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Difficult Math Problem #76 - Probability

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #76 - Probability

Post Fri Dec 15, 2006 9:31 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

    A. 1/336
    B. 1/120
    C. 1/56
    D. 1/720
    E. 1/1440

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    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Sun Dec 31, 2006 12:25 pm
    OA coming when a few people reply

    yuseop Newbie | Next Rank: 10 Posts Default Avatar
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    Post Sun Jan 07, 2007 5:51 am
    It's simply 3C3 / 8C3, which is 1/56.

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    Yuseop

    Chrystelle Junior | Next Rank: 30 Posts Default Avatar
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    Post Fri Jan 12, 2007 4:35 am
    The probability is

    3/8*2/7*1/6 = 1/56

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Jan 17, 2007 11:31 am
    OA:

    Since each draw doesn't replace the cards:
    Prob. of getting an ace in the first draw = 3/8
    getting in the second, after first draw is ace = 2/7
    getting in the third after the first two draws are aces = 1/6
    thus total probability for these mutually independent events = 3/8*2/7*1/6 = 1/56

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