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Math Help - Card Probability

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    Two cards are drawn successively and without replacement from an ordinary deck of cards. Find the probability of drawing:

    a) Two hearts 13/52 *12/52 = 3/52
    b) A heart on the 1st draw and a club on the 2nd draw
    13/52 * 13/52 = 1/16
    c) A heart on the first draw, and an ace on the second draw

    I really don't know how to do c because the heart could be an ace, so I have 13/52 for the heart and then I'm not sure where to go from there. Hopefully I have the other two parts correct.


    Corection Edit: For a and b I think I should have done 13/52 *12/51 = 1/17 and 13/52 * 13/51 = 13/204?
    Last edited by mr fantastic; March 28th 2009 at 02:42 PM. Reason: Merged posts
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  2. #2
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    Quote Originally Posted by antman View Post
    Two cards are drawn successively and without replacement from an ordinary deck of cards. Find the probability of drawing:
    [snip]

    c) A heart on the first draw, and an ace on the second draw

    I really don't know how to do c because the heart could be an ace, so I have 13/52 for the heart and then I'm not sure where to go from there. Hopefully I have the other two parts correct.
    One approach to see the answer might be to draw a tree diagram:

    The first three branches from the first drawing are 'Ace of Hearts', 'Heart other than ace', 'No heart'.

    The two branches from the second drawing are 'Ace', 'No ace'.

    I get  \left(\frac{1}{52}\right) \cdot \left(\frac{3}{51}\right) + \left(\frac{12}{52}\right) \cdot \left(\frac{4}{51}\right).

    Quote Originally Posted by antman View Post
    Two cards are drawn successively and without replacement from an ordinary deck of cards. Find the probability of drawing:

    a) Two hearts 13/52 *12/52 = 3/52
    b) A heart on the 1st draw and a club on the 2nd draw
    13/52 * 13/52 = 1/16

    [snip]
    Corection Edit: For a and b I think I should have done 13/52 *12/51 = 1/17 and 13/52 * 13/51 = 13/204?
    The correction you posted is correct.
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