Question

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.)

Answer 1

The probability that a randomly selected carton has a puncture or a smashed corner is 0.143

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