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Probability:sampling without relacement

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data

    say i have fifty keys and 1 is the correct key.With replacement i can modell this as a geometric distribution to find the expectation of the number of bernouli trials required until i pick the correct key.But if there is no replacemnt then the probabilities change so how do i model this scenario in order to find quantities such as expoectation and variance?
    we are asked to find the expectation when the keys are replaced and then the keys are not replaced. I have done the first bit but cant figure out the second when there is no replacement.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 6, 2010 #2

    HallsofIvy

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    The probability that the first key chosen is correct is 1/50. If it is not and is not replaced, then there are 49 keys left to choose from, 1 is the correct key so the probability of getting the correct key on the second choice is 1/49. But in order to get to the second choice, the first choice must have been wrong- that had probability 49/50 so the probability of getting the correct key on the second choice is (49/50)(1/49)= 1/50.

    If you do not get the correct key on the first or second choice, there will be 48 keys left. The probability of getting the correct key on the third choice is 1/48 but to get to the third choice you had to get the wrong key on both first and second trials- which have probabilities 49/50 and 48/49. The probability of getting the correct key on the third trial is (49/50)(48/40)(1/48)= 1/50.

    Get the idea? One important difference between choosing with replacement and choosing without replacement is that when choosing with replacement, it is possible you will never get the right key! But when choosing without replacement, you reduce the number of keys so, at worst, you would get the right key by the 50th trial.

    The probability distribution is the uniform distribution with P(i)= 1/50 for all i from 1 to 50.
     
  4. Aug 6, 2010 #3
    ok thanks
     
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