a. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
b. If this experiment is repeated, with replacement, what is the probability that both parts are defective?

Question

02-03-2020
Mathematics

contestada

a. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
b. If this experiment is repeated, with replacement, what is the probability that both parts are defective?

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dfbustos dfbustos · Answer 1 · 2020-03-03T11:59:15+01:00

Answer:

a) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

b) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

Step-by-step explanation:

For this case we know that we have a batch of 30 parts with 5 defective.

Part a

If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

Part b

If this experiment is repeated, with replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

a. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective? b. If this experiment is repeated, with replacement, what is the probability that both parts are defective?

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Otras preguntas

a. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
b. If this experiment is repeated, with replacement, what is the probability that both parts are defective?