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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?

Respuesta :

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

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