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ad-work [718]
3 years ago
11

A company that makes cartons finds that the probability of producing a carton with a puncture is 0.03​, the probability that a c

arton has a smashed corner is 0.08​, and the probability that a carton has a puncture and has a smashed corner is 0.002. Answer parts​ (a) and​ (b) below. ​(a) Are the events​ "selecting a carton with a​ puncture" and​ "selecting a carton with a smashed​ corner" mutually​ exclusive? Explain. A. ​No, a carton cannot have a puncture and a smashed corner. B. ​No, a carton can have a puncture and a smashed corner. Your answer is correct.C. ​Yes, a carton cannot have a puncture and a smashed corner. D. ​Yes, a carton can have a puncture and a smashed corner. ​(b) If a quality inspector randomly selects a​ carton, find the probability that the carton has a puncture or has a smashed corner. The probability that a carton has a puncture or a smashed corner is nothing.
Mathematics
1 answer:
matrenka [14]3 years ago
4 0

Answer:

(a) Having a smashed corner need to now not depend on whether or not the carton had a puncture or not,

so these occasions have to be independent.

One manner to check it the usage of the given probabilities:

If P(A)P(B) = P(A and B), then the events are independent.

Here we have : (0.05)(0.08) which is 0.004, that is P(A and B), so the events are  independent.

(b) P(A or B) = P(A)+P(B)-P(A and B) = 0.05 + 0.08 – 0.004 = 0.126  

Step-by-step explanation:

Please mark me brainliest!

I can help with any other questions you may have!

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