Questions: Of the cartons produced by a company, 4% have a puncture, 6% 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. The probability that a randomly selected carton has a puncture or a smashed corner %. (Type an integer or a decimal. Do not round.)

Of the cartons produced by a company, 4% have a puncture, 6% 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.

The probability that a randomly selected carton has a puncture or a smashed corner %.
(Type an integer or a decimal. Do not round.)
Transcript text: Of the cartons produced by a company, $4 \%$ have a puncture, $6 \%$ 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. The probability that a randomly selected carton has a puncture or a smashed corner $\square$ $\%$. (Type an integer or a decimal. Do not round.)
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Solution

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Solution Steps

Step 1: Identify Given Probabilities
  • Probability of a carton having a puncture, P(A)=0.04 P(A) = 0.04 .
  • Probability of a carton having a smashed corner, P(B)=0.06 P(B) = 0.06 .
  • Probability of a carton having both a puncture and a smashed corner, P(AB)=0.004 P(A \cap B) = 0.004 .
Step 2: Apply the Formula for the Union of Two Events

Use the formula for the probability of the union of two events: P(AB)=P(A)+P(B)P(AB) P(A \cup B) = P(A) + P(B) - P(A \cap B)

Step 3: Substitute the Given Values into the Formula

Substitute the given probabilities into the formula: P(AB)=0.04+0.060.004 P(A \cup B) = 0.04 + 0.06 - 0.004

Final Answer

0.096\boxed{0.096}

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