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Simple combinatorics q

  1. Jul 28, 2005 #1
    The question goes something like this...

    What is the probability of an E being one of the 4 randomly chosen letters from the word ENERGISE?

    This is how i did it (the book says its wrong):

    ENERGISE ==> 3 Es, 5 Non-Es (partitioning)

    hence: (3c1*5c3)/(8c4)

    the book has 55/56....

  2. jcsd
  3. Jul 28, 2005 #2
    I don't see how they came up with that answer....I think your answer is correct.

    (note 8C3 is 56...why would the book have that?)
  4. Jul 28, 2005 #3


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    Catalyst : I suggest you write down the question exactly as it appears. There are two similar looking possibilities : (i) the probability that there is EXACTLY 1 E among 4 randomly chosen letters , and (ii) the probability that there is AT LEAST 1 E among the 4 letters.

    PS : You have answered assuming the first case. If that is the right case, then your answer is correct. Notice that the book's answer = 1 - your answer. If it is the second case, the book's answer is still wrong.
  5. Jul 28, 2005 #4

    Why do you say that? catalyst55's answer is 3/7....1-3/7=4/7. :confused: Am I missing something here?????? :redface:
  6. Jul 28, 2005 #5


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    No, you're not. I'm just losing it slowly... :biggrin:

    My bad there. I must have forgotten how to multiply ! :redface:
  7. Jul 29, 2005 #6
    Hey Gokul43201,

    I've thought about that and ive concluded that its definitely the former -- either way the answer on the back is wrong.

    I've also asked my teacher and he's confirmed this.

    Thanks a lot for your help guys.
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