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Determining probability

  1. Oct 23, 2012 #1
    There are 15 CDs in a box. On 10 of the CDs there are saved data files,
    and the other CDs have no data files saved on them.

    i) Suppose that 12 CDs are randomly selected. Determine the probability that exactly 9 of these CDs selected have saved data files.

    I'm not sure how to do this. First I thought some kind of conditional probability? But I'm confused about that.

    P(A|B) = P(A intersection B)/P(B)?

    Is that simply [P(A)*P(B)]/P(B) ? Wouldn't the P(B)'s cancel out?

    If not, then if 12 are selected. The probability of any of them having data is (10/15)*(12/15) = 8/15

    Now,

    (9/12)*(8/15) = 40% [probability of 9 of those having any data from the 12 selected]
     
  2. jcsd
  3. Oct 23, 2012 #2

    haruspex

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    No, P(A intersection B) is the probability of A and B occurring together. It only reduces to P(A)*P(B) if they're independent. In another extreme, A might imply B, in which case it reduces to P(A).
    How many equally likely ways are there of picking 12 of the 15? In how many of these do you get exactly 9 with data?
     
  4. Oct 23, 2012 #3
    As long as the CDs aren't being placed back in the box after each selection, it should be a binomial distribution.
     
  5. Oct 23, 2012 #4

    Hmm...15C12 = 455?

    Then 9/455?
     
  6. Oct 23, 2012 #5

    haruspex

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    455 is right but 9 is wrong. Need to pick 9 of the ten and one of the 5.
     
  7. Oct 23, 2012 #6
    I don't quite follow, do you mean 15C12 * (9/10)*(1/5) ?
     
  8. Oct 23, 2012 #7

    haruspex

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    No. Want
    number of ways of choosing 9 from the 10 and 1 from the 5
    Since these are independent, that's
    (number of ways of choosing 9 from the 10) * (number of ways of choosing 1 from the 5)
    right?
     
  9. Oct 23, 2012 #8
    9 is right, but "one" is wrong. You need to pick 3 of the 5 blank CDs.

    (edit: btw, the problem is somewhat easier to grasp, if you replace it with drawing 3 CDs, and ask about the probability of exactly one having data)
     
    Last edited: Oct 23, 2012
  10. Oct 23, 2012 #9
    Now I'm super confused.

    What exactly is happening here? Is this some kind of binomial distribution like someone posted earlier.
     
  11. Oct 23, 2012 #10

    haruspex

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    Sorry - got confused between the number of CDs with data and the number to be chosen.
    So it's:
    number of ways of choosing 9 from the 10 and 3 from the 5
    Since these are independent, that's
    (number of ways of choosing 9 from the 10) * (number of ways of choosing 3 from the 5)
     
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