Full Question
<em>Of the cartons produced by a company, 10% 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 nothing ____%. (Type an integer or a decimal. Do not round.)</em>
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Answer:
Step-by-step explanation:
Given
Required
For non-mutually exclusive event described above, P(Punctured or Smashed Corner) can be calculated as thus;
Substitute:
10% for P(Puncture Corner),
6% for P(Smashed Corner) and
0.4% for P(Punctured and Smashed Corner)
Convert % to fraction
Convert to decimal
