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  1. #1
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    Conditional Probability




    I have some ambiguity in the following problem
    Three cards are drawn in succession from a deck without replacement. find the probability distribution for the number of spades.
    I have come with this solution.
    Let S1: appearance of spade on first draw S2: appearance of spade on 2nd draw S3: appearance of spade on 3rd draw
    S01: No spade on 1st draw S02: No spade on 2nd draw
    P(S1)=13/52
    P(S2)=P(S2/S1)+P(S2/S0)=12/51+13/51
    P(S3)=P(S3/S01*S02)+P(S3/S1*S02)+P(S3/S1*S2)=13/50+12/50+11/50

    Need your opinion Please!

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    Re: Conditional Probability

    I think the problem is more complicated than that. Your answer should be a probability distribution like this -
    P(0 spades after 3 cards) = ...
    P(1 spades after 3 cards) = ...
    P(2 spades after 3 cards) = ...
    P(3 spades after 3 cards) = ...
    Here is one way of attacking the problem.
    Let's just look at P(2 spades after 3 cards) = ... The possibilities are SSO, SOS, and OSS. These need to be calculated and added.
    Just looking at one of them SOS. The probability is 13/52x39/51x12/50.
    You can also draw a probability tree.

  3. #3
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    Re: Conditional Probability

    You can also solve this problem using the hypergeometric distribution.

    For example, P(1S in 3 draws)= [(13C1)(39C2)]/(52C3)
    Last edited by Buckeye; 03-29-2017 at 04:35 PM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

  4. The Following User Says Thank You to Buckeye For This Useful Post:

    Roohul Amin (03-29-2017)

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    Re: Conditional Probability

    Thanks Buckeye, your approach fit will the situation, that is the condition of without replacement.

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    Re: Conditional Probability

    Of course. The old hypergeometric trick.

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    Re: Conditional Probability

    Do you not agree upon the Old Hypergeometric trick? if there is anything wrong with it, please tell.

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    Re: Conditional Probability


    No. it's perfect. (I was just quoting Maxwell Smart.) I wish I had thought of it before I went to first principles but it gives the same answers.

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