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17 December, 18:20

Of the cartons produced by a company, 8% have a puncture, 7% have a smashed corner, and 0.7% 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.)

+1
Answers (1)
  1. 17 December, 21:01
    0
    The probability that a randomly selected carton has a puncture or a smashed corner is 0.143

    Step-by-step explanation:

    we know that P (A∪B) = P (A) + P (B) - P (A∩B).

    since given that P (A) = Probability of getting puncture = 0.08

    P (B) = probability of getting smashed corner = 0.07

    P (A∩B) = probability of getting both puncture and smashed corner = 0.007

    P (A∪B) = probability of getting any of them

    so P (A∪B) = 0.08 + 0.07 - 0.007 = O. 143
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