Bayes Theorem Problem

Nov 2019
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0
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Greetings. I am looking for some help with a series of problems that I have been assigned. I have worked the problem out a few times but am unable to get the correct answer.

Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301

1 is answered correctly to my knowledge.
Work for the rest is: D = defective, ND = non-defective, M= Minneapolis, P = philly
P(M) 0.7 P D/M = 1/70 P(M AND D) (0.7)(1/70) = 0.01 P M/D = .01/1
P(M) 0.7 P ND/M = 69/70 P (M AND ND) (0.7)(69/70) = 0.69 P M/ND = 0.69/1
P(P) 0.3 P D/P = 0.5/30 P(P AND D) (0.3)(0.5/30) = 0.005 P P/D = .005/1
P(P) 0.3 P ND/P = 29.5/30 P(P AND ND) (0.3)(29.5/30) = 0.295 P P/ND = 0.295/1
p(C)= 1.0

I have done something wrong here. Can someone please tell me what I am missing? I'm not asking for an answer. I am asking for how I should set this up.
 

Debsta

MHF Helper
Oct 2009
1,313
600
Brisbane
From the statement:
"A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective"
and using your abbreviations we know that:

P(M)=0.7
P(P) =0.3
P(D|M) = 0.01
P(D|P) = 0.005


Q1 is not quite correct.
Let's say that 100 bikes are produced in total, 70 in M and 30 in P.
So number of defective bikes is 1% of 70 + 0.5% of 30 = 0.7 + 0.15 = 0.85 bikes
That is 0.85 out of 100 bikes are defective.
0.85% of bikes are defective (Note: as a probability this is 0.0085)
So P(D) =0.0085 (given as 0.85% in B)

Q2 "A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant?"
So, you are given that it is defective.
You need to find P(M|D)
Bayes Theorem
:
1574298561371.png
So, P(M|D) = ...... you can finish it . You have all the info you need above.

Give it a go. (BTW I can't follow your reasoning as I don't know if there should be multiplication signs or full stops in each row.)
 
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Plato

MHF Helper
Aug 2006
22,474
8,643
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: \(\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)\)
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
Thus \(\displaystyle \mathcal{P}(M|D)=\dfrac{\mathcal{P}(M\cap D)}{\mathcal{P}(D)}=\dfrac{\mathcal{P}(M\cap D)}{\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)}\)
 
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