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Math Help - probability question, cards

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    probability question, cards

    With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?

    How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?
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    Quote Originally Posted by khor View Post
    With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?

    How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?
    1. The chance from 1 pack is \frac{13}{52} = \frac{1}{4}. -facepalm- . EDIT: As below do 1-\left(1-\frac{1}{4}\right)^2

    2. Impossible to calculate unless it's known which cards remain/have been taken out
    Last edited by e^(i*pi); December 27th 2009 at 01:45 PM.
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    Quote Originally Posted by khor View Post
    With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?
    There is a probability of \frac{1}{4} of drawing a diamond from a deck.
    So what is the probability of not drawing a diamond?
    Square that number.
    That will be the probability of not drawing at least one diamond when drawing one card from two different decks.
    So what is your answer?
    Last edited by Plato; December 27th 2009 at 01:50 PM.
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    43.75%. Okay I see. Thanks.
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    Quote Originally Posted by khor View Post
    [snip]

    How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?
    Since you don't know which cards were removed, the probability of drawing at least one diamond is the same as if you were drawing from a full deck.

    P(\text{drawing exactly one diamond}) = \frac{\binom{13}{1} \binom{39}{1}}{\binom{52}{2}}

    P(\text{drawing exactly two diamonds}) = \frac{\binom{13}{2}}{\binom{52}{2}}

    Add these to get the probability of drawing at least one diamond.
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  6. #6
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    Hello, khor!

    The second one has an incredibly long and tedious solution.


    How about with one deck with 21 cards removed (you don't know which ones),
    and then 2 cards are drawn. What is the chance at least one is a diamond?
    Since it depends on how many diamonds were removed,
    . . we must consider all possible cases.


    Suppose no \diamondsuit is removed. . This probability is: . \frac{{39\choose21}}{{52\choose21}}
    The deck contains: 13 \diamondsuits and 18 Others.
    The probability that no \diamondsuit is drawn is: . \frac{{18\choose2}}{{31\choose2}}
    The probability that no \diamondsuit is removed and at least one \diamondsuit is drawn is: . \frac{{39\choose21}}{{52\choose21}}\left(1 - \frac{{18\choose2}}{{31\choose2}} \right)

    Suppose 1 \diamondsuit is removed. . This probability is: . \frac{{13\choose1}{39\choose20}}{{52\choose21}}
    The deck contains 12 \diamondsuits and 19 Others.
    The probability that no \diamondsuit is drawn is: . \frac{{19\choose2}}{{31\choose2}}
    The probabiity that 1 \diamondsuit is removed and at least one \diamondsuit is drawn is: . \frac{{13\choose1}{39\choose20}}{{52\choose21}}\le  ft(1 - \frac{{19\choose2}}{{31\choose2}}\right)

    . . . \vdots

    We must extend this list up to:

    Suppose 11 \diamondsuits are removed. . This probability is: . \frac{{13\choose11}{39\choose10}}{{52\choose21}}<br />
    The deck contains 2 \diamondsuits and 29 Others.
    The probability that no \diamondsuit is drawn is: . \frac{{29\choose2}}{{31\choose2}}
    The probability that 11 \diamondsuits are removed and at least one \diamondsuit is drawn is: . \frac{{13\choose11}{39\choose10}}{{52\choose21}} \left(1 - \frac{{29\choose2}}{{31\choose2}}\right)


    The final probability is the sum of these twelve probabilities.


    You go on ahead . . . I'll wait in the car.
    .
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    Quote Originally Posted by Soroban View Post
    Hello, khor!

    The second one has an incredibly long and tedious solution.


    Since it depends on how many diamonds were removed,
    . . we must consider all possible cases.


    Suppose no \diamondsuit is removed. . This probability is: . \frac{{39\choose21}}{{52\choose21}}
    The deck contains: 13 \diamondsuits and 18 Others.
    The probability that no \diamondsuit is drawn is: . \frac{{18\choose2}}{{31\choose2}}
    The probability that no \diamondsuit is removed and at least one \diamondsuit is drawn is: . \frac{{39\choose21}}{{52\choose21}}\left(1 - \frac{{18\choose2}}{{31\choose2}} \right)

    Suppose 1 \diamondsuit is removed. . This probability is: . \frac{{13\choose1}{39\choose20}}{{52\choose21}}
    The deck contains 12 \diamondsuits and 19 Others.
    The probability that no \diamondsuit is drawn is: . \frac{{19\choose2}}{{31\choose2}}
    The probabiity that 1 \diamondsuit is removed and at least one \diamondsuit is drawn is: . \frac{{13\choose1}{39\choose20}}{{52\choose21}}\le  ft(1 - \frac{{19\choose2}}{{31\choose2}}\right)

    . . . \vdots

    We must extend this list up to:

    Suppose 11 \diamondsuits are removed. . This probability is: . \frac{{13\choose11}{39\choose10}}{{52\choose21}}<br />
    The deck contains 2 \diamondsuits and 29 Others.
    The probability that no \diamondsuit is drawn is: . \frac{{29\choose2}}{{31\choose2}}
    The probability that 11 \diamondsuits are removed and at least one \diamondsuit is drawn is: . \frac{{13\choose11}{39\choose10}}{{52\choose21}} \left(1 - \frac{{29\choose2}}{{31\choose2}}\right)


    The final probability is the sum of these twelve probabilities.


    You go on ahead . . . I'll wait in the car.
    .
    After you carry out this computation, I think you will find the result is the same obtained by the simpler method I previously posted.

    Look at it this way: You shuffle the deck and deal out the top two cards. Or you shuffle the deck, cut 21 cards from the top to the bottom, and deal out the top two cards. What's the difference? In terms of the distribution of the two cards dealt, none, really.
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