Question

Can Defects: A company that makes cans finds the probability of producing a can without a puncture is 0.96, the probability that a can does not havesmashed comer is 0.93, and the probability of a can not being punctured or smashed is 0,893. If a quality inspector randomly selects a can, find theprobability that the can does not have a puncture or smashed corner?

A) 0.96
B) 0.93
C) 0.893
D) 0.07.

Answer 1

After calculating the probability that a can does not have a puncture or a smashed corner using the formula for the union of two events, the result is 0.997, which does not match any of the provided answer choices.

Step-by-step explanation:

The question involves calculating the probability that a can selected at random is not defective in either of the two ways specified: no puncture and no smashed corner. The probability of a can not having a puncture is 0.96, the probability of it not having a smashed corner is 0.93, and the probability of a can not having either defect is 0.893. To find the probability that a can does not have a puncture or a smashed corner, we can use the formula for the probability of the union of two events: P(A or B) = P(A) + P(B) - P(A and B). Substituting the given probabilities into the formula, we get P(not punctured or not smashed) = 0.96 + 0.93 - 0.893 = 0.997. As none of the provided answer choices match this result, it seems there might be an error in the question or answer choices.