Farhanthestudent005
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- May 22, 2021
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- HSC
- 2022
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Just note that the probability of a defect when the manager is on duty is 0.02, otherwise it is 0.03.This requires use of Bayes' Theorem I think.
P(Mgr on Duty|Defective) = 0.39 x 0.03 / (0.39 x 0.03 + 0.61 x 0.02) = 0.4895 . . .
You can draw a tree diagram: 1st branch for Manager(prob: 0.39) vs No Manager(prob: 0.61), and from each branch, consider Defective vs Non-Defective.
Thanks. I mixed up the 0.02 and 0.03.Just note that the probability of a defect when the manager is on duty is 0.02, otherwise it is 0.03.
Therefore P(M|D) = 0.39 x 0.02 / (0.39 x 0.02 + 0.61 x 0.03) = 0.2988...
Yes, I gathered you’re very strong at maths & it was a ’typo’.Thanks. I mixed up the 0.02 and 0.03.
It is in the new Advanced syllabus that started with the 2020 HSCI would not expect Bayes Thm to be in 2U Advanced.
Screenshot?It is in the new Advanced syllabus that started with the 2020 HSC
Oh really? Thanks for pointing this out.It is in the new Advanced syllabus that started with the 2020 HSC
It’s always good to ask questions when you don’t understand something.@Eagle Mum
This is gonna sound hella dumb but no dumb questions right?
If the P(AnB) = P(A) x P(B)
and P(A|B) = P(AnB)/P(B)
then substituting P(A|B) = P(A) x P(B)/P(B) which is just P(A) as it cancels out???
I tried to google the shaded regions of what the venn diagrams look like for P(AnB) and P(A|B) but got nowhere so any help would be great.
Also that stuff on Bayes theorem, is that syllabus or nah? TY