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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?

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  1. 6 May, 12:00
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    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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