Question

Of the cartons produced by a company, 5% have a puncture, 8% 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.

Answer 1

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:

$P(AUB)=P(A)+P(B)-P(A$∩$B)$

5% have a​ puncture, hence $P(A)=0.05$

8% have a smashed​ corner, hence $P(B)=0.08$

0.4% have both a puncture and a smashed corner, hence$P(AUB)=0.004$

Then:

$P(AUB)=0.05+0.08-0.004= 0.126$

Therefore, The probability that a randomly selected carton has a puncture or a smashed corner is 12.6%.

Learn more about probability here brainly.com/question/20344684

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