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Math Help - Expected Value (Probability) Help Pls

  1. #1
    Junior Member Ruichan's Avatar
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    Expected Value (Probability) Help Pls

    Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

    Please kindly help.
    My professor just gave the definition of expected value in yesterday's lecture then he assigned 5 questions for us to hand in tomorrow. Out of the 5 questions, I have no idea how to start or do this question.
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    Quote Originally Posted by Ruichan View Post
    Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

    Please kindly help.
    My professor just gave the definition of expected value in yesterday's lecture then he assigned 5 questions for us to hand in tomorrow. Out of the 5 questions, I have no idea how to start or do this question.
    Here is the simplest way to answer the question.

    Let the random variable X1 = 1 if an ace turns up on the first card, 0 otherwise. E(X1) = 4/52 = 1/13. Likewise let X2 =1 when an ace turns up on the second card and E(X2) = 1/13. The expected number of aces is E(X1 + X2) = E(X1) + E(X2) = 1/13 + 1/13 = 2/13. The X1 and X2 are dependent random variables but when summing expected values it does not matter.
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  3. #3
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by JakeD View Post
    Here is the simplest way to answer the question.

    Let the random variable X1 = 1 if an ace turns up on the first card, 0 otherwise. E(X1) = 4/52 = 1/13. Likewise let X2 =1 when an ace turns up on the second card and E(X2) = 1/13. The expected number of aces is E(X1 + X2) = E(X1) + E(X2) = 1/13 + 1/13 = 2/13. The X1 and X2 are dependent random variables but when summing expected values it does not matter.
    Thank you very much. I was stuck at E(X1)=1/13 coz I was thinking that without replacement, E(X2)=3/51=1/17.

    Just a question on the side, how does the formula np(x) work around here or is it redundant? Thanks
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    Quote Originally Posted by Ruichan View Post
    Thank you very much. I was stuck at E(X1)=1/13 coz I was thinking that without replacement, E(X2)=3/51=1/17.

    Just a question on the side, how does the formula np(x) work around here or is it redundant? Thanks
    Sorry, I don't know what you mean by the formula np(x).
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  5. #5
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by JakeD View Post
    Sorry, I don't know what you mean by the formula np(x).
    Sorry, wasn't clear.
    It's the formula
    E[X] = Summation k=1 to infinity xkp(xk)

    Sorry, I can't copy and paste from my Microsoft Word equation editor. Keeps turning up blank on this. The ks are all subscripts.
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    Quote Originally Posted by Ruichan View Post
    Sorry, wasn't clear.
    It's the formula
    E[X] = Summation k=1 to infinity xkp(xk)

    Sorry, I can't copy and paste from my Microsoft Word equation editor. Keeps turning up blank on this. The ks are all subscripts.
    That formula applies.

    E(X_1) = \sum_{X_1=0}^1 X_1 P(X_1) = 1 \times 4/52 + 0 \times 48/52 = 4/52.
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  7. #7
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by JakeD View Post
    That formula applies.

    E(X_1) = \sum_{X_1=0}^1 X_1 P(X_1) = 1 \times 4/52 + 0 \times 48/52 = 4/52.
    Thank you very much. Now I get it. I was wondering how do you apply that formula. I didn't realized that  0 \times 48/52 has to be included.
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