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Thread: Joint Probability problem

  1. #1
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    Joint Probability problem




    Hi again! Have a bit of a problem with this question and would gratefully appreciate some help.
    There are 4 cards in a game, numbered 1 to 4. A player randomly draws one card at a time (and without replacement). The game ends when the numbers on the drawn cards sum to 3 and above.
    Let X be a random variable - The sum of numbers on the drawn cards.
    Let Y be a RV - The number of cards the player had drawn.
    Find the joint probability of X & Y.

    Well, my struggle is mostly in which method do I calculate the probabilities. I know that X can take the values 3,4,5,6 and Y can take 1,2.
    The sample space is 8 I think, because there are 8 possible scenarios of taking cards out until we stop. Am I right to think this way?
    However when putting the probabilities in the table, they sum up to 6/8 only. What am I overlooking?

    Thanks a bunch!

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    Re: Joint Probability problem

    Hard for us to say when you don't show us what you came up with for the joint probability distribution.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Joint Probability problem

    It is a good start to know that the support of the joint distribution is a subset of the product set of individual supports.

    So in your case there are at most 2 \times 4 = 8 support points. As what Dason suggested, show us the joint pmf calculated by you on those points (list it or tabulate it) and we can check which part goes wrong.

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    Re: Joint Probability problem


    My apologies. Here goes-

    P(x=3,y=1)= 1/8 (because that would mean choosing the 3 card so there's only one way to do that)
    P(x=3,y=2)=2/8 (either taking card '1' out and then card '2' or the other way around)
    P(x=4, y=2)=2/8
    P(x=5,y=1);P(x=6,y=1)=0
    P(x=5,y=2)=2/8
    P(x=6,y=2)=1/8

    And I just realized I found my mistake. Thank you all!

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