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Probability question

  1. Jan 8, 2009 #1
    Let's say we have 10 boxes and I open each of them one by one...

    I open the 1st box, there is a toy car in it.
    I open the 2nd box, there is also a toy car in it.
    I open the 3rd box, there is also a toy car in it.
    ...
    I open the 9th box, there is also a toy car in it. :) wow, I got 9 toy cars in 9 boxes...

    what is the probability that 10th box also has a toy car in it?

    also generalize 10 to any number...
    what is the probability that nth box also has a toy car in it, if all n-1 boxes each have a toy car in them.
     
  2. jcsd
  3. Jan 8, 2009 #2

    CRGreathouse

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    There are lots of ways to do this binomial confidence interval problem. One common way is prob = (# successes + 1) / (# trials + 2), which would suggest a 91% chance.
     
  4. Jan 9, 2009 #3
    Depends on where you're getting the boxes.
     
  5. Jan 9, 2009 #4

    uart

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    Yes this is a silly question, unless you give some more information there is no well defined answer.

    About the best answer I could give (without any additional information) would be,

    [tex]P = \frac{m-9}{n-9}[/tex]

    Where n is the number of "boxes" in the universe and m<n is number of boxes in the universe that contain toy cars. I know that's not a very useful answer, but you know that if you want a useful answer you have to ask a sensible question right.
     
  6. Jan 10, 2009 #5

    HallsofIvy

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    What CRGreathouse is suggesting is to use the sample data to estimate the probability that a single box contains a car. Of course, if you have gotten a car in every box so far, the "maximum likelihood" estimate of that probability is 1 but you can use the sample size to put bounds on a confidence interval for it.
     
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