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-   -   Probabilities in a deck of cards (https://www.askmehelpdesk.com/showthread.php?t=395910)

  • Sep 12, 2009, 10:18 PM
    merliot
    Probabilities in a deck of cards
    What are the probabilities of getting an a's in a deck of standar cards?
  • Sep 12, 2009, 10:37 PM
    ohsohappy

    It's not that hard if you think about it, it's a fraction. 4/52 right? 4 Aces, 52 cards in a pack.
  • Sep 12, 2009, 10:44 PM
    merliot
    Ino and the answer is 1/13 right? But in my math but at the ack of the book it says that the answer is 10/13 and not one? How come? Because I no that if I put 40 divide it by 52 its 10/13 so that's why I don't get it
  • Sep 12, 2009, 10:47 PM
    ohsohappy

    Well crap, that's weird. I wish my boyfriend was here, cause he'd know off the top of his head.
  • Sep 12, 2009, 11:00 PM
    merliot
    Quote:

    Originally Posted by ohsohappy View Post
    Well crap, that's weird. I wish my boyfriend was here, cause he'd know off the top of his head.


    Or maybe the book is wrong!
  • Sep 13, 2009, 06:24 PM
    ohsohappy

    Your book was definitely wrong.
  • Sep 14, 2009, 07:24 AM
    ebaines
    Quote:

    Originally Posted by merliot View Post
    what are the probabilities of getting an a's in a deck of standar cards?

    Merliot - please write the question out exactly as it appears in your book. Based on the discussion in this thread there seems to be confusion over the wording of the problem.
  • Sep 14, 2009, 08:26 AM
    ballengerb1

    Based on what you ask the chances are 1 in 13. If the books says something else the question is faulty or the answer is wrong. Just because something is in print doesn't isure it correct. I visited the Chicago museum of Science and Industry yesterday and saw many examples of errors printed in their displays.
  • Sep 14, 2009, 08:55 AM
    Capuchin

    I don't know, the answer is effectively 1, but there are probably some freak manufacturing faults where no aces are included in the deck. I would suspect an answer much higher than 10/13 though.
  • Sep 14, 2009, 09:55 AM
    Unknown008

    I think too it's an error in the book. The 1/13 is very similar to 10/13, in the way that only the zero is present. Such can very easily arise when the book was written.

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