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Probability of choosing 2 items randomly

  1. Jul 24, 2014 #1
    1. The problem statement, all variables and given/known data
    i am having problem with part ii) a and b . i have no idea how to start. can someone please help me?:smile:

    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Jul 24, 2014 #2

    Simon Bridge

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    The probability tree only works for a single component selected.
    It will let you find out, for instance, the probability that component comes from factory A and it is defective.
    What you need to start with is a mathematical version of the word problem.
    i.e. can you write out the information given in maths?
    (Hint: conditional probability)

    P(D|A)=0.05, P(D|B)=0.02, P(A)=0.4, etc.
     
  4. Jul 24, 2014 #3
    P(D|C)= 0.01
    P(B) =0.35
    what else info needed?
     
  5. Jul 24, 2014 #4

    Simon Bridge

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  6. Jul 24, 2014 #5

    Ray Vickson

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    It is a shame that you persist in submitting thumbnails (instead of properly typed work)---in violation of PF policies---because it means that I am unable (and unwilling) to offer any assistance. I cannot read your submissions on some media.
     
  7. Jul 25, 2014 #6

    Simon Bridge

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    Here - let me help:
    The supplied working involved a probability tree in two stages - the first one involves the supplier probabilities and the other the defective probabilities.

    The probability that the first selected component is both defective and comes from supplier A would be written down as...?
     
  8. Jul 25, 2014 #7

    Ray Vickson

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    Thanks for that, although it should be the OP that writes it out, not you.

    Anyway, the OP may find it helpful to use a tabular presentation instead of a tree. So, imagine 1,000,000 parts. How many come from suppliers A, B and C? Of the ones from A, how many are defective? Non-defective? Ditto for suppliers B and C. So, altogether, how many of the million parts are defective? All this can be nicely summarized in a table.

    The answers to (i) and (ii) (a) are easily obtained from this, but (ii) (b) involves a bit of extra work (or, rather, consulting a different part of the table).
     
  9. Jul 25, 2014 #8

    Simon Bridge

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    That would work - though I suspect the exercise is in using the equations.
    Mind you - I have not done the work myself.
     
  10. Jul 25, 2014 #9
    someone told me that the working is 1 - (1-.0285)^2 = 0.0562

    this is 1-(all not defective)

    but the working is based on after one electronic component is picked , then the electronic componenet is put back which is with replacement. but why it is with replacement? the question didnt state that.
     
  11. Jul 27, 2014 #10

    Simon Bridge

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    The question does not state how many components were obtained either - which you need to know if you want to use sampling without replacement.
    What they have done is suppose that the number of components is so large that removing one from the collection does not significantly affect the odds. It's like taking a drop of water from the ocean - technically that should affect the sea-level by some small amount right? But you don't bother taking that into account in day-to-day calculations.

    That structure from the example working is because they have used an equation.
    You know P(D|A) so probability not defective P(notD|A)=1-P(D|A) and you know P(A) and the equation for conditional probability.
     
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