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Math Help - small easy problem

  1. #1
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    small easy problem

    You have a deck of cards. You random take 2 cards from it (without putting the first one back). Now i need to know what's the chance that one of those 2 cards is a spade.
    maybe very easy, but some people doubted my answers and I would like to know the thruth
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  2. #2
    Moo
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    Hi

    There are 13 spades in a deck of 52 cards.

    Therefore, the probability of picking a spade the first time is \frac{13}{52}.

    Then, there remain 12 spades in a deck of 51 cards.
    So the probability that the second card you pick is a spade is \frac{12}{51}

    Thus, the probability of picking two spades is :

    \mathcal{P}=\frac{13}{52} \cdot \frac{12}{51}=\frac 14 \cdot \frac{4}{17}=\frac{1}{17}


    I hope I'm not wrong
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by Moo View Post
    Hi

    There are 13 spades in a deck of 52 cards.

    Therefore, the probability of picking a spade the first time is \frac{13}{52}.

    Then, there remain 12 spades in a deck of 51 cards.
    So the probability that the second card you pick is a spade is \frac{12}{51}

    Thus, the probability of picking two spades is :

    \mathcal{P}=\frac{13}{52} \cdot \frac{12}{51}=\frac 14 \cdot \frac{4}{17}=\frac{1}{17}


    I hope I'm not wrong
    I don't know probability yet, taking that next semester, but what if you pull a not spade the on the first card?

    Following your lovely example I get:
    \frac {39}{52} * \frac {13}{51} = \frac {13}{68}
    Last edited by angel.white; June 22nd 2008 at 05:14 AM.
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  4. #4
    Moo
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    Quote Originally Posted by angel.white View Post
    I don't know probability yet, taking that next semester, but what if you pull a not spade the on the first card?
    You're supposing you have taken one, that's why you multiply the probabilities.

    ---------------------------------
    Further explanation :
    Let's denote the event A : I pick a spade the first time.
    Let's denote the event B : I pick a spade the second time.


    We want the probability of \{A \text{ and } B\} :

    \mathcal{P}(A \cap B)=\mathcal{P}(B/A)\mathcal{P}(A).
    This is the formula of conditional probability (Conditional probability - Wikipedia, the free encyclopedia).

    \mathcal{P}(B/A) is the probability that : "B happens when A has happened". This is why you have to consider that you pick a spade the first time
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  5. #5
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    Quote Originally Posted by angel.white View Post
    I don't know probability yet, taking that next semester, but what if you pull a not spade the on the first card?

    Following your lovely example I get:
    \frac {13}{52} * \frac {13}{51} = \frac {13}{204}
    Not a spade has probability 39/52 on the first draw. Then:

    \frac{39}{52}\times \frac{13}{51}

    is the probability of drawing anything but a spade on the first card and a spade on the second.

    RonL
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  6. #6
    Moo
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    Quote Originally Posted by angel.white View Post
    Following your lovely example I get:
    \frac {13}{52} * \frac {13}{51} = \frac {13}{204}
    This would be for example, the probability of getting a heart on the first picking, and a spade on the second picking

    \frac{13}{52} : probability to get one type of cards (heart, spade, etc.. i don't know the other names )

    \frac{13}{51} : probability of getting one type of cards, where there are 13 cards of this type, among the remaining 51 cards...

    I don't know if it's clear enough ?
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  7. #7
    Super Member angel.white's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Not a spade has probability 39/52 on the first draw. Then:

    \frac{39}{52}\times \frac{13}{51}

    is the probability of drawing anything but a spade on the first card and a spade on the second.

    RonL
    Fixed it, thank you.
    Quote Originally Posted by Moo View Post
    This would be for example, the probability of getting a heart on the first picking, and a spade on the second picking

    \frac{13}{52} : probability to get one type of cards (heart, spade, etc.. i don't know the other names )

    \frac{13}{51} : probability of getting one type of cards, where there are 13 cards of this type, among the remaining 51 cards...

    I don't know if it's clear enough ?
    Since the question is "what's the chance that one of those 2 cards is a spade."

    It seems to me that the situation of a ~spade on the first card is a valid scenario.

    If we interpret the question to mean exactly one card is a spade then we would have to calculate getting:
    S, ~S and ~S, S

    Or if we interpret it to mean at least one card is a spade, then we would have to calculate:
    S, S and S, ~S and ~S, S
    (which I suspect is 100% minus probability of ~S,~S)

    If this is wrong than I am misunderstanding the question.
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  8. #8
    Moo
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    Quote Originally Posted by angel.white View Post
    If this is wrong than I am misunderstanding the question.
    That's because I misread the question


    Ok, let A be the number of spades that are picked (A is 0,1 or 2).

    We want P(A \ge 1), that is to say 1-P(A<1)=1-P(A=0).

    So it's 1-\frac{39}{52} \cdot \frac{38}{51}=\frac{15}{34}


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  9. #9
    Super Member angel.white's Avatar
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    Quote Originally Posted by Moo View Post
    P(A \ge 1), that is to say 1-P(A<1)=1-P(A=0).
    Hey, I got that right ^_^ Guess some of that set theory in discrete math stuck afterall :P
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  10. #10
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    thx for the help, it was indeed 1 spade or more
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