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anyone? :(Quote:
50%, 30% and 20% of the products from company A, B, C respectively are
brought in. The defective rate of the products from the 3 companies are 1%,
3% and 3% respectively. 1 defective product has been picked up randomly.
What is the probability that the defective product picked is from company A?
a) .005
b) 0.25
c) 0.3
d) 0.45
Probability that product comes from A given that it is defective is:
P(A|Def) = P(A n Def) / P(Def)
probability that it comes from A and is defective / probability that it is defective.
P(Def) is just the multiplication of the relevant proportions to get the overall probability that you pick a defective product. P(A n Def) is the A bit of this.
So the answer is....
...see if you can do it now (because it's much more satisfying that way)....
I believe it's A although it has been a few years since high school ;)
EDIT: Didn't see zaza's post....
It is not a).
a) gives you the probability that it comes from company A and is defective, not that it comes from company A given that you already know it is defective.
There are some assumptions made in this question though, particularly since actual numbers of products are not given. For example, if company B only made 3 products, company C made 5 and company A made 200 million, then you'd get something of a different answer. You have to assume, in calculating the probability that you pick a defective product, that you can give equal weighting to the three companies involved.
It's not A??? visualAd thought it is A...
hmmm
you presented the conditional theorem hmm i'll try to work on that. Thanks zaza! :D
was not able to see this thread at work because of some freakin virus! but now that i am home i have access now hehehehe. brb
The question is badly worded.
I interpreted it as follows:
50% of the products are from company A, 30% from company B and 20% from company C.
you'd need to know the number s of products from each to calculate it otherwise. :(
It should be badly worded... the original exam was from a japanese set of questions which was translated by my boss to then aspiring passers which typed the translated english into notepad or something...
so it's twisted. :( huhuhu
I remember about this now. Just imagine that there are 1,000 products from company A, B and C. Now do the math and calculate. Or you can do it with 10,000 products. Making it divisible by 100 is easier.
i'll stick with 100 if i have to do it manually...
there should be a shortcut.. wait TV is too much of a distraction.
Stop watching TV. That is the solution to this problem.
Well, I was being stupid... anyways
With zaza's prodding and with Hairball's help, the formula finally yielded some results that fall in one of the choices...
Is it 0.25? B?
It is, although you don't need to assume numbers if you assume that they are all the same.
P(A|Def) = (0.5*0.01)/((0.5*0.01)+(0.3*0.03)+(0.2*0.03)) = 0.25
It is important to remember that you have already been told that one of the items is defective. This means that you have narrowed down your field of possibilities substantially. 0.05 is the probability of picking, from all possible items, a defective one which came from company A.
zaza
Now this makes sense! at least i don't have to quantify!
thanks zaza!