Question

Of the cartons produced by a​ company, 33​% have a​ puncture, 44​% have a smashed​ corner, and 0.60.6​% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

Answer 1

Answer:

0.71

Step-by-step explanation:

Let A be the event of the cartons produced by a company that have a puncture

and B have a smashed corner

Given that P(A) = 33% or 0.33 and

P(B) = 44% =0.44

P(both) = P(AB) = 0.06

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

=P(AUB)

= $P(A)+P(B)-P(AB)\\=0.33+0.44-0.06\\=0.71$