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Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors. The first inspector detects 90$\%$ of all defectives that are present, and the second inspector does likewise. At least one inspector fails to detect a defect on 20$\%$ of all defective components. What is the probability that the following occur?

(a) A defective component will be detected only by the first inspector? By exactly one of the two inspectors?

(b) All three defective components in a batch escape detection by both inspectors (assuming inspections of different components are independent of one another)?

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Okay. We're giving the probability of being detected the first time as 90% on all. Still it a second time. Okay. Thank you. So you also have the probability that at least one grounds that they got to 20% So that means the probability of no bailing physical to it. Absent, Get a rest. Um, when the probability of our first detecting and our second killing that's equal to probably of our. Given that our second fails our first detect times of probability over second feeling that's equal to just you. See, that's approximately 10%. Yeah, that's this. 90 centimes. 10% Which gives you points. You're lying, which is approximately your man for part B. We have are actually have another part. What is the probability of its first being detected? Are the first Inspector detecting it and then the second failing? We're actually we're asked by exactly one of the two inspectors. So that means when the text one fails or about their way around the first round and the second text that's equal to there's 10% plus 10%. They think it's Tony for that. All right, we got 10% over here. and he started saying Basically an implement, be were asked, probably of neither of them detecting the defects when it's one minus, both of them protecting it or just exactly one detecting it. So that's 80% 20% And that's actually, though.

University of California, Berkeley

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