The probability that a randomly selected carton has a puncture or a smashed corner is 12.6%.
In this problem, the events are:
Event A: Puncture.
Event B: Smashed corner.
The "or" probability is given by:
[tex]P(AUB)=P(A)+P(B)-P(A[/tex]∩[tex]B)[/tex]
5% have a puncture, hence [tex]P(A)=0.05[/tex]
8% have a smashed corner, hence [tex]P(B)=0.08[/tex]
0.4% have both a puncture and a smashed corner, hence[tex]P(AUB)=0.004[/tex]
Then:
[tex]P(AUB)=0.05+0.08-0.004= 0.126[/tex]
Therefore, The probability that a randomly selected carton has a puncture or a smashed corner is 12.6%.
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