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    Probability with sampling question




    i have been having problems with what should be a trivial probability problem

    Suppose there are 130 patients. Each patient was administered one of 6 drugs as follows:

    Drug A: 25 patients
    Drug B: 31 patients
    Drug C: 13 patients
    Drug D: 17 patients
    Drug E: 28 patients
    Drug F: 16 patients

    What is the probability that in a random sample of 4 patients, all 4 selected were administered drug A?

    My answer is (25/130)^4
    The answer in the back is 0.0823 but something tells me it is wrong.

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    Re: Probability with sampling question

    Doesn't (25/130)^4 assume replacement?

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    Dason (08-05-2012)

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    Re: Probability with sampling question

    Quote Originally Posted by derksheng View Post
    Doesn't (25/130)^4 assume replacement?
    I am not sure. Suppose one case I assume replacement and another without replacement, what happens then?

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    Re: Probability with sampling question

    Well think of a bag with 2 marbles where one is white and one is black. If you draw 2 marbles without replacement then with probability 1 you'll have 1 white and 1 black. If you draw with replacement then there is probability .25 of getting 2 white, probability .5 of getting 1 white and 1 black, and probability .25 of getting 2 black.

    So clearly it makes a difference which one you assume is occurring. But in a random sample of patients I would assume that it is without replacement.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Probability with sampling question

    Quote Originally Posted by Dason View Post
    Well think of a bag with 2 marbles where one is white and one is black. If you draw 2 marbles without replacement then with probability 1 you'll have 1 white and 1 black. If you draw with replacement then there is probability .25 of getting 2 white, probability .5 of getting 1 white and 1 black, and probability .25 of getting 2 black.

    So clearly it makes a difference which one you assume is occurring. But in a random sample of patients I would assume that it is without replacement.
    So I should use hypergeometric distribution?

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    Re: Probability with sampling question

    Well you could. But you could just use basic probability rules to figure this one out.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Probability with sampling question

    Quote Originally Posted by Dason View Post
    Well you could. But you could just use basic probability rules to figure this one out.
    I am drawing a blank on how to use basic probability to figure this out. I would appreciate something to get me started

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    Re: Probability with sampling question

    Think about sequentially sampling the patients. What is the probability that the first patient you select was administered drug A? Now - given that the first patient was given drug A what is the probability that the second patient was administered drug A? So on and so forth.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Probability with sampling question

    ok I think I may have to use a bunch of bayes forumula?
    P(1st given drug A) * P(2nd given drug A | 1st given drug A) * P(3rd given drug A | (1st given drug A and 2nd given drug B))

    can I assume independence?

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    Re: Probability with sampling question

    You don't need Bayes formula for this and you don't need to worry about independence. What you have is definitely one way to do it though (if you continue it to the 4th person).

    You should probably think about why you aren't actually using Bayes formula for this (what are you actually doing?) and why it doesn't matter if things are independent.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Probability with sampling question

    The formula you gave is actually assuming dependence (P(B \cap A) = P(B|A)*P(A)).

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    Re: Probability with sampling question

    Quote Originally Posted by derksheng View Post
    The formula you gave is actually assuming dependence (P(B \cap A) = P(B|A)*P(A)).
    That formula works regardless of if A and B are dependent or independent.
    I don't have emotions and sometimes that makes me very sad.

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    derksheng (08-05-2012)

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    Re: Probability with sampling question

    A: 1st person in sample gets drug A
    B: 2nd person in sample gets drug A
    and so on

    P(A)P(B|A)
    =P(A)P(A and B)/P(A)
    =P(A and B)

    ???

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    Re: Probability with sampling question

    ok so I am now completely lost. In my post ^ what would the next logical step be? I don't have P(A and B)

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    Re: Probability with sampling question


    Well what you're actually looking for is P(Person1 had drugA AND Person2 had drugA AND Person3 had drugA AND Person4 had drugA) = P(Person1 had drugA)*P(Person2 had drugA | Person1 had drugA)*P(Person3 had .... | ...)*P(Person4 ... | ...) where you can fill in the ...s
    I don't have emotions and sometimes that makes me very sad.

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