if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random, a 1 defective b none defective c at most 2 bolts will be defective think i have it just not sure?
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Originally Posted by question if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random, a 1 defective b none defective c at most 2 bolts will be defective think i have it just not sure? A)
this is a binomial distribution problem. see if post #2 here helps
Originally Posted by colby2152 A) could u possibly post the formula in its pure form with its components explained
Originally Posted by question could u possibly post the formula in its pure form with its components explained i did that in the post i directed you to
Originally Posted by question if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random, a 1 defective b none defective c at most 2 bolts will be defective think i have it just not sure? for part c i have the following: (4c2).(0.2)^2.(0.8^2) = 0.1536 this is the answer that we were supposed to arrive at, but just looking at the wording of the question im not sure if its correct. Can anyone verify???
Originally Posted by question for part c i have the following: (4c2).(0.2)^2.(0.8^2) = 0.1536 this is the answer that we were supposed to arrive at, but just looking at the wording of the question im not sure if its correct. Can anyone verify??? no. that is if EXACTLY two are defective. "at most two" means 2 or less. meaning 0 or 1 or 2 (or more easily, not(3 or 4))
Originally Posted by question if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random, a 1 defective b none defective c at most 2 bolts will be defective think i have it just not sure? BTW, the solution to c will start off like this: P(at most 2)
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