Maths Help (1 Viewer)

[Farhanthestudent005]

Please help me with this question somebody thank you. Will greatly appreciate it.

 

[Drongoski]

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.

Edit
I mixed up the 0.02 with the 0.03.
So, as shown by Eagle Mum, answer should be: 0.39 x 0.02 /(0.39 x 0.02 + 0.61 x 0.03) = 0.2988 . . .

 

[Eagle Mum]

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.

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...

 

[Drongoski]

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...

Thanks. I mixed up the 0.02 and 0.03.

 

[Eagle Mum]

Thanks. I mixed up the 0.02 and 0.03.

Yes, I gathered you’re very strong at maths & it was a ’typo’.
My post was just in case there are viewers who don’t have a good grasp of the subject and might become shakier if unable to work out what’s going on if they come across this thread.

 

[hiiitsme]

@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

 

[Drongoski]

P(AnB) = P(A)P(B|A) = P(B)P(A|B)

P(AnB) = P(A) x P(B) only if events A and B are independent.


I would not expect Bayes Thm to be in 2U Advanced.

 

[CM_Tutor]

I would not expect Bayes Thm to be in 2U Advanced.

It is in the new Advanced syllabus that started with the 2020 HSC

 

[hiiitsme]

It is in the new Advanced syllabus that started with the 2020 HSC

Screenshot?

 

[Drongoski]

It is in the new Advanced syllabus that started with the 2020 HSC

Oh really? Thanks for pointing this out.

 

[Eagle Mum]

@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

It’s always good to ask questions when you don’t understand something.
I note that Drongoski has already replied, but since you addressed me, my additional cent worth, is that a probability tree diagram is quite useful in questions like this.


For this particular question:
P(A) is the probability that the manager is present (0.39)
P(Ā) is the probability that the manager is absent (0.61)
P(B) is the probability of a defect
P(B̅) is the probability of no defect

Therefore, in the tree diagram:
In the first branch, P(A) = 0.39, P(B|A) = 0.02, P(A ∩ B) = 0.0078
In the second branch, P(A) = 0.39, P(B̅|A) = 0.98, P(A ∩ B̅) = 0.3822
In the third branch, P(Ā) = 0.61, P(B|Ā) = 0.03 P(Ā ∩ B) = 0.0183
In the fourth branch, P(Ā) = 0.61, P(B̅|Ā) = 0.97, P(Ā ∩ B̅) = 0.5917

Edited: note that the sum of all branches should add up exactly to 1.

The question states that the condition is there is a defect, so the denominator is the sum of the first plus third branches: (0.39 x 0.02 + 0.61 x 0.03), whilst the numerator is the first branch (the manger was present when the defect occurred) 0.39 x 0.02.
Therefore, the answer is 0.0078 / (0.0078 + 0.0183) = 26/87 = 0.2988...

WRT your question, the problem with using a Venn diagram, as Drongoski has replied, is that B is not independent of A (when the manager is present, the defect rate fell from 0.03 to 0.02), so in a Venn diagram, P(B) within the shaded area would be 0.02, whereas P(B) in the rest of the circle B outside of the shaded area would be 0.03. In theory, the four regions of a Venn diagram (the fourth being the area outside both circles) could represent the four branches of the probability tree, but the tree is much more suitable for visualising all the possible cases and the probability of each case.

A probability tree can have many more branches than the four shown here from two binary possibilities. To apply Bayes theorem, all you need to be able to do is pick the appropriate branch(es) for the denominator and for the numerator from the stated information (ie. conditions).