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Conditional probability problem - good solution?

  1. Jun 30, 2015 #1
    1. The problem statement, all variables and given/known data
    We have 6 new and 4 used products in a box.
    We draw randomly 2 products and used them for a while. then return them back in to the box.
    After that, again we draw 2 products.

    a) What is probability that at least one is new?
    b) What is probability that at least one is new if we found out that in first draw two used were not drawn out.

    3. The attempt at a solution
    a)
    A={ we drawn out 2 used products in second draw}
    H1={ we drawn out 2 new products in first draw}
    H2={ we drawn out 1 new and 1 used product in first draw}
    H3={ we drawn out 1 used and 1 new product in first draw}
    H4={ we drawn out 2 used product in first draw}

    P(H1)=6/10 * 5/9 = 1/3
    P(H2)=6/10 * 4/9 = 24/90
    P(H3)=4/10 * 6/9 = 24/90
    P(H4)=4/10 * 3/9 = 12/90

    P(A|H1)=6/10 * 5/9 = 1/3
    P(A|H2)=5/10 * 4/9 = 2/9
    P(A|H3)=4/10 * 5/9 = 2/9
    P(A|H4)=4/10 * 3/9= 12/90

    P(A) = 1/3 * 1/3 + 24/90 * 2/9 + 24/90 * 2/9 + 12/90 * 12/90
    P(A) = 0.24

    P(at least one new) = 1- P(A) = 0.76

    b)
    A={ we drawn out 2 used products in second draw}
    H1={ we drawn out 2 new products in first draw}
    H2={ we drawn out 1 new and 1 used product in first draw}
    H3={ we drawn out 1 used and 1 new product in first draw}

    P(H1)=6/10 * 5/9 = 1/3
    P(H2)=6/10 * 4/9 = 24/90
    P(H3)=4/10 * 6/9 = 24/90

    P(A|H1)=6/10 * 5/9 = 1/3
    P(A|H2)=5/10 * 4/9 = 2/9
    P(A|H3)=4/10 * 5/9 = 2/9

    P(A) = 1/9 + 2/9 * 24/90 + 2/9 * 24/90
    P(A) = 0.2296

    P(at least one new) = 1- P(A) = 0.7703


    Is this correct solution?
    My apologies if bad english, and if unpractical in absence of latex.

    Thanks!
     
  2. jcsd
  3. Jun 30, 2015 #2

    RUber

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    The first part looks good since you covered all possibilities.
    On the second part, your scale factor for the condition has to account for the fact you know that H4 didn't happen. So you need to multiply by a factor so that P(H1|~H4) + P(H2|~H4) + P(H3|~H4) = 1. Then you can multiply what you found for P(A|H1), etc. by that scaling factor. This will ensure you are preserving the known information.
     
  4. Jun 30, 2015 #3
    I am not sure I understand exactly.
    I am supposed to find factor which preserves P(H1|~H4) + P(H2|~H4) + P(H3|~H4) = 1 and later on multiply every P(A|H1) with it.
    What does tilda ~ means?
     
  5. Jun 30, 2015 #4

    RUber

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    The tilde means "not"
    Essentially, you know that H4 didn't happen. So the sum of the other probabilities has to add up to 1. As it stands, your probabilities for H1 to H3 sum to 78/90=13/15, so if you multiplied each one by 15/13, they would sum to one. Then use that scaled probability for the calculation of P(A).
     
  6. Jun 30, 2015 #5
    Oh I get it now. Thanks!
     
  7. Jul 1, 2015 #6

    haruspex

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    2016 Award

    I make it closer to 0.25.
     
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