Evaluate cot (\pi )/(7)
1 answer:
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Answer:
a) C. No, a carton can have a puncture and a smashed corner.
b) The probability that a carton has a puncture or a smashed corner is P(X ∪ Y) = 0.104.
Step-by-step explanation:
To be mutually exclusive, the probability of the two events happening at the same time should be 0. But the probability that a carton has a puncture and has a smashed corner is 0.004 and not 0.
Then, we can conclude the events "selecting a carton with a puncture" and "selecting a carton with a smashed corner" are not mutually exclusive.
The answer is "C. No, a carton can have a puncture and a smashed corner."
We can calculate the probability that the carton has a puncture <em>or </em>has a smashed comer simply by adding the probability of each event:
P(X ∪ Y): probability that a carton has a puncture or a smashed corner.