spin spin sugar
it's gotta be big
yeah he does that in all the ecmt threads ive noticed... gay1Time4thePpl said:
ecmt1010 ..QUiZ next week? (1 Viewer)
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i don't think working on these practice questions with other people would violate academic honesty. if you want my answers, just tell me and i'll email them to you. but i'd rather not keep them up here where any random can get them
matt#1
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could you please send me the answers sarevok...please?sarevok said:
matt#1
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thankyousarevok said:
repped
matt#1
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*must pass rep around before giving it to 1time again*1Time4thePpl said:
matt#1
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anyone worked out q 11 yet?
I've kinda got 11. I've got the figures required for it. I'm just not sure which ones answer the question as I really don't understand the question.
The prior probability is 0.95 for S (Satisfactory condition of the machine)
The prior " is 0.05 for S' (the unsatisfactory condition of the machine)
The conditional probability of P(D|Si) - probability of a defect conditional on Si:
P(D|S) = 0.02
P(D|S') = 0.15
The joint probability of P(D|Si)P(Si):
P(D|S)P(S) = .019
P(D|S')P(S') = .0075
P(D|S)P(S)+P(D|S')P(S') = 0.0265
The revised probability is:
P(S|D) = 0.72
P(S'D) = 0.28
The answers will be in there somewhere. I just really don't get the Q
The prior probability is 0.95 for S (Satisfactory condition of the machine)
The prior " is 0.05 for S' (the unsatisfactory condition of the machine)
The conditional probability of P(D|Si) - probability of a defect conditional on Si:
P(D|S) = 0.02
P(D|S') = 0.15
The joint probability of P(D|Si)P(Si):
P(D|S)P(S) = .019
P(D|S')P(S') = .0075
P(D|S)P(S)+P(D|S')P(S') = 0.0265
The revised probability is:
P(S|D) = 0.72
P(S'D) = 0.28
The answers will be in there somewhere. I just really don't get the Q
"I don't think that's fair. I posted up my answers. This is a help-forum."
Yeah, I guess you're right. Answers attached.
For question 11), if you look at the phrasing, it's actually quite similar to the phrasing of 4.34a) in the textbook, in which they applied Bayes' Theorem to find P(H|F), P(M|F) etc. I think the problem with question 11 is that it's just phrased vaguely, I think they just want you to find the probability of the machine being in good condition given that there is a defect and the probability that it is in bad condition given that there is a defect..which I don't think is too difficult to do.
Yeah, I guess you're right. Answers attached.
For question 11), if you look at the phrasing, it's actually quite similar to the phrasing of 4.34a) in the textbook, in which they applied Bayes' Theorem to find P(H|F), P(M|F) etc. I think the problem with question 11 is that it's just phrased vaguely, I think they just want you to find the probability of the machine being in good condition given that there is a defect and the probability that it is in bad condition given that there is a defect..which I don't think is too difficult to do.
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i still dont know how to do question 11. considering hte mix up with the questions and no one mentioned in our lect of that question 11 was amended (some random from another lec told me) what are the chances of a question like that tom?
When they say 'based on this new sample information' in 11.b), what information are they exactly referring to!? This is too ambiguous... Are we to assume that the manufacturer only had knowledge of the '95 %' part? Grr it makes me so angry...
Rorix
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Guys, the meaning of the question is quite obvious.
What is the new sample information? That we have found a defect.
So, if we have randomly sampled and found a defect, what is the probability that another random sample taken will be a defect?
What changes when we sample and find a defect? P(W) and P(W'), where P(W)= probability of machines working properly.
So, obviously we need to find P(W|D) (or P(W'|D) as they sum to one) and then reapply the method of a) (that is, reapply P(W)*P(D) + P(W')*P(d) with P(W|D) and P(W'|D), where P(d) is the probability of a defect when not working properly)
P(W|D) can either be found using Bayes therom or a contigency table, and is approx 0.716 from memory, making P(W'|D) approx .284. When you put that into the aforementioned probability equation [as P(D) and P(d) remain constant] you get the new probability of defect as .0568 approx.
What is the new sample information? That we have found a defect.
So, if we have randomly sampled and found a defect, what is the probability that another random sample taken will be a defect?
What changes when we sample and find a defect? P(W) and P(W'), where P(W)= probability of machines working properly.
So, obviously we need to find P(W|D) (or P(W'|D) as they sum to one) and then reapply the method of a) (that is, reapply P(W)*P(D) + P(W')*P(d) with P(W|D) and P(W'|D), where P(d) is the probability of a defect when not working properly)
P(W|D) can either be found using Bayes therom or a contigency table, and is approx 0.716 from memory, making P(W'|D) approx .284. When you put that into the aforementioned probability equation [as P(D) and P(d) remain constant] you get the new probability of defect as .0568 approx.
