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Basic probability question

  1. Jul 5, 2008 #1
    I was playing Scrabble with a few friends last night and, as none of us have quite yet learned the ways of prob & stats, we were debating how to calculate the probability of this scenario: I had just formed a word and had to pull three letters out of a bag containing seven letters. One of the letters was surely a Z. What was my probability of pulling out that Z (which, by the way, I did)?

    As I've never covered any prob & stats in years (and even then it was covered in just one class day), I wasn't sure how to even begin this. Thanks in advance for helping solve this crucial problem!
  2. jcsd
  3. Jul 5, 2008 #2


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    Since there was exactly one z in the bag, the probability is simply 3/7.
  4. Jul 5, 2008 #3


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    If exactly one tile in the bag (out of 7) was a Z, then 3/7.

    But if there were, say, 21 tiles you hadn't seen (2 other players), and you only knew that one of those was a Z, then it would be 3/21 = 1/7. Similarly with other numbers of tiles.

    If you had more than one Z, it would get slightly more complicated.
  5. Jul 5, 2008 #4
    Wow, that's a lot easier than we thought it would be. Originally I thought maybe 3/7, but my friends stated that perhaps it would alter it because technically if you take out one at a time it would be 1/7, then 1/6, then 1/5, but they didn't know what to do with those. Definitely overcomplicated it! Thank you very much :)
  6. Jul 6, 2008 #5

    matt grime

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    Since you wanted to break it down, let's do it.

    You can produce it on the first go: 1/7 chance, and then afterwards doesn't matter what you draw.

    You can get it on the second: to do this you need to draw any of the other 6 first,a 6/7 chance, then draw the z which is now 1/6, so the probability is (6/7)*(1/6)=1/7.

    You get it with the third tile: 6/7 * 5/6 *1/5 = 1/7.

    Add 'em all up and you get 3/7.
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