Of the cartons produced by a company, 3% have a puncture, 5% 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 corm The probability that a randomly selected carton has a puncture or a smashed corner %. (Type an integer or a decimal. Do not round.)

MATLAB: An Introduction with Applications
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**Question:**

Of the cartons produced by a company, 3% have a puncture, 5% 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.

**Solution:**

To find the probability that a randomly selected carton has a puncture or a smashed corner, we need to apply the principle of inclusion-exclusion. The formula to find the probability of the union of two events A and B is:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Where:
- \( P(A) \) is the probability of having a puncture = 3% = 0.03
- \( P(B) \) is the probability of having a smashed corner = 5% = 0.05
- \( P(A \cap B) \) is the probability of having both a puncture and a smashed corner = 0.4% = 0.004

Plug these values into the formula:

\[ P(A \cup B) = 0.03 + 0.05 - 0.004 = 0.076 \]

Convert the decimal back to a percentage:

\[ P(A \cup B) = 7.6\% \]

Thus, the probability that a randomly selected carton has a puncture or a smashed corner is **7.6%**.
Transcribed Image Text:**Question:** Of the cartons produced by a company, 3% have a puncture, 5% 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. **Solution:** To find the probability that a randomly selected carton has a puncture or a smashed corner, we need to apply the principle of inclusion-exclusion. The formula to find the probability of the union of two events A and B is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Where: - \( P(A) \) is the probability of having a puncture = 3% = 0.03 - \( P(B) \) is the probability of having a smashed corner = 5% = 0.05 - \( P(A \cap B) \) is the probability of having both a puncture and a smashed corner = 0.4% = 0.004 Plug these values into the formula: \[ P(A \cup B) = 0.03 + 0.05 - 0.004 = 0.076 \] Convert the decimal back to a percentage: \[ P(A \cup B) = 7.6\% \] Thus, the probability that a randomly selected carton has a puncture or a smashed corner is **7.6%**.
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