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A-Level Maths Question

I just had this question on an assessment (it’s from one of Edexcel’s Past AS Papers in 2018 so is available publicly). In the mark scheme it talks about tree diagrams and Venn diagrams. I got the right answer but went about it in a different way. I saw assume they buy 100 components, and found the percentage that were and were not faulty. I found that 4.2 components from Factory C must have been faulty so divided this by the number of components brought from them to get 0.07, so a percentage of 7%. The mark scheme makes no mention of this method so will I not score the marks?
Reply 1
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Reply 2
Original post by satsun
I just had this question on an assessment (it’s from one of Edexcel’s Past AS Papers in 2018 so is available publicly). In the mark scheme it talks about tree diagrams and Venn diagrams. I got the right answer but went about it in a different way. I saw assume they buy 100 components, and found the percentage that were and were not faulty. I found that 4.2 components from Factory C must have been faulty so divided this by the number of components brought from them to get 0.07, so a percentage of 7%. The mark scheme makes no mention of this method so will I not score the marks?

Can you upload a pic of your working?
Reply 3
Original post by mqb2766
Can you upload a pic of your working?

Something like this:

Assume factory buys 100 components.

Supplier A supplies 10, 0.9 of those are faulty.
Supplier B supplies 30, 0.9 of those are faulty
Supplier C supplies 60, X of those are faulty.

6% are faulty so 6 - 1.8 = X (2.4)
4.2/60 (this is how many C supply) to get 0.07, then times by 100 to get 7%.
Reply 4
Original post by satsun
Something like this:

Assume factory buys 100 components.

Supplier A supplies 10, 0.9 of those are faulty.
Supplier B supplies 30, 0.9 of those are faulty
Supplier C supplies 60, X of those are faulty.

6% are faulty so 6 - 1.8 = X (2.4)
4.2/60 (this is how many C supply) to get 0.07, then times by 100 to get 7%.

Im presuming the model solution did Bayes,, and it looks like you've done pretty much the same thing, but not called it Bayes.
p(F|C)p(C) = p(C|F)p(F)
Where F is fault and C is supplier
p(F|C)*0.6 = (1 - (p(A|F)+p(B|F)))*0.06

So your method is right, ,but its "obviously" a question about Bayes and I dont know whether they'd dock a mark for not saying that (or not). Bayes isn't too hard
https://www.mathsisfun.com/data/bayes-theorem.html
and if you've not covered it yet, the tree/venn diagram is fairly obvious.

Gut feeling, you should get the marks.
(edited 2 years ago)
Reply 5
Original post by mqb2766
Im presuming the model solution did Bayes,, and it looks like you've done pretty much the same thing, but not called it Bayes.
p(F|C)p(C) = p(C|F)p(F)
Where F is fault and C is supplier
p(F|C)*0.6 = (1 - (p(A|F)+p(B|F)))*0.06

So your method is right, ,but its "obviously" a question about Bayes and I dont know whether they'd dock a mark for not saying that (or not). Bayes isn't too hard
https://www.mathsisfun.com/data/bayes-theorem.html
and if you've not covered it yet, the tree/venn diagram is fairly obvious.

Gut feeling, you should get the marks.

Ah okay! Yeah that is what the mark scheme looks like online. Is it possible that I’ll get 0 marks? The question didn’t mention to use Bayes so surely another method is acceptable. Thank you for your help!
(edited 2 years ago)
Reply 6
Original post by satsun
Ah okay! Yeah that is what the mark scheme looks like online. Is it possible that I’ll get 0 marks? The question didn’t mention to use Bayes so surely another method is acceptable. Thank you for your help!

Its hardly another method, your calculation is fine. Honestly, I'd spend less time worrying about whether you'd get marks and a bit more time into making sure you learn/understand conditional probabilities and Bayes. The way the question is phrased, its clearly a conditional probability/Bayes question. Sometimes the questions leave it to you to work out how to solve the problem.

The advantage for saying its Bayes is that you can clearly write down the associated conditional probabilities and have a pre-canned method to solve them.
(edited 2 years ago)

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