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Standard Card Question - Probability

  1. Oct 30, 2008 #1
    A standard 52 deck contains 4 different suits with 13 cards in each suit
    A player is allowed to draw 2 cards at a time. If a pair of matching in suit is drawn the player recieves 1 point, if a pair matching in number is drawn, the plaeyer recieves 2 points, otherwise the player gets no points. After the 2 cards are drawn, they are replaced before the next draw of 2 cards. A player is allowed two turns at this game. Let X be the discrete random variable; number of poitns obtained. What is the probability distribution of X and what are its mean and standard deviation?

    HELP!!! PLEASE AND THANK U! the answer is P(0) = .498 p(1) = .332 p(2)=.138 P(3)=.028 p(4) = .004
    .708 and .836

    I dont even know were to begin!
     
  2. jcsd
  3. Oct 30, 2008 #2

    HallsofIvy

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    Drawing two cards, whatever suit is drawn first, there are 51 cards remaining in the deck and 12 of those are the same suit as the first: the probability that they agree in suit is 12/51. Whatever numerical value is drawn, there are 51 cards remaining in the deck and 3 of those are the same number as drawn: the probability that they agree in numerical value is 3/51. The probability that they do not agree in either suit or numerical value is 1- 12/51- 3/51= (51-5)/51= 46/51.
    In order to get 0 points, a player must draw two cards and have them agree in neither suit nor value, a probability of 46/51, and then do the same thing again: probability (46/51)(46/51)= 0.814, not .498. I don't know what you mean by "p(4)= .004 .708 and .836".
     
  4. Oct 30, 2008 #3

    mathman

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    You made an arithmetical error in your 0 point calculation. It should be 36/51=12/17, so p(0)=.498.
    My calculation gives p(1)=.332, p(2)=.138, p(3)=.028, p(4)=.0034.
     
  5. Oct 31, 2008 #4
    Thank u very much! i understand this all now, excecet can someone please explain further I understand P(1) is saying what is the probability that they get 1 point? Would you not take the probability of getting one point which is .24 (12/51) * .24? I do not udnerstand this part!

    THANK U!
     
  6. Nov 1, 2008 #5

    mathman

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    To get exactly one point, you need to get one point on either try and zero on the other try. Probability to get one point on a given try is 4/17, probability to get no points is 12/17. Net probability therefore is 2x(4/17)x(12/17).
     
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