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Why are those two events not independent?

  1. Aug 21, 2013 #1
    Provided I know that after 10 attempts I will get 8 successes with a probability of a success per try being 'p', does that affect the success of the first try?

    My first instinct was to say it's independent because each attempt is independent but now I know this is a special case where "I know the future" so to speak hence the first try can be expressed conditionally based on that information after 10 tries.

    But what I am asking is, what is the formal way to express that dependency, is it only per definition a vague "the occurrence of those 10 tries affect the 1st" or something more explicit?
     
  2. jcsd
  3. Aug 21, 2013 #2

    Mark44

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    No. What you're describing is some event in which "success" has a probability of .8. This says that if you run a large number of trials, about 80% of them will result in "success", but that doesn't tell you what the outcome will be on any individual trial.

    For example, flipping a fair coin should result in heads half the time and tails the other half, so p = .5. This doesn't give you information about an individual coin flip, though.
    You don't really know the future, other than to be able to say something about the number of successes in the long run.
     
  4. Aug 21, 2013 #3

    haruspex

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    The OP can be read another way. Maybe it has been reported after the fact that 8 of the 10 trials were successful, but we'e not told which. Each try is known to have had an a priori probability of p. Assuming the trials were independent, we can now say the probability that the first trial was successful was .8, regardless of p.
    cdux, which is the correct interpretation?
     
  5. Aug 21, 2013 #4

    Ray Vickson

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    You need the conditional probability that the first trial is a success, given 8 successes in 10 trials. That is, if##X_1 = 1## is the event that the first outcome is a success (with ##P\{X_1=1\} = p##) and if ##\{S_{10} = 8\}## is the event that the sum of the first 10 outcomes is 8 (the other 2 being 0), you want ##P\{X_1 = 1|S_{10} = 8\},## which you can get via Bayes' formulas.
     
  6. Aug 21, 2013 #5

    haruspex

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    but it's much simpler just to ask what is the probability that the first trial is one of the eight successes.
     
  7. Aug 21, 2013 #6

    Ray Vickson

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    I agree---it is simpler---after one is sure that that is a correct way of viewing the problem. Sometimes, to gain familiarity with a method/result/insight, it helps to start with formalism. (One can come up with somewhat similar scenarios in which that view would be questionable, so the binomial nature of the current problem plays an important role.)
     
    Last edited: Aug 22, 2013
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