - **1.2%** have **both a puncture and a smashed corner**. To determine the **probability** that a randomly selected carton has either a puncture or a smashed corner, apply the principle of inclusion-exclusion: \[ P(\text{puncture or smashed corner}) = P(\text{puncture}) + P(\text{smashed corner}) - P(\text{both}) \] By substituting the given percentages: \[ P(\text{puncture or smashed corner}) = 9\% + 3\% - 1.2\% \] Therefore, the probability that a carton has a puncture or a smashed corner is calculated as: \[ 9 + 3 - 1.2 = 10.8\% \] This analysis helps in understanding and improving product quality by identifying and reducing defects.
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